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Fiitjee Work Energy

The document provides a comprehensive overview of work in physics, including definitions, calculations, and examples related to work done by constant and variable forces. It discusses the scalar nature of work, the conditions under which work is positive, negative, or zero, and includes illustrations to demonstrate these concepts. Additionally, it covers the integration of variable forces and various scenarios involving work done by different forces.

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Fiitjee Work Energy

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FIITJEE

PHYSICS
FIITJEE
PINNACLE
For – JEE (Main/Advanced)
FIITJEE Ltd. Material Provided by - Material Point Available on - Learnaf.com

Syllabus for IITJEE and Maharashtra Board:

Work done by constant and variable forces, Conservative and nonconservative forces,
Potential and Kinetic energy, Conservation of energy, Work energy theorem, Power.

WORK
Work is said to be done by a force when the point of application is displaced under the influence of the
force. Work is a scalar quantity and it is measured by the product of the magnitude of force and the
component of displacement along the direction of force. In fact, work is the scalar product (dot product) of
the force vector and the displacement vector.
 
Thus, W  F  S  FScos , where F and S are the magnitudes of force and displacement vectors and  is the
angle between them.
For 0    /2, work done is positive.
For  = /2, work done is zero
For /2 <  < 3/2, work done is negative.

For example,
(a) When a person lifts a body from the ground, the work done by the lifting force is positive but the
work done by the gravitational force is negative.
(b) When a body slides on a fixed rough surface, work done by the pulling force is positive while
work done by the force of friction is negative. The work done by normal reaction is zero.

Note:
1. Work done by a constant force is path independent, i.e. it depends on initial and final positions
only.
2. Work done depends on the frame of reference. For example, if a person is pushing a box inside a
 
moving train, the work done in the frame of reference of train will be F  S . While work done in the
   
 
frame of earth will be F  S  S0 , where S0 is the displacement of the train relative to the ground.
3. Work done by friction may be zero, positive or negative depending upon the situation. When force
applied on a body is insufficient to overcome the friction, work done by the frictional force is zero.
When this force is large enough to overcome the friction, then work done by the frictional force is
negative. When force is applied on a body, which is placed above another body, the work done by
the frictional force on the lower body may be positive.

Work Done By a Variable Force

  
The equation W  F  S  FScos  is applicable when F remains constant, but when the force is variable
 
work is obtained by integrating F.dS
 
Thus, W   F  dS
An example of a variable force is the spring force in which force depends on the extension x,
i.e. F  x
When the force is time dependent, we have,
    dx  
W   F.dx   F    dt   F.v dt
 dt 
 
where F and v are force and velocity vectors at an instant.

Geometrically, the work done is equal to the area between the F(x) curve and the xaxis, between the limits
xi and xf, e.g. Consider spring force:
k F
In the given figure, one end of a spring is k
attached to a fixed vertical support and other
end to a block which can move on a x=0 x
x x
horizontal frictionless table.
At x = 0, the spring is in its natural length. When the block is displaced by an amount x, a restoring force
(F) due to elasticity is applied by the spring on the block.
i.e. F = kx . . . (1) F
where k is the force constant of the spring which depends on
the nature of the spring.
From equation (1), we can observe that x=x

(i) F is a variable force x

(ii) Fx graph is a straight line which passes through
origin with slope k.
(iii) Work done by the force F when block is displaced
from x = 0 to x is
x x
1
W   Fdx =  kxdx =  kx 2 = area under F  xgraph.
0 0 2
Since force (F) and displacement are opposite in direction, hence work done by restoring force is negative,
but work done by the external agent is equal to the potential energy stored in the system.

Illustration 1. A block moves up a 30 incline under the action of certain F2 F3

forces which are shown in the figure. F1 is horizontal and of
F1
magnitude 40 N. F2 is normal to the plane and of magnitude
30
20 N. F3 is parallel to the plane and of magnitude 30 N. 30
Determine the work done by each force as the block (and
point of application of each force) moves 80 cm up the
incline plane.

Solution: Component of F1 along the direction of displacement

3
= F1 cos 30  40  =34.6 N
2
Hence, the work done by F1  F 1s  34.6 0.8 = 28 J
Work done by F2 is zero because it has no component in the direction of the
displacement.
Component of F3 in the direction of displacement = 30 N
Hence, work done by F3 = 300.80 = 24 N

Illustration 2. A block of mass 10 kg slides down on an incline 5 m long and 3 m high. A man pushes up
on the ice block parallel to the incline so that it slides down at constant speed. The
coefficient of friction between the ice and the incline is 0.1. Find:
(a) the work done by the man on the block.
(b) the work done by gravity on the block.
(c) the work done by the surface on the block.
(d) the work done by the resultant forces on the block.

Solution: From the figure, sin = 3/5

and cos = 4/5 5m
F = force by the man 3m
 = frictional force 
N = normal reaction of the surface,
mg = force of gravity
Since block slides with constant speed, Hence, F.B.D. of the block
mg sin = F +  F
N
 F = mg sin   
3 4
= 10  10   0.110 10   52 N ,
5 5
as f =  mg cos
mgsin mgcos

(a) Wman = F.S = FScos180 = FS
Here F = 52 N and S = 5 m.
 Wman =  52  5 J =  260 J
(b) Wgravity = mg S sin = 10  10 5  (3/5) J = 300 J
(c) Wsurface = WN + Wfriction = 0 + fS cos180o = 0  mg cos S
= 0.1  10  10  (4/5)  5 J = 40 J
(d) Work done by the resultant force is given by
W = Wm + Wg + WN + Wf = 260 J + 300 J + 0  40 J = 0.

Illustration 3. The force acting along xaxis on an

Fx(N)
object as a function of x is shown in the
figure. Find the work done by the force 5
in the interval. 4

(a) 0  x 3 cm , 3
2
(b) 3  x 5 cm,
1
(c) 0  x 6 cm.
O 1 2 3 4 5 6 7 8 x (cm)
1
2
3

Solution: We know that, work done = Area under the curve

(a) For the interval 0  x 3 cm
1
W =   0.03m   5N  0.075J
2
(b) For the interval 3  x 5 cm
1
W =   0.02 m   3 N  0.03J
2
(c) For the interval, 0  x 6 cm
Work done between 5 and 6 cm = 0.01 3 = 0.030 J
Adding this and (a) & (b), we will get total work done in the 6 cm interval
= 0.075  0.030 + 0.030) = 0.075 J.

FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi -16, Ph 26515949, 26854102, Fax 26513942
FIITJEE Ltd. Material Provided by - Material Point Available on - Learnaf.com
Pinnacle Study Package-68
PH-WEP-4

Illustration 4. ˆ 4 ˆj ) N from point (2m, 3m) to (3m, 0m) in xy
A particle is moved by a force F = ( 3i+
plane. Find the work done by the force on the particle.

Solution : Displacement of the particle is


 
S   3  2  î   0  3 ˆj m  î  3jˆ m

 
 W = F.S  3iˆ  4jˆ . î  3jˆ = 9 J 
Illustration 5. A force of magnitude 26 N, along the direction of a vector 5iˆ  12ˆj displaces a body of
mass 2 kg from a point (2, 3, 4) to a point (4, 3, 2). What is the work performed by the
force ?
 
Solution :   
W  F  S  26   5i  12 j  4iˆ  3 ˆj  2 kˆ  2 î  3 ˆj  4 kˆ 
 
 2 5iˆ  12 ˆj  2iˆ  2 kˆ  20 J
Illustration 6. A force F = kx acting on a particle moves it from x = 0 to x = x1. The work done in the
process is
1 2
(A) kx12 (B) kx1
2
(C) zero (D) kx13

Solution: (B).
x1 x1 kx 2 x1 kx12
W  0 F.dx  0 kxdx  |
2 0 2

Illustration 7. A force F = 9  4x + 6x2 N acts on a body of mass 5 kg and displaces it from x = 1 m to

x = 3 m. What is the work done by the force ?

3 4x 2 6x 3 3
Solution: W   F.dx  1 (9  4x  6x 2 )dx  9x   1
2 3
3
 9x  2x 2  2x 3 |  (27  18  54)  (9  2  2)
0

= 54 J.

Illustration 8. A block of mass m is suspended by a light thread from an elevator. T

The elevator is accelerating upward with uniform acceleration a.
a
Find the work done during the first ‘t’ seconds by the tension in the
thread. m

Solution: Let the block moves up with an acceleration a T

 Fnet = T – mg = ma
a
 T = m (g + a) …(1)
Now the work done W by the tension T in displacing the block mg
through a distance x is given as,
W = T.x …(2)

1 2
where x = at …(3)
2
Putting x and T from (1) and (3) in (2), we obtain
1 m
W = m(g + a) ( at2)  W = (g + a) at2.
2 2
Illustration 9. A block of mass m is attached rigidly with a light spring of m m
force constant k. The other end of the spring is fixed to a
wall. If block is displaced by a distance x, find the work x
done on the block by the spring for this range.
(The spring force is given by F= kx, where k is spring constant and x is displacement of
the block from its free length.)

Solution: Since F = kx, Therefore, force varies with

m
displacement. This force has tendency to bring the F kx
block to its equilibrium point Hence, it is opposite to
the displacement.
For infinitesimal displacement (dx) this force is supposed to be constant. Therefore,
Work done by this force for the displacement dx is given by
 
dW = F.dx  kxdx cos 
x
1
 W =  dW    kxdx  W =  kx 2 .
0 2


Illustration 10. A body is subjected to a constant force F in Newton given by F = – ˆi  2jˆ  3kˆ
where ˆi, ˆj and k̂ are unit vectors along x, y and z axes respectively of a co-ordinate
system..
(i) What is the work done by this force in moving the body through a distance of
(a) 4 m along the zaxis, and (b) 3 m along the yaxis?
(ii) What is the total work done by the force in moving the body through a distance of 4 m
along the zaxis and then 3 m along the yaxis?

Solution : (i) The force is given by


F  î  2ˆj  3kˆ
(a) Displacement along zaxis S = 4 k̂ metres, Therefore, work done is

  
W1 = F.S = î  2jˆ  3kˆ  4kˆ  4iˆ  kˆ  8jˆ  kˆ  12kˆ  kˆ
Now, î  kˆ  0 and ˆj  kˆ = 0 because î and ˆj are perpendicular to k̂ , but k̂ . k̂ = 1
Hence, W1 = 12 J.

(b) Displacement along yaxis is S = 3 ˆj metres. Hence, work done in this case is
  
W  F  S  66iˆ  2ˆj  3kˆ  3jˆ  6 J
2

(ii) Since work is a scalar quantity, the total work done is the algebraic sum of W1
and W2,
i.e W = W1 + W2 = 12 + 6 = 18 J

Illustration 11. The displacement of a particle of mass 1 kg on a horizontal smooth surface is a function of
1 3
time is given by x = t . The work done by external agent for first one second is
3
(A) 0.5 J (B) 2 J
(C) 0.60 J (D) none of these

Solution: (B).
1 dx 2
x  t3  v  t
3 dt
dv
Acceleration, a   2t
dt
f = ma = 2mt
dW = Fdx = 2mt  t2dt
W 1

0
dW  0 2mt 3 dt
1
2m  t 4 m 1
   0.5 J
4 0 2

Illustration 12. A force F  k(xiˆ  yj)
ˆ , where k is a positive constant, acts on a particle moving in the
xy plane. Starting from the origin, the particle is taken along the positive xaxis to the
point (a, 0) and then parallel to the yaxis to the point (a, a). Total work done by the

force F on the particle is
(A) 2ka2 (B) – ka2
2
(C) ka (D) 2ka2

Solution : (B).
 
The displacement occurs in two slabs S1 and S2 , the (a, a)

corresponding amount of work done W1 and W2, are given

by S2
x a
ka 2
W1    k xdx  
x 0 2
S1
y a
ka 2 O (a, 0)
W2  k  y dy  
y 0 2
W = W1 + W2 =  ka2

Illustration 13. A stone of mass 10 kg and specific gravity 2.5 lies at the bed of a lake 15 m deep. The
work required to be done to bring it to the top of the lake is
(consider g = 10 m/s2)
(A) 1000 J (B) 1200 J
(C) 750 J (D) 900 J

Solution : (D).
Forces acting on the stone are shown in the figure. To move the
15 m
stone up without acceleration F
F + B = mg
F = mg  B B

W = Fh = (mg  wVg)h
 Vg 
  mg  h
 2.5  mg

 1   1.5 
 mgh 1    mgh  
 2.5   25
10  10  15  15
  900 J
25

Exercise 1:
(i) Springs A and B are identical except that A is stiffer than B, i.e., force constant kA > kB. On
which spring, more work will be done, if
(a) they are stretched by same amount ?
(b) they are stretched by the same force ?

(ii) The sign of work done by a force on a body is important to understand. State carefully if the
following quantities are positive or negative.
(a) Work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) Work done by the gravitational force in the above case.
(c) Work done by friction on a body sliding down an inclined plane.
(d) Work done by an applied force on a body moving on a rough horizontal plane with uniform
velocity.
(e) Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Work depends on the frame of reference

For example, if a person is pushing a box inside a moving train, the work done in the frame of train will
    
be F.s . while work done in the frame of earth will be F.  s  s0  , where s0 is the displacement of the train

relative to the ground and s is the displacement of box relative to the train.
Work done by friction may be zero, positive or negative depending upon the situation. When force applied
on a body is insufficient to overcome the friction, work done by the friction force is zero. When this force
is large enough to overcome the friction then work done by the friction force is negative. When force is
applied on a body, which is, placed above another body the work done by the friction force on the lower
body is positive.
Let us consider the situation in which a horizontal rough trolley, on which a block and a man is standing, is
accelerating along the horizontal direction. The block is not slipping on the trolley. The following
conclusions can be drawn from above:
(i) In this case work done by friction (between trolley and the block) is zero as observed by the man
on trolley.
(ii) Work done by friction (between trolley and the block) is positive as observed by an observer on
the ground.
(iii) Work done by friction is negative as observed by observer who is moving along the direction of
motion of trolley with higher speed.

CONSERVATIVE AND NON-CONSERVATIVE FORCES

A force is said to be conservative if the work done by the force along a closed path is zero. Work done by
the conservative forces depends only upon the initial and final positions and is path independent.
A force is said to be non-conservative if the work done by the force along a closed path is not zero.
Conservative forces are non-dissipative whereas non-conservative forces are dissipative.
Examples of conservative forces are gravitational force, electrostatic force, etc.
Examples of non-conservative forces are frictional force, viscous force, etc.

Q
Illustration 14. A particle is taken from point P to point Q via the path
PAQ and then placed back to point P via the path A
h
QBP. Find the work done by gravity on the body over B

this closed path. The motion of the particle is in

P
the vertical plane.

Solution: Here, displacement of the particle is PQ , gravity is acting vertically downward. The

vertical component of PQ is h upward. Hence,
W(PAQ) = mgh …(1)
For the path QBP, component of the displacement along vertical is h(downward)
In this case, W(QBP) = mgh
 Total work done = WPAQ + WQBP = 0.

B
Illustration 15. A particle of mass m is moved on a rough horizontal 1 2
surface on a closed path as shown in the figure. If
A
C
coefficient of friction between the particle and the
surface is  then find the work done by frictional force on 4 3
the particle over closed path ABCDA. D

Solution: Since friction force is always opposite to the motion, hence work done by it is given by
W =  fd  mg   d
ve sign indicates that force is opposite to displacement
B C D A

 W = mg   d   d   d   d 
A B C D 
 mg  1   2   3   4 

Mechanical Energy [Kinetic Energy + Potential Energy]

It is the capability of doing mechanical work.
Mechanical energy possessed by a body is of two types, kinetic and potential

Kinetic Energy
The capacity of a body to do work by virtue of its motion is known as kinetic energy of the body. Kinetic
energy is equivalent to work done by an external force on a body of mass ‘m’ to bring the body from rest
upto its velocity v in absence of dissipative forces.

Mathematical expression
Consider a body of mass m initially at rest. Let us consider that an external constant force F acts on the
body to bring its velocity to v. If s be the displacement, then
v2 = 2aS and F = ma
Now, work done by the constant force, W = FS
 v2  1
  ma     mv2
 2a  2
Therefore, according to the definition
1
K.E.  mv 2 .
2
Note:
1. Both m and v2 are always positive, Therefore, K.E. is also always positive.

2. Like work-done, K.E. is also frame dependent. For example, the kinetic energy of a person of
1
mass m in a frame moving with velocity v is zero in the same frame but it is mv 2 in a stationary
2
frame.

Exercise 2:
(i) Is work done by a non- conservative force always negative.

(ii) A man in an open car moving with high speed, throws a ball with his full capacity along the
direction of the motion of the car. Now the same man throws the same ball when the car is not
moving. In which case the ball possesses more kinetic energy in
(a) ground frame,
(b) in car frame.

Potential Energy:
The energy possessed by a body by virtue of its position is called its potential energy.
The change in potential energy produced by a conservative force is defined as the negative of the work
done by the conservative force.

rf
 
 Uf – Ui =  F.dr , where Ui = Potential energy at the initial reference position,

ri

Uf = Potential energy at the final position.

Usually, the initial reference position is taken as infinity and the potential energy at infinity is assumed to
be zero.

r
 
Then, we get the potential energy of a body as, U =   F.dr .

r 

The negative derivative of the potential energy function with respect to the position gives the conservative
dU
force acting on the particle. Mathematically, F   .
dr

Illustration 16.  
A 150 g mass has a velocity v = 2iˆ  6jˆ m / s at a certain instant. What is its kinetic
energy?
1
Solution: We know that, K = mv 2
2
1   
or, K  m  v.v  [ v . v = v2]
2
1
   0.150 kg   2 2  6 2  = 3.0 J
2
Illustration 17. The potential energy of a spring when stretched through a distance S is 10 J. The amount
of work (in J) that must be done on this spring to stretch it through an additional distance
S will be
(A) 30 J (B) 40 J
(C) 10 J (D) 20 J

1 2 1 1 
Solution: (A). u1  kS ; u 2  k(2S) 2  4  kS2   4u 1
2 2 2 
u  u 2  u1  4u1  u1  3u1  30 J

Illustration 18. The momentum of a body is increased by 20%. The percentage increase in its kinetic
energy is
(A) 36% (B) 44%
(C) 20% (D) 50%

Solution: (B).
20
mv  mu  mu  1.2 mu
100
 v  1.2 u
 v 2  1.44u 2
1 1
mv2  m(1.44 u 2 )
2 2
1 1
 mv 2  1.44( mu 2 )
2 2
 KE final = 1.44 KEinitial

Illustration 19. A uniform chain is held on a frictionless table with

onefifth of its length hanging over the edge. If the /5
chain has a length  and a mass m, how much work is
required to pull the hanging part back on the table?

Solution: Mass of hanging part = m/5.

mg
The weight acts at the centre of gravity of the hanging chain, i.e. at a distance /10
5
below the surface of the table. The gain in potential energy in pulling the hanging part on
the table.
mg  mg
W  
5 10 50
 Work done = W = mg/50

Illustration 20. The kinetic energy of a particle moving along a circle of radius R depends upon the
distance s as K = as2. The force acting on the particle is
1/ 2
2as 2  s2 
(A) (B) 2as  1  2 
s  R 
(C) 2 as (D) 2 a

Solution: (B).
It is given that, K = as2
1
 mv 2  as 2
2
Differentiating w.r.t time
1 dv ds
 m  2v  2a s
2 dt dt
ds
But v
dt
dv
 m  2a s
dt

dv
The tangential force, Ft = m  2a s
dt
mv 2 2as 2
The radial or centripetal force, Fr  
R R
2 2
2 2  2as 2
F  Ft  Fr  (2as)   
 R 
1/ 2
 s2   s2 
 2as 1  2   2as 1  2 
 R   R 

Work-Energy Theorem
This theorem states that work done by all the forces acting on a particle or body is equal to the change in its
kinetic energy.
Let us take an example shown in Figure (a), in which a block of mass v
m kept on a rough horizontal surface is acted upon by a constant F

force F parallel to the surface. The corresponding F.B.D. is shown in m F
  
Fig.(b) which gives F  f k  ma ...(1) A B
and N = mg …(2)
x
(a)

Initially, while the force F is just applied, the block is at the position N y
A and has a velocity v0. The force acts on it for some interval of time 
a
‘t’ so that the block reaches to position B at a distance x from A.
fk F x
mg
(b)
Now, the work done by the net external force is maximum along the surface and is given by
  

W  F  fk  x . . . (3)
 
Since cos  = cos 00 = 1,  being the angle between a and x .
Therefore, W = m×a×x . . . (4)
Again from the kinematics equation for the velocities at A and B, we have,
v 2  v02  2ax.
where v is the velocity of the block at position B.
v2  v02
Thus, a x  …(5)
2
Putting the value of ‘ax’ from equation (4) in equation (3), we have,
 v 2  v20 
W  m 
 2 
1 1
W mv 2  mv20 …(6)
2 2
  
The work done by the other two forces in F.B.D., for the displacement x , are zero because N  x  0 and
 
also mg  x  0 .
1
Considering mv02  k i (initial kinetic energy)
2

1
and mv 2  k f (final kinetic energy)
2
Equation (6) becomes
W = kf – ki …(7)
Now, equation (7) can be explained as: the net work done by all the forces on a system gives the change in
kinetic energy of the system. This is known as workenergy theorem.

Thus, the change in kinetic energy of the body equals the total work done by all the forces (conservative
and non conservative).

Illustration 21. The block of mass M shown in the V0

figure initially has a velocity V0 to the k 
right and its position is such that the M  M
spring exerts no force on it, i.e. the
spring is not stretched or compressed.
The block moves to the right a distance
 before stopping in the dotted position
shown. The spring constant is k and the
coefficient of friction between block
and table is . As the block moves the
distance ,
(a) What is the work done on it by the frictional force?
(b) What is the work done on it by the spring force?
(c) Are there other forces acting on the block, and if so, what is the work done by these
force on the block?
(d) What is the total work done on the block?

Solution: (a) Work done by friction =  Mg.

1
(b) Work done by spring force   kt 2
2
(c) Gravitational force and normal reaction of the table do not work as they act in
direction perpendicular to displacement.
1
(d) Total work done on block = (mg + k 2 )
2

Illustration 22. A truck is moving with constant acceleration a0. A block of mass m is kept on the rough
trolley of the truck and is observed stationary w.r.t. truck. Using work energy theorem,
find the velocity of block: (a) relative to ground, and (b) relative to truck, when truck
moves a distance x from the starting point.

Solution: (a) From the ground frame of reference

Since there is no relative motion between the block and the truck, m
therefore, the force of friction on the block is f = ma in forward
f = ma0
direction as shown in the figure
If the truck moves a distance x on the ground, the block will also move the same distance
x as there is no slipping between the two. Hence, work done by friction on the block
(w.r.t. ground) is Wf  f  x  ma x
0
From workenergy theorem,
 K.E. = Wf
1 1
mv 2  ma x or v = 2a x =  Mv 20
2 0 0 2
where v is the velocity of block w.r.t. ground.

(b) Relative to the truck accelerating (noninertial)

frame ma0
F.B.D. in truck frame: Here fP = pseudo force = ma0
Work done by all forces
W  Wf  Wfp  0 fp = ma0
From work energy theorem,
1
mv 2r  W  0 or vr = 0
2
where vr is relative velocity of block w.r.t. truck

Illustration 23. The displacement of a body in metre is a function of time according to x  2t 4  5 . Mass
of the body is 2 kg. What is the increase in its kinetic energy one second after the start of
motion?
(A) 8 J (B) 16 J
(C) 32 J (D) 64 J

Solution: (D).
x = 2t4 + 5
dx
v  8t 3
dt
dv
a  24t 2
dt
 F  ma  48t 2
dW  Fdx  48t 2  8t 3 dt  48  8t 5 dt
Increase in the kinetic energy results from the work done by the applied force
1 48  8 6 1 48  8
KE  0 48  8t 5 dt  t |  64 J
6 0 6

Illustration 24. A bullet having a speed of 153 m/s crushes through a plank of wood. After passing
through the plank, its speed is 130 m/s. Another bullet, of the same mass and size but
travelling at 92 m/s is fired at the plank. What will be the second bullet’s speed after
tunneling through? Assume that the resistance of the plank is independent of the speed of
the bullet.

Solution: Since plank does the same amount of work on the two bullets, therefore, decreases their
kinetic energies equally
1 2 1 2 1 2 1
 m 153  m 130   m  92   mv 2
2 2 2 2
or, v 2 = 1955
 v = 44.2 m/s

Illustration 25. A ball of mass m is thrown in air with speed v1 from a height h1 and it is caught at a
height h2 > h1 when its speed becomes v2. Find the work done on the ball by the air
resistance.

Solution: Work done on the ball by gravity is

Wg = mg(h2  h1)

Work done on the ball by air resistance is Wair = ?

As Wg + Wair = K.E.
1
 mg(h2  h1) + Wair = m  v 22  v12 
2
1
 Wair = mg(h2  h1) + m  v 22  v12 
2

Illustration 26. A particle slides along a track with A D

elevated ends and a flat central part as h

shown in the figure. The flat part has a B E C

length  = 3m. The curved portions of the 
track are frictionless.
For the flat part the coefficient of kinetic friction is K = 0.2. The particle is released at
point A which is at height h = 1.5m above the flat part of the track. Where does the
particle finally come to rest?

Solution: The particle will finally come to rest on the flat part. Hence, displacement of the particle
along vertical is h. If Wg be the work done on the particle by the gravity then
Wg = mgh where m = mass of the particle … (1)
If distance travelled by the particle on the flat part is x, the work done on the particle by
the friction is
Wf = mgx … (2)
Since initially particle was at rest and finally it comes to rest again. Hence change in its
K.E. is zero.
From work energy theorem
Wg + Wf = K.E.
 mgh  mgx = 0
h 1.5
 x= = m  x = 7.5 m
 0.2
Since x > , the particle will reach C and then will rise up till the remaining KE at C is
converted into potential energy. It will then again descend to C and it will have the same
kinetic energy as it had when ascending but now will move from C to B; At B , the same
will be repeated (because 7.5 > 2) and finally, the particle will stop at E such that
BC + CB + BE = 7.5
 BE = 7.5  6 = 1.5 m

Conservation of Energy and Conservation of Mechanical Energy

Conservation of energy means conservation of all forms of energy together. Accounting all forms of energy
within an isolated system, the total energy remains constant.
The mechanical energy accounts for only two forms of energy, namely kinetic energy, K and potential
energy, U.
If only conservative forces acts on a system then total mechanical energy of the system remains constant.
i.e. K + U = const. …(1)
Therefore,
K + U = 0 …(2)
or, K =  U …(3)

Note:
1. In case of any conservative force, the potential energy is a function of its position.
i.e. U = U(x)
2. For conservative force field, the negative of differentiation of U(x) with respect to x, gives the
dU  x 
force acting on the system, i.e. F(x)  
dx
 
Illustration 27. An object is acted upon by the forces F1  4iˆ N and F2  (iˆ  ˆj) N. If the displacement of
  
the object is ( i  6 j  6k ) metre, the kinetic energy of the object
(A) remains constant (B) increases by 1 J
(C) decreases by 1 J (D) decreases by 2 J
Solution : (C).
  
   
F  F1  F2  4 iˆ  î  j N  5 î  ˆj N
 
  
W  F  S  5iˆ  ˆj  î  6 ˆj  6 kˆ  5  6  1 J
Work done = change in KE =  1 J

Illustration 28. A 2 kg block is placed on a frictionless horizontal F(N)

5
surface. A force shown in the Fx graph is applied
to the block horizontally. The change in kinetic 4

energy is 3

(A) 15 J (B) 20 J 2

2 4 6 8 10 x (m)

Solution : (B).
Work done = Area under Fx graph
W = ½  (10  2)  5 = 20 J
Work done = change in kinetic energy = 20 J

Illustration 29. Just before striking the ground, a 2.0 kg mass has 400 J of kinetic energy. If friction can
be ignored, from what height was it dropped?(g = 9.8 m/s2)
Solution: By conservation of mechanical energy,
U f  K f  Ui  K i
or, 0  K f  mgh  0
Kf 400
 h= =  20.4 m
mg 2  9.8

Illustration 30. A body dropped from height h acquires momentum p just as it strikes the ground. What is
the mass of the body ?
1
Solution: By conservation of energy, mgh = mV 2
2
 2m 2 gh  p 2
p2 p
m 
2gh 2gh

Illustration 31. A 40 g body starting from rest falls through a vertical distance of 25 cm before it strikes
to the ground. What is the
(a) kinetic energy of the body just before it hits the ground?
(b) velocity of the body just before it hits the ground? (g = 9.8 m/s2)

Solution: (a) As the body falls, its gravitational potential energy is converted in kinetic energy
 K.E. = P.E. = mgh = (0.040 kg)(9.8 m/s2)  0.25 m = 0.098 J
1
(b) As K.E. = mv 2
2
2  0.098
 v2 
0.040
 v = 2.21 m/s

Illustration 32. A plate of mass m, length b and breadth ‘a’ is initially

lying on a horizontal floor with length parallel to the b
floor and breadth perpendicular to the floor. Find the a
work done to erect it on its breadth.

Solution: Gravitational potential energy of the block at the

position 1, b
b
U1 = mgh1 …(1)
Gravitational potential energy of the block at the h2 = b/2
a/2 h1
position 2 is given as
Position (I) Position (II)
U2 = mgh2 …(2)
 Work done by the external agent = W = U = (U2 – U1)
= mgh2 – mgh1
where h2 = b/2 and h1 = a/2
 b a  mg(b  a)
 W = mg     .
 2 2 2

Illustration 33. A block of mass 'm' is pushed against a spring of spring k m

constant k fixed at other end to a wall. The block can
slide on a frictionless table as shown. The natural
length of the spring is L0. It is compressed to half its
L0/2
natural length and released. Find the velocity of the
block as a function of its distance from the wall.

Solution: Here initial compression = L0/2

2
1  L0  1
 Energy stored in the spring = k   = kL20
2  2  8
As it moves a small distance x, say the velocity it has is v and so
2
1  L0  1 1
k  x   mv 2 = kL20
2  2  2 8
1 2 1 1
 kx  k(L0 x) + mv2 = 0
2 2 2
 kx2 – kL0x + mv2 = 0
[L x  x 2 ]k
v2 = 0
m
k(L 0 x  x 2 )
or v=
m

Illustration 34. Two blocks of masses m1 and m2 are connected by m3

an ideal spring of spring constant k. This system is m2
placed on a smooth inclined surface of inclination k
. Initially when the system is released, block of
mass m3 is in contact with m2 and the spring is in
its natural length. Then the maximum compression m1

in the spring is
m m g sin  m g sin 
(A) 1 2 (B) 2
m1  m 2 k k
2(m 2  m 3 )g sin  (m 2  m3 )g
(C) (D)
k k

Solution: (C).
At the position of maximum compression x,
m2
the masses m1 and m2 are displaced vertically m1
x
by x sin .
Loss of PE by the mass m2 and m3 = gain in
m3
KE of the spring m2
1 2
kx   m 2  m3  gx sin 
2
2  m 2  m3  g sin  m1
 x 
k

Illustration 35. An ideal massless spring 'S' can be compressed 1.0 m by a

100 N force. It is placed as shown at the bottom of a
frictionless inclined plane which makes an angle of s
 = 300 with the horizontal. A 10 kg block is released  = 300
from rest from the top of the incline and is brought to rest
momentarily after compressing the spring 2.0 m. Through
what distance does the mass slide before coming to rest ?

Solution: Spring constant = 100 N/m

Conserving energy, s
1 2
mg (x + 2)sin30 = K.  2   = 300
2
1 1
1010(x + 2)    100  4
2 2
x+2=4
x = 2 m along the inclined surface.

Illustration 36. A block of mass m is attached to two unstretched k1 k2

springs of spring constants k1 and k2 as shown in m
figure. The block is displaced towards right
through a distance 'x' and is released. Find the
speed of the block as it passes through a distance
x/4 from its mean position.
Solution: Applying conservation of energy
1 1 1 1 1
k1 x 2  k 2 x 2  mv2  k1 (x / 4)2  k 2 (x / 4) 2
2 2 2 2 2
x
v= 15(k1  k 2 ) .
4m

Illustration 37. A body starts moving from the highest point O

of the smooth curved surface horizontal at the

end as shown in figure without losing contact. P
Find out the horizontal distance moved by the H =5m
body after breaking off at point P from the h =1m

curved surface.
x

1
Solution: 0 + mgH = mv 2 + mgh
2
1
mg (H  h) = mv 2
2
 v2 = 2g (H  h)
 v= 2 10  5  1 = 80 m/s
For time:
1
h=0+ gt 2
2
2h 2 1 1
t=   sec
g 10 5
Now x = v. t
1
x= 80 .  16
5
x = 4 m.

Exercise 3.
(i) A lorry and a car moving with the same kinetic energy are brought to rest by the application of
brakes, which provide equal retarding forces. Which of them will come to rest in a shorter
distance?
(ii) Can a body have energy without having momentum? Can a body have momentum without
having energy?

(iii) In a ballistics demonstration, a police officer fires a bullet of mass 50 gm with speed 20 m/s on
soft plywood of thickness 2 cm. The bullet emerges with only 10 % of its initial K.E. What is the
emergent speed of the bullet ?

(iv) A body of mass 5 kg initially at rest is moved by a horizontal force of 20 N on a frictionless

table. Calculate the work done by the force in 10 second and prove that this equals to the
change in kinetic energy.

(v) A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N
on a table with coefficient of kinetic friction = 0.1. Calculate the
(a) work done by the applied force in 10 sec.
(b) work done by friction in 10 sec.
(c) work done by the net force on the body in 10 sec.
(d) change in K.E. of the body in 10 sec.

POWER

Power is defined as the rate of doing work.

dW
Mathematically, P  …(1)
dt
 
As dW  F  dx

dW  dx
Therefore, P   F
dt dt
 
P  Fv …(2)
If the force is variable, we calculate the average power as
t

W  P dt
0
Pav   t
…(3)
t
 dt
0
Power can also be expressed as the rate of change of kinetic energy.
Let a body of mass m moves with a velocity v. Then, its kinetic energy is,
1
K  mv2
2
dK 1 d
Now, 
dt 2 dt
 mv 2 

  dv 
 mv   
 dt 
 
 Fext  v
dK
Therefore, P 
dt

Illustration 38. An advertisement claims that a certain 1200 kg car can accelerate from rest to a speed of
25 m/s in a time of 8s. What average power must the motor produce to cause this
acceleration ? (Ignore friction losses)

Solution: The work done in accelerating the car is given by

1 1
W = K = m  v 2f  vi2   (1200)[(25) 2  0] = 375 kJ
2 2
W 375
Power =   46.9 kW
t 8

Illustration 39. A hoist powered by a 15 kW motor is used to raise a 500 kg bucket to a height of 80 m. If
the efficiency is 80%, find the time required.

Solution: Upward force needed = bucket’s weight (mg)

= 500  9.8 = 4900 N
Power available (P) = 0.80  15  103 W = 1.2  10 4 W
Since, P = W/t = Fs/t ,
4900  80
 t= = 32.7 sec.
1.2  104

Illustration 40. A block of mass m is pulled up on a smooth incline of angle  with the horizontal. If the
block moves with an acceleration of a m/s2, find the power delivered by the pulling force
at a time t after the motion starts. What is the average power delivered during the t
seconds after the motion starts ?

N F
Solution: The forces acting on the blocks are shown in figure.
Resolving the forces parallel to the incline, we get
F  mg sin  = ma
F = ma + mg sin 
mg
The velocity at t sec is 
v = at
The power elivered by the force at t is

P = F.v
= (ma + mg sin ) v = mat (a + g sin)
The displacement during the first t seconds is
1
x = at 2
2
The work done in these t seconds is Therefore,
 
W = F x
1
= ma(a  g sin  t2
2
1 ma(a  g sin t 
The average power delivered =
2 t
1
= ma(a  g sin t
2

Illustration 41. A particle of mass m at rest is acted upon by a force P for a time t. Its kinetic energy after
an interval t is
P2 t 2 P2 t 2
(A) (B)
m 2m
P2 t 2 Pt
(C) (D)
3m 2m

1
Solution: (B). S  ut  at 2
2
1 P  2
 0  t
2m
P2 t 2
KE  F.S 
2m

Illustration 42. A particle of mass m is acted upon by a constant power P. The distance travelled by the
particle when its velocity increases from v1 to v2 is
m 2 2m
(A)  v 2  v12  (B)  v 2  v1 
3P 3P
3P 2 m 3
(C)  v 2  v12  (D)  v 2  v13 
m 3P

Solution: (D)
P = F v = mav
P
 a
mv
dv P
v 
ds mv
P
 v2 dv  ds
m
v2
P s
  ds   v 2 dv
m0 v1

P 1
 s   v32  v13 
m 3
m 3
s   v 2  v13 
3P

Exercise 4:
(i) An elevator having mass 1200 kg with a passenger of mass 50 kg is moving up with an
acceleration of 2 m/s2. A friction force of 2000 N opposes its motion. Determine the minimum
power delivered by the motor to the elevator.

(ii) A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 min. A second crane does
the same job in 2 min. Do the cranes consume the same or different amounts of fuel ? What is
the power supplied by each crane. Neglect power dissipation against friction.

(iii) A pump on ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15
min. If the tank is 40 m above the ground and the efficiency of the pump is 30%, how much
electric power is consumed by the pump.

(iv) An elevator weighing 500 kg is to be lifted at a constant velocity of 0.4 m/s. What should be the
minimum horse power of motor to be used ? Take g = 10 m/s2.

Motion in a vertical circle

A particle of mass m is attached to a light and inextensible string. The other end of the string is fixed at O
and the particle moves in vertical circle of radius r equal to the length of the string as shown in the figure.
Consider the particle when it is at the point P and the string makes an
angle  with vertical. Forces acting on the particle are: O
T
T = tension in the string along its length,  P
 mgcos
mg = weight of the particle vertically downward.
mgsin
Hence, net radial force on the particle is FR = T – mg cos mg
2
mv
Since FR = , where v = speed of the particle at P, and
R
R = radius of the circle, Here, R = (length of the string)
mv 2 mv 2
 T  mg cos =  T= + mg cos
R R

Since speed of the particle decreases with height, hence, tension is maximum at the bottom, where
cos  = 1 (as = 0)
mv 2 mv ' 2
 Tmax =  mg ; Tmin =  mg at the top
R R
Here v = speed of the particle at the top.

Critical Velocity
It is the minimum velocity given to the particle at the lowest point to complete the circle. The tendency of
the string to become slack is maximum when the particle is at the topmost point of the circle.
mvT2
At the top, tension is given by T =  mg ; where vT = speed of the particle at the top.
R
mv2T
  T  mg
R
For vT to be minimum, T  0  vT = gR
If vB be the critical velocity of the particle at the bottom, then from T+mg
conservation of energy
1 1
Mg(2R) + mvT2  0  mv B2
2 2 T
1 1
As vT = gR  2mgR + mgR = mv 2B
2 2 mg
 vB = 5gR
Note: In case the particle is attached with a light rod of length , at the highest point its minimum velocity
may be zero. Then the critical velocity is 2 g

Illustration 43. A pendulum bob has a speed u m/s while passing through its lowest position. What is its
speed when it makes an angle of  with the vertical? The length of the pendulum is .

Solution: As Tension  velocity

 Work done by tension = 0
 Total mechanical energy will remain conserved.
Increase in P.E. = mg (1  cos )
Increase in P.E. = decrease in K.E.
1 1
mg (1  cos ) = mu 2  mv2
2 2
v= u 2  2g(1  cos 

Illustration 44. A heavy particle hanging from a fixed point by a light inextensible string of length  is
projected horizontally with speed g  . Find the speed of the particle and the
inclination of the string to the vertical at the instant of the motion when the tension in the
string is equal to the weight of the particle.

Solution: Let tension in the string becomes equal to the weight of the
particle when particle reaches the point B and deflection of O
T
the string from vertical is . Resolving mg along the string  B
and perpendicular to the string, we get net radial force on  mgcos
the particle at B, i.e. A
mgsin
mg
FR = T  mg cos (i)
If vB be the speed of the particle at B, then
mv2B
FR = (ii)

From (i) and (ii), we get,
mv2B
T  mg cos = (iii)

Since at B, T = mg
mv2B
 mg(1  cos  )=

2
 v B = g (1  cos) (iv)
Applying conservation of mechanical energy of the particle at points A and B, we have
1 1
mv A 2  mg 1  cos    mvB2 ;
2 2

where vA= g and vB= g 1  cos  

 g = 2g(1  cos) + g (1  cos)
2
 cos = (v)
3
g
Putting the value of cos  in equation (iv), we get : v =
3
Illustration 45. A mass ‘m’ is revolving in a vertical circle at the end of a string of length 20 cm. By how
much does the tension of the string at the lowest point exceed the tension at the topmost
point?
(A) 2 mg (B) 4 mg
(C) 6 mg (D) 8 mg
Solution: (C).
At the lowest point A v m
2 B
mu
TA  mg  …(1)
r mg
At the highest point B TB 2r
2
mv TA
TB  mg  …(2)
r
m
Gain in PE from A to B = 2mgr = loss in KE u
A
 2 mgr = (1/2) m(u2  v2)
 u2  v2 = 4gr …(3) mg
From (1), (2) and (3)
m 2
TA  TB  2mg   u  v2 
r
m
TA  TB  2mg   4gr  6 mg
r

Illustration 46. A particle of mass m is attached to the ceiling of a cabin with an

 
inextensible light string of length . The cabin is moving upward T
a
with an acceleration a. The particle is taken to a position such that m
string makes an angle  with vertical. When string becomes
vertical, find the tension in the string.

Solution: In a frame associated with cabin:

 
Work done on the particle when it comes in the vertical position = T
a
mg(1  cos) + ma(1 cos)
By work energy theorem, m(g+a)
mv 2
= (mg + ma)(1  cos)
2
v2
= (g+a)(1  cos)
2
At vertical position,
mv 2
T  (mg + ma) =

T = (mg + ma) +2m(g + a)(1  cos)

Equilibrium
As we have studied in the chapter of LOM, a body is said to be in translatory equilibrium if net force acting
on the body is zero, i.e.

Fnet  0
If the forces are conservative,
dU
F=
dr
dU
and for equilibrium F = 0  0
dr
i.e. at equilibrium position slope of U  r graph is zero or the potential energy is optimum (maximum or
dU
minimum or constant). Equilibria are of three types, i.e., the situation where F = 0 and  0 can be
dr
obtained under three conditions. These are stable equilibrium, unstable equilibrium and neutral
equilibrium. These three types of equilibria can be better understood from the following three figures.

(a) (b) (c)

Three identical balls are placed in equilibrium position as shown in figures (a), (b) and (c), respectively.
In figure (a), ball is placed inside a fixed smooth spherical shell. The ball is in stable equilibrium position.
In figure (b), the ball is placed over a fixed smooth sphere. This is unstable equilibrium. In figure (c), the
ball is placed on a smooth horizontal ground. This ball is in neutral equilibrium position.

Table: Types of equilibria

S. No. Stable equilibrium Unstable equilibrium Neutral equilibrium

1. Fnet = 0 Fnet = 0 Fnet = 0
2. dU dU dU
0 0 0
dr dr dr
or slope of U r graph is zero. or slope of Ur graph is zero. or slope of Ur graph is zero.

3. When displaced from its When displaced from its When displaced from its
equilibrium position, a net equilibrium position a net equilibrium position the body
restoring force starts acting on force starts acting on the body has neither the tendency to
the body which has a in the direction of come back nor to move away
tendency to bring body back displacement or away from from the original position.
to its equilibrium position. the equilibrium position.
4. Potential energy in Potential energy in Potential energy remains
equilibrium position is equilibrium position is constant even if the body is
minimum as compared to its maximum as compared to its displaced from its.
neighbouring points or neighbouring points or equilibrium position
d2U d2U d2U
= +ve =  ve or =0
dr 2 dr 2 dr 2
5. When displaced from When displaced from When displaced from
equilibrium position the equilibrium position the equilibrium position the
centre of gravity of the body centre of gravity of the body centre of gravity of the body
goes up. comes down. remains at the same level.
a b
Illustration 47. The potential energy function of a diatomic molecule is given by U =  , where a
r12 r 6
and b are constants. The equilibrium point for the potential field is at
1/ 6 1/ 6
 2a  a
(A) r =   (B) r =  
 b  b
1/ 6
 a 
(C) r =   (D) r = (ab)1 / 6
 2b 
Solution: (A)
a b
U 12
 6
r r
For equilibrium
dU a 6b 12a 6b
 12 13  7  0  13  7
dr r r r r
6 12a
r 
6b
1

 2a  6
r  
 b 

Different forms of energy

Some other forms of energies are also present in nature.
(a) Internal Energy:
Internal energy of a body is possessed because of its temperature. A body can be supposed to be
made of molecules. The sum of the kinetic and potential energies of all the molecules constituting

the body is called the internal energy. If the temperature of a body increases, this change cause
increase in the kinetic and potential energy and Hence, in the internal energy.
(b) Heat Energy:
Due to the disordered motion of molecules of a body, it possesses heat energy.
(c) Chemical Energy:
Due to the chemical bonding of its atom, a body possesses chemical energy.
(e) Electrical Energy:
In order to move an electric charge from one point to the other in an electric field or for the
transverse motion of a current carrying conductor in a magnetic field, work has to be done. This
work done appears as the electrical energy of the system.
(f) Nuclear Energy
When a heavy nucleus breaks up into lighter nuclei on being bombarded by a neutron, a large
amount of energy is released. This energy is called nuclear energy.

SUMMARY

 
1. When a constant force F acts on a particle that undergoes a
 F

displacement S , the work done by the force on the particle is defined as  F cos
 
the scalar product of F and S . The unit of work in SI units is Joule = 
F
Newtonmeter (1J = 1 Nm). Work is a scalar quantity; It has an 
algebraic sign (positive or negative) but no direction in space. 
S

  
W = F.S = F S cos , where  = Angle between F and S .

2. The kinetic energy of a particle equals the amount of work required to v

accelerate the particle from rest to speed v. It is also equal to the amount of m
work a particle can do in the process of being brought to rest. Kinetic energy
is a scalar quantity that has no direction in space, it is always positive or zero.
Its unit is the same as the unit of work.
1
K= mv 2
2
3. When forces act on a particle while it undergoes a displacement, the k m
particle’s kinetic energy changes by an amount equal to the total work 
v1
done on the particle by all the forces. This relation, called the work
energy theorem, is valid whether the forces are constant or varying
and whether the particles move along a straight line or curved path.
Wtot = K2 – K1 = K
4. When a force varies during a straight line motion, the work done by N
the force is given by an integral:
x2
x
W=  F dx
x Fspring
x1
Mg

5. If the particle follows a curved path, then the P

 
work done by a force F is given by an integral F F
that involves the angle  between the force and
the displacement. This expression is valid even if 
the force magnitude and the angle  vary during O
 d F|| = F cos 
the displacement.
W =  F cos  d
6. The work done on a particle by a constant
Fother
gravitational force can be represented as a change Fother
in the gravitational potential energy. motion
motion
U = mgy. This energy is a shared property of
particle and earth.
Wgravity = mgy1 – mgy2 y2
U1 – U2 =  U y1 mg
mg

7. An ideal stretched or compressed spring exerts an elastic force 

x1 m S
Fx =  kx on a particle, where x is the amount of stretch or k
compression. The work done by this force can be represented
as a change in the elastic potential energy of the spring,
1 2 Fspring m
U= kx k
2

8. The total potential energy is the sum of gravitational and elastic potential energy. If no forces other
than the gravitational and elastic force do work on a particle, the sum of kinetic and potential energy is
conserved.
K1 + U1 = K2 + U2

9. When forces other than the gravitational and elastic forces do work on a particle, the work Wother done
by these other forces equals the change in total mechanical energy, considering that the work done by
gravitational and elastic forces has already been taken into account as P.E. of the system.
K1 + U1 + Wother = K2 + U2

10. All forces are either conservative or nonconservative. A conservative force is one for which the
workenergy theorem relation is completely reversible. The work of a conservative force can always
be represented by a potential energy function, but the work of a nonconservative force cannot.

11. The work done by non-conservative forces manifests itself as changes in internal energy of bodies.
The sum of kinetic, potential and internal energy is always conserved.
K + U + Uint = 0

12. For motion along a straight line, a conservative force Fx(x) is the negative derivative of its associated
potentialenergy function U. In three dimensions, the component of a conservative force is negative
partial derivative of U.
dU(x) U U U
Fx(x) =  , Fx =  , Fy =  and Fz  
dx x y z
  U ˆ U ˆ U ˆ 
 F   i j k
 x y z 

13. Power is the time rate of doing work. The average power (Pavg) is the amount of work W done in
time t divided by that time. The instantaneous power (P) is the limit of average power as t goes to
 
zero. When a force F acts on a particle moving with velocity v , the instantaneous power is the scaler
 
product of F and v . Like work and kinetic energy, power is also a scalar quantity.
W
Pavg 
t
W dW 
P = Limit  and P = F.v
t  0 t dt

MISCELLANEOUS EXERCISE

1. A force F = 9  4x + 6x2 N acts on a body of mass 5 kg and displaces it from x = 1 m to x = 3 m. Find

the work done by the force ?

2 An inelastic ball is dropped from a height of 100 cm. Due to collision with the ground, 20% of its
energy is lost. To what height will the ball rise?

3. A body of 3 kg initially at rest is subjected to a force of 15 N. Find the kinetic energy acquired by the
body at the end of 10 sec.

4. An object of mass 10 kg falls from rest through a vertical distance of 10 m and acquires a velocity of
10 m/s. Find the work done by the push of air on the object?

5. Water is falling at the rate of 100 kg/sec on the blades of a turbine from a height of 100 m. Find the
power delivered to the turbine.

6. A constant force of 2.50 N accelerates a stationary particle of mass 15 g through a displacement of

2.50 m. Find the work done and the average power delivered.

7. A car of mass 1200 kg going with a speed of 30 m/s applies its brakes and skids to rest. If the frictional
force between the sliding tyres and the road is 6000 N, how far does the car skid before coming to rest?

8. A block of mass 2kg initially at rest is subjected to a force of 20 N. Calculate the kinetic energy
acquired by the body at the end of 10 sec. Assume gravity free space.

9. A driver of a 1200 kg car notices that the car slows down from 20 m/s to 15 m/s as it moves a distance
of 130 m along the level ground. How large a force opposes the motion?

10. A motor having an efficiency of 90% operates a crane having an efficiency of 40%. With what
constant speed does the crane lift a 500 kg weight if the power supplied to the motor is 5 kW?

ANSWERS TO MISCELLANEOUS EXERCISE

1. 54 J 2. 80 cm
3. 3750 J 4.  500 J
5. 100 kW 6. 6.25 J, 36.1 W
7. 90 m 8. 10000 J
9. 807.7 N 10. 0.36 m/s

SOLVED PROBLEMS

Subjective:

BOARD TYPE

Prob 1. With what minimum speed v must a small ball should be pushed
d
inside a smooth vertical fixed tube from a height h so that it may
reach the top of the tube? Radius of the tube is R. (d <<R) R
h v
Sol. Applying conservation of energy
1
mv2 + mgh = mg . 2R
2
v = 2g(2 R- h)

Prob 2. A projectile is fired from the top of a tower 40 meter high with an initial speed of 50 m/s at an
unknown angle. Find its speed when it hits the ground.

1 1
Sol. Initial K.E. = m.u 2  m.502
2 2
1 2
Final K.E. = mv
2
Work done by gravity = +mgh = mg. 40
From w~E principle
1
mg . 40 = kf  kI = m(v2  502)
2
 v = 57.4 m/s.

Prob 3. A body of mass m is thrown vertically upward into air with initial velocity v0 . A constant
force F due to air resistance acts on the body opposite to the direction of motion of the body.
What is the velocity of the body at a height h above the surface during ascent?

Sol. Using COE,

1 1 2Fh
mv20 = mgh + F.h + mv2  v 02  v 2 = 2gh +
2 2 m
 2Fh 
 v = v20   2gh  
 m 

A
Prob 4. Two bodies A and B having masses 100 gm each
are allowed to move on a frictionless path as M
shown in the figure. What is the initial velocity 8m
given to B such that each body have same kinetic v 2m
B
energy at M. Body A starts from rest?

Sol. K.E. of A at M = decrease in Potential energy = mg  (8 2)

= (100  103 kg)(10 m/s2)  (82) = 6 J

K.E. of B at M = 6 J
1
Potential of B at M = 10  2  2 J
10
By energy conservation at B and M
1
mv 2B   6  2   v 2B  160
2
 vB = 4 10 m/s

Prob 5. A bob of mass 'm' is suspended by a light inextensible string of length 'l' from a fixed point.
The bob is given a speed of 6gl . Find the tension in the string when string deflects through
an angle 1200 from the vertical.

Sol. By C.O.E .theorem B

1 1 v
mu 2  mgl(1  cos120)  mv 2
2 2 T
 v = 3gl mg
0
120
At point B,
mv 2
T + mg cos 600 = l
l A
5
By putting v = 3gl we get, T = mg .
2 u= 6gl

IITJEE TYPE

Prob 6. Two bodies m1 and m2 are kept on a table with

coefficient of friction '' and are joined by a spring. m2 m1
Fmin
Initially, the spring is in its relaxed state. Find the
minimum force F which will make the other block m2 
move. (k is the spring constant).

Sol. Motion of m2 starts when, kx = .m2.g, where x = elongation in the spring

m 2 g
x=
k
The minimum force will be such that m1 has no kinetic energy.
Applying work energy principle for m1
x
  F  m g  kx  dx  0
0 1

1 2
F x   m1gx  kx = 0
2
 1   m 2 g 
 F  m1g  kx   m1g 
 2   2 
m 2 g
 Fmin  m1g 
2

Prob 7. A small metallic sphere is suspended by a light spring of force constant k from the ceiling of a
cage, which is accelerating uniformly by a force F. The ratio of mass of the cage to that of
the sphere is ‘n’. Find the potential energy stored in the spring.

Sol. The force equation for the cage and the sphere are
F – (T + Mg) = Ma …(1) F
And, T – mg = ma  T = m (g + a) …(2)
Adding (1) and (2)  (M + m)a = F – (M + m)g /
 F = (M + m) (g + a) …(3) T
dividing (2) by (3) /
T a
mF a
 T = [m(g + a) / (M + m) (g + a)] F = mg
(M  m)
mF
where T = kx, Therefore, x = .
(M  m)k
Mg
1 m 2 F2
Therefore, P.E. stored in the spring = kx2 =
2 2(M  m)2 k
As M/m = n,
F2
P.E. = .
2(n  1)2 k

Prob 8. An ideal massless spring can be compressed 1 m by a force of

100 N. This same spring is placed at the bottom of a
frictionless inclined plane which makes an angle  = 30o
with the horizontal. A 10 kg mass is released from rest at the
o
top of the incline and is brought to rest momentarily after 30
compressing the spring 2 meters.
(a) Through what distance does the mass slide before coming to rest?
(b) What is the speed of the mass just before it reaches the spring?

Sol. (a) Let total distance moved by the block is

S = ( + 2)m, where  is the distance moved by the 
2m
block before touching the spring.
Now, work done by gravity on the block is
30o
Wg = mg S sin = 10  10  S sin30 J
 Wg = 50 S J . . . (1)
Work done by spring on the block is
1
WS =  kx 2
2
Here, k = 100 N/m and x = 2 m
1
 WS =   100  4 J = 200 J . . . (2)
2
Total work done W = Wg + Ws = (50 S  200) J
Since change in K.E. of the block is zero as W = K.E.
 50 S  200 = 0  S = 4 m

(b) As S=+2 =S2=2m

Work done by gravity over this path length is Wg = mg  2 sin
1
As Wg =  K.E.  100 = mv 2  0
2

100  2
 v2 =  20
10
 v = 2 5 m/s .

Prob 9. A body of mass 100 g is attached to a hanging spring whose

force constant is 10 N/m. The body is lifted until the spring is in
its unstretched state. The body is then released. Calculate the
speed of the body when it strikes a table 15 cm below the release m
point. h

Sol. By conservation of mechanical energy,

Kf  Uf  Ki  Ui
1 1
mvf2  kh i2 = 0 + mgh
2 2
Since vi = 0
2
kh 2 10  0.15
 vf  2gh  = 2  9.80  0.15  = 0.831 m/s
m 0.1

Prob 10. A heavy particle is suspended by a string of length  from a fixed point O. The particle is
given a horizontal velocity vo. The string slacks at some angle and the particle proceeds on a
projectile path. Find the value of vo, if the particle passes through the point of suspension.

Sol. Let at P, the string slacks when it makes an angle  with the
vertical, Hence, at the point P the centripetal force is only due to  P
mgcos
the component of the gravitational force. cos

mv 2 O
mg
mg cos  = , where v = velocity of the particle at P.

v 2  g. cos  . . . (i) 
Conserving energy at initial point and at P , we get
1 1
mvo2  mv2  mg 1  cos   . . . (ii)
2 2
1 1
From (i) and (ii), we get, mvo2  mgcos + mg(1+cos)
2 2
vo2 = g [2 + 3 cos] . . . (iii)
Now particle will pass through the point of suspension, if
sin = (v cos)t . . . (iv)
1
And  cos = (vsin )t  gt 2 . . . (v)
2
Eliminating t from (iv) and (v) we get .
2
  sin   1   sin  
 cos   (vsin )   g  
 v cos   2  v cos  
Substituting v2 = g cos, and simplifying tan = 2
1
Therefore, cos= .
3

Substituting the value of cos in equation (iii), we get

v = [g (2 + 3 )]1/2 .

Probs 11. In the figure shown, stiffness of the spring is k and mass of the block is
m. The pulley is fixed. Initially, the block m is held such that, the
elongation in the spring is zero and then released from rest. Find :
(a) the maximum elongation in the spring, k
(b) the maximum speed of the block m. Neglect the mass of the spring m
and that of the string and the friction.

Sol. Let the maximum elongation in the spring be x, when the block is
at position 2.
(a) The displacement of the block m is also x. m
1
If E1 and E2 are the energies of the system when the block
is at position 1 and 2, respectively, then h1
m 2
h2
E1 = U1g + U1s + T1 S

where U1g = gravitational P.E. with respect to surface S,

U1s = P.E. stored in the spring, and T1 = initial K.E. of the block.
 E1 = mgh1 + 0 + 0 = mgh1 . . . (1)
1 2
And E2 = U2g + U2s + T2= mgh2 + kx + 0 . . . (2)
2
From conservation of energy, E1 = E2
1 1 2
 mgh1 = mgh2 + kx2  kx  mg  h1  h 2   mgx
2 2
 x = 2mg/k

(b) From work energy theorem:

1 1
mgx  kx2 = mv2  0
2 2
For maximum v,
dv mg
 0 x 
dx k
2
 mg  1  mg  1 2
So, mg    k   mv max
 k  2  k  2
 m
 vmax =   g
 k 

Prob 12. Given k1 = 1500 N/m, k2 = 500 N/m, m1 = 2 kg, m2 = 1 kg. Find
(a) potential energy stored in the springs in equilibrium. k1
(b) work done in slowly pulling down m2 by 8 cm. m1

Sol. Let the initial extension in the springs of force constants k1 k1x10
and k2, at equilibrium position, be x20 and x10. Then k1
mg (m1  m 2 )g
x20 = 2 , x10 =
k2 k1 m1 m1g
(a) Potential energy stored in the springs in equilibrium
k2 k2x20
position is
1 2
1 k2x20
U1 = k1 x10 + k2 x 220 m2
2 2 m2g
Putting values of x10 , x20 from above we get U1 =0.4 J
(b) Let x1 and x2 be additional elongations due to pulling m2 by  = 8 cm. Additional
forces on m1 are equal and in opposite direction.
 k1x1 = k2x2 . . . (i)
Also x1 + x2 =  . . . (ii)
x1 , x2 can be found from (i) and (ii)
wg +wP + ws = 0 (where wp is the work done by the pulling force)
 wp = ws – wg= (U2 – U1) – [m1gx1 + m2g(x1 + x2)]
1 1
where U2 = k1(x10 + x1)2 + k2 (x20 + x2)2
2 2
Putting the values
 wp = 1 J

Prob 13. A pendulum with a bob of mass m is suspended from a horizontal

platform. The platform is given a horizontal uniform acceleration. The
2
breaking tension in the light string of the pendulum is mg. Find the T
3
work done by the extreme tension T on the bob in the first one sec.
m
Sol. Work done, W = (F) (S).

Force F and displacement S are parallel. Here, the bob does not move in y axis.
 The work done by the tension T along y axis is zero.
Since the bob moves horizontally (along x – axis) with an acceleration, a (say)
 Fx = ma T cos
T
 T sin = ma  y
Tsin
 a = …(1) x
m
a
For equilibrium of the bob along Yaxis
Fy = 0  T cos = mg …(2) mg
 mg  sin 
From (1) and (2), a =     g tan 
 cos  m 
 a = gtan …(3)
Since the breaking strength,
2mg mg 2mg
T=  
3 cos 3
  = 30, putting  = 30 in (3).
g
We obtain a = g tan 30 = .
3

Hence, required work done = W = FS= (T sin) (1/2) at2

2 g
where T = mg,  = 30, a = , t = 1 sec.
3 3
 2  1  1  g  2 mg 2
 W=  mg        (1) = .
 3  2  2  3  6

Prob 14. A point mass m starts from rest and slides down the m
surface of a fixed frictionless solid hemisphere of radius R

as shown in figure. Measure angles from the vertical and R
potential energy from the top. Find
(a) the change in potential energy of the mass with angle
(b) the kinetic energy as a function of angle,
(c) the radial and tangential acceleration as a function of angle,
(d) the angle at which the mass flies off the sphere.

Sol. (a) Consider the mass when it is at the point B. A

R(1  cos
UA (P.E. at A) = 0
 B
UB(P.E. at B)= mgR (1  cos ) R
 U = UB  UA
 U =  mgR(1  cos)
Negative sign indicates that P.E. decreases as particle slides down.

(b) Conserving energy at points A and B.

UA + TA = UB + TB
where UA = P.E. at A, UB = P.E. at B
TA = K.E. at A, TB = K.E. at B
 0 + 0 = mgR(1 cos) + TB
 T = mgR(1  cos)
1
(c) Since T = mv 2
2
1
 mv 2 = mgR(1  cos = 2mgR sin2( )
2
 v = 2 gR sin(/2)  ar = v2/R
dv
 ar = 4g sin2(/2) As at =
dt
d
 at = (gR) cos(/2) = (gR) cos(/2)
dt
 at = (g/R) v cos (/2) , as R = v
(d) For circular motion A N
mv 2
mg cos N =
R mgcos mgsin
at the moment when the particle breaks off the
sphere N = 0.
mv 2 v2
 mg cos =  g cos =
R R

As v = 2 gR sin( )
2

 g cos = 4g sin2(/2) = 2g(1  cos)

2 2
 cos =   = cos1 ( )
3 3

Prob 15. A small disc of mass m slides down a smooth hill m

of height h without initial velocity and gets onto a

plank of mass M lying on the horizontal plane at
h
the base of the hill, as shown in the figure. Due to
friction between the disc and the plank, the disc
M
slows down and ultimately, both move together.
(a) Find the common velocity of the disc and the plank.
(b) Find the work done by the friction
(c) Find the distance moved by the disc with respect to the planck before they start moving
together.
N
Sol. (a) Velocity of the disc at the foot of
the hill is v0 = 2gh . When it
f
slides on the plank, friction ms
opposes its motion and favours N
f
the motion of the plank as shown
in the figure.
Let their common velocity be V, then
for m, V = v0  gt
mgt
for M, V =
M
Eliminating t from the above two equations,
mv0
V=
mM
(b) Using Work  Energy theorem
Wf = Kf  Ki
1 1
Wf = (m + M)V2  mv02
2 2
1 mMv20 mMgh
= =
2 mM mM
(c) Acceleration of the disc w.r.t the plank is
m mM
arel = g +  g = g  
M  M 
u2 Mv 02 Mh
xrel = rel  
2a rel 2g  m  M    m  M 

Objective:

Prob 1. A force of magnitude 2mg (1  ay) starts acting in the vertically upward direction on a body
of mass m placed on earth’s surface, where y is the height of the object above the ground
during ascent and ‘a’ is a positive constant. Total height through which the body ascends is
(A) 1/a (B) 1/2a
(C) 2/3a (D) 2/a

Sol. Initial and final kinetic energy of the body is zero. F

Applying work energy principle

h
 
 Fnet .dy = 0 where Fnet = F  mg
0
h
or   mg  2mgay  dy = 0,
0
mg

h
2 h
 1  2ay  dy  0   y  ay
0
  0
0

1
 h
a
Hence, (A) is correct.

Prob 2. A block of mass m = 0.1 kg is released

from a height of 4 m on a curved smooth 4m
surface. On the horizontal surface path
AB is smooth and path BC offers A B C
c o e f f i c i e n t o f f r i c t i on  = 0. 1. 1m 2m
If the impact of block with the vertical wall at C be perfectly elastic, the total distance covered
by the block on the horizontal surface before coming to rest will be (take g = 10ms2)
(A) 29 m (B) 49 m
(C) 59 m (D) 109 m

Sol. KE attained by block at B = mgh = 4J

Work done by friction force on path BC = (mg)(BC) = 0.2 J.
On the other horizontal surface block can make total forward and backward
 4 
   20 trips. but initially it will stop at B.
 0.2 
 Distance covered = 20  (AB + BC)  AB = 59 m.
Hence, (C) is correct.

Prob 3. A particle of mass m is projected with velocity u at an angle  with horizontal. During the
period when the particle descends from highest point to the position where its velocity vector
makes an angle /2 with horizontal, the work done by the gravity force is
1 1
(A) mu2 tan2 /2 (B) mu2 tan2 
2 2
1 2 1
(C) mu cos2  tan2/2 (D) mu2 cos2 /2 sin2
2 2

Sol. As horizontal component of velocity does not u u cos 

change,  /2
v
v cos /2 = u cos 
u cos 
v=
cos 
1 1
Wgravity = K = mv2  m(ucos )2
2 2
1 2 2 2
= mu cos  tan /2
2
Hence, (C) is correct.

Prob 4. A body of mass 1 kg thrown upwards with a velocity of 10 m/s comes to rest (momentarily)
after moving up 4 m. The work done by air drag in this process is
(Take g = 10 m/s2)
(A) 20 J (B) –10 J
(C) 30 J (D) 0 J

Sol. According to work energy theorem

Wgravitational force + Wair drag = change in KE.
 
 mg. r + Wair drag = change in KE.
  mgh + Wair drag = change in KE
1
 Wair drag = change in KE + mgh = [0  mu2 ] + mgh
2
= 50 + 40 = 10 Joules.
Hence, (B) is correct.

Prob 5. A vehicle is driven along a straight horizontal track by a motor, which exerts a constant
driving force. The vehicle starts from rest at t = 0 and the effects of friction and air
resistance are negligible. If kinetic energy of vehicle at time t is E and power developed by the
motor is P, which of the following graph is/ are correct?
(A) (B)
P E

O t O t

Sol. Since force on the vehicle is constant, therefore, it will move with a constant acceleration. Let
this acceleration be ‘a’.
Then at time t, its velocity will be equal to v = a.t

1 1
Hence, at time t, the kinetic energy, mv 2  ma 2 t 2
E=
2 2
The power associated with the force is equal P = F.v = Fat. Hence, the graph between power
and time will be a straight line passing through the origin.
Hence, (A) is correct.

Prob 6. A block is suspended by an ideal spring of force constant K. If the block is pulled down by
applying a constant force F and if maximum displacement of block from its initial position of
rest is , then
F 2F
(A) < <
K K
2F
(B)  =
K
(C)  = F/K
1
(D) Increase in potential energy of the spring is K 2
2

Sol. Let mass of the block hanging from the spring be m. Then, initial elongation of the spring will
be equal to mg/K. When the force F is applied to pull the block down, then work done by F
and further loss of gravitational potential energy of the block is used to increase the potential
energy of this spring.
Hence, (F. + mg.)
2 2 2
1  mg  m g
= K   
2  K  2 K
2F
From this equation,  = ,
K
Hence, (B) is correct.

Prob 7. A stone is projected at time t = 0 with a speed V0 at an angle  with the horizontal in a
uniform gravitational field. The rate of work done (P) by the gravitational force plotted
against time (t) will be as
(B)
(A)
P
P

O t O t

O O
t t

Sol. Rate of work done is the power associated with the force. It means rate of work done by the
gravitational force is the power associated with the gravitational force.

Gravitational force acting on the block is equal to its weight mg which acts vertically
downwards.
Velocity of the particle (at time t ) has two components,
(i) a horizontal component v0 cos, (ii) a vertically upward component (v0 singt)
Hence, the power associated with the weight mg will be equal to

P = m g.v  mg  v0 sin   gt 
This shows that the curve between power and time will be a straight line having positive slope
but negative intercept on Yaxis.
Hence, (D) is correct.

Prob 8. A 5 kg block is kept on a horizontal platform at rest. At time t = 0, the platform starts moving
with a constant acceleration of 1 m/s2. The coefficient of friction  between the block and the
platform is 0.2. The work done by the force of friction on the block in the fixed reference
frame in 10 s is
(A) +250 J (B) 250 J
(C) +500 J (D) 500 J

mg
Sol. Assuming that the block does not slide on the platform a
FF = ma = 5(1) = 5N ; N – mg = 0 Ff
 N = mg = 50 N, As N = 10 N
Ff < N N
The block will remain at rest relative to the platform.
1
Displacement D relative to the ground = (1)(10) 2 = 50 m
2
 Work done by friction = FfD cos 0 = +250 J,
Hence, (A) is correct.
Prob 9. In the previous Problem, if  = 0.02, the work done by the force of friction on the block in the
fixed reference frame in 10 seconds is
(A) +10 J (B) 10 J
(C) +250 J (D) 250 J
mg
a
Sol. Limiting force of friction = N = 0.02 (50) = 1N
 The block will slide on the platform. Ff
1
Ff = ma = 1 ; a = m/s2
5 N
11
 Displacement D =   (10) 2  10m
25
Work done by Ff = 1N (10 m) = +10 J
Hence, (A) is correct.

Prob 10. A particle of mass m starts from rest and moves in a circular path of radius R with a uniform
angular acceleration . The kinetic energy of the particle after n revolutions is
(A) nmR2 (B) 2 nmR2
(C) (1/2) mnR2 (D) mnR2

Sol. For uniform angular acceleration in a circular path, the angular speed is given by
2 = 02  2 = 0 + 2 (2n)

Kinetic energy of the particle

1 1
KE = m(R)2 = mR2 (4n) = 2nm R2
2 2
Hence, (B) is correct.

Prob 11. A particle of mass m attached to an inextensible light string is moving in a vertical circle of
radius r. The critical velocity at the highest point is v0 to complete the vertical circle. The
tension in the string when it becomes horizontal is
3mv02 9mv02
(A) (B)
r r
(C) 3mg (D) both (A) and (C) are correct.

Sol. At point A, in the horizontal direction B

mv2A
T= . . . (1) T
r A
At the highest point B,
mv 20
mg  . . . (2)
r
By conserving energy at points A and B,
1 1
mv02  mg  2r   mv2A  mgr . . . (3)
2 2
From equations (2) and (3)
v A  3gr
Hence, the tension in the string at the point A,
m  3gr 
T=
r
or, T = 3 mg
Hence, (C) is correct.

Prob 12. A block of mass 10 kg accelerates uniformly from rest to a speed of 2 m/s in
20 sec. The average power developed in time interval of 0 to 20 sec is
(A) 10 W (B) 1 W
(C) 20 W (D) 2 W

Sol. Average power

Net work done
Pav =
Total time taken
Net work done = change in kinetic energy
1
= 10  22 = 20J . . . (1)
2
20
Average power = = 1 watt
20
Hence, (B) is correct.

Prob 13. A pumping machine pumps water at a rate of 60 cc per sec at a pressure of 1.5 atm. The
power delivered by the machine is
(A) 9 watt (B) 6 watt
(C ) 9 kW (D) None of these

Sol. Power = F.v Where F = force imparted by the machine , v = velocity of the liquid
P = p.A.v, Where p = pressure & A = effective area
dV  3 
P=p =  105  (60  106 ) = 9 watt. ( 1 atm  105 N/m2 )
dt 2 
Hence, (A) is correct.

Prob 14. A spring, placed horizontally on a rough surface is compressed by a block of mass m, placed
on the same surface so as to store maximum energy in the spring. If the coefficient of friction
between the block and the surface is  , the potential energy stored in the spring is
2 m 2 g 2 2m 2 g 2
(A) (B)
2k k
2 2
mg 3 mg 2
2
(C) (D)
2k k

Sol. For equilibrium of the block Fmax

m
Fmax  mg = 0  Fmax = mg
F2 2 m 2 g 2
 U = max  mg
2k 2k
Hence, (A) is correct.

FILL IN THE BLANKS IN THE FOLLOWING QUESTIONS.

Prob 1. A stone is projected up with a velocity u. It reaches to a maximum height of h. When it is at a

height of 3h/4 from the ground, the ratio of K.E. and P.E. at that point is ……..

Sol. 1 : 3.

Prob 2. The negative of the work done by the conservative forces on a system equals the change in
…….. energy.

Sol. Potential

Prob 3. The work done by all the forces (external and internal) on a system equals the change in
……….…..energy.

Sol. Kinetic

Prob 4. In the stable equilibrium position, a body has …….. potential energy.

Sol. Minimum.

Prob 5. A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient
of friction between the tyres and the road be , then the stopping distance is ……..

P2
Sol.
2m 2 g

STATE WHETHER THE FOLLOWING QUESTIONS ARE TRUE OR FALSE.

Prob 1. Kinetic energy of an object can have negative value.

1
Sol. False. K.E.  mv 2 , v2 and m are always positive and hence KE is always positive
2

Prob 2. A block starting from rest reaches the bottom of a rough inclined plane with velocity u. Now if
velocity u is given at the bottom, up the incline plane, the block will reach the top.

Sol. False. Some energy will be spent against frictional force

Prob 3. When a body moves in a circular path with uniform speed, no work is done by the force.

Sol. True. This is because force and displacement are always perpendicular to each other.

Prob 4. Power is a vector quantity,

False. It is a dot product of two vector quantities, force and velocity.

Prob 5. Work done by frictional force is always negative because the force always acts opposite to
displacement.

Sol. False.

ASSIGNMENT PROBLEMS

Subjective:

Level - O

1. How much work is done on a body of mass M in moving one round on a horizontal circle of radius r
with constant speed ? Give reason.

2. A lorry and a car moving with the same kinetic energy are brought to rest by the application of brakes,
which provide equal retarding forces. Which of them will come to rest in a shorter distance?

3. A car of mass m starts with acceleration ‘a’ along a straight level road against a constant external
resistive force R. When the velocity of the car is v, what is the rate at which the engine of the car will be
doing work ?

4. If a body of mass m accelerates uniformly from rest to the velocity v1 in time t1 , then find out the
instantaneous power delivered to the body as a function of time t.

5. A block of mass m is pulled along a horizontal surface by applying a force at an angle  with the
horizontal. If the block travels with a uniform velocity and has a displacement d and the coefficient of
friction is F, then find the work done by the applied force.

6. A coin of mass m slides a distance D along a tabletop. If the coefficient of friction between the coin and
table is , find the work done on the coin by friction.

7. A ball is thrown at an angle of 60 with the horizontal with initial kinetic energy K. What is the kinetic
energy at the highest point of its flight ?
1
8. A bird is flying at a speed of 10 m/s. The kinetic energy of the bird is times its potential energy
10
(with respect to the ground). What is the height above the ground at which the bird is flying ?

9. Explain why friction is a nonconservative force.

10. Prove the workenergy principle for a particle moving with constant acceleration (under a constant
force) along a straight line.
11. State and explain the law of conservation of energy. Show that the energy in case of free fall of a body
is conserved.

12. A bucket tied to a string is lowered at a constant acceleration of g/4. If the mass of the bucket is m and
it is lowered by a distance , then find the work done by the string on the bucket.

13. Can you associate potential energy with a nonconservative force?

14. In a tug of war, one team is slowly giving way to the other. What work is being done and by whom?

15. If the simple pendulum shown in the figure is released from point A, what A
75 cm
will be the speed of the ball as it passes through point C? 37

B
C
16. A spring which stretches 10 cm under a load of 200 g requires how much work to stretch it 5 cm from
its equilibrium position?

17. When a ball is thrown up, the magnitude of its momentum first decreases and then increases. Does this
violate the conservation of momentum principle?

18. Top views of two horizontal forces pulling a box along the floor is shown 85 N
in the figure. 60 N 45 30
(a) How much work does each forces do as the box is displaced 70 cm
along the broken line?
(b) Calculate the total work done by the two forces in pulling the box
through this distance.
19. A thin uniform rectangular slab of area (3.4  2.0) m2 has a mass of 180 kg. It is lying on the flat
ground. Calculate the minimum amount of work needed to stand it over its larger edge.

20. A man weighing 60 kg climbs up a stair case carrying a 20 kg load on his head. The stair case has 20
steps and each step has a height of 20 m. If he takes 10 sec to climb, calculate the power.

Level - I
 
1. A particle moves from a point r1   2m  î   3m  ˆj to another point r2   3m  î   2m  ˆj during which a

certain force F   5N  î   5N  ˆj acts on it. Find the work done by the force on the particle during the
displacement.

2. A force F = a + bx acts on a particle in the x direction, where a and b are constants. Find the work done
by this force during a displacement from x = 0 to x = d.

3. A small block of mass m is kept on a rough inclined plane of inclination  fixed in an elevator going up
with uniform velocity v and the block does not slide on the wedge. Find the work done by the force of
friction on the block in time t.

O
4. The figure shows a smooth circular path of radius R in the vertical plane which B


subtends an angle at O. A block of mass m is taken from position A to B
2
under the action of a constant horizontal force F. F
A

(a) Find the work done by this force.

(b) In part (a), if the block is being pulled by a force F which is always tangential to the surface, find
the work done by the force F between A and B.

5. A block of mass M is kept on a platform which is accelerating upward from rest with a constant
acceleration a, during the time interval from t = 0 to t = t0. Find
(a) the work done by gravity, (b) the work done by normal reaction.

6. Two cylindrical vessels of equal crosssectional area A contain water upto heights h1 and h2. The
vessels are interconnected so that the levels in them become equal. Calculate the work done by the force
of gravity during the process. The density of water is .

7. A body of mass m is thrown at an angle  to the horizontal with a velocity u. Find the mean power
developed by gravity over the whole time of motion and the instantaneous power of gravity as a
function of time.

8. Two blocks are connected by a string as shown in the figure. They are `
m1

released from rest. If the coefficient of friction between m1 and the

surface is , find the common speed of the blocks at the instant when
they have moved a distance . m2

9. A 263 gm block is dropped onto a vertical spring with force constant

k = 2.52N/cm. The block sticks to the spring, and the spring compresses
11.8 cm before coming momentarily to rest. While the spring is being
compressed, how much work is done
(a) by the force of gravity, and
(b) by the spring?
(c) What was the speed of the block just before it hits the spring?

(d) If this initial speed of the block is doubled, what is the maximum
compression of the spring? Ignore friction.

10. The displacement x of a particle moving in one dimension under the action of a constant force is related
to time ‘t’ by the equation t = x  3 , where x is in meter and t in sec. Calculate
(a) the displacement of the particle when its velocity is zero.
(b) the work done by the force in the first 6 sec.

11. Two smooth wedges of equal mass m are placed as shown in figure. All
surfaces are smooth. Find the velocities of A & B when A hits the
ground.
A B

h


Level - II

1. Two blocks of different masses are hanging on two ends of a string passing over a frictionless pulley.
The heavier block has mass twice that of the lighter one. The tension in the string is 60 N. Find the
decrease in potential energy during the first second after the system is released.

2. A block of mass 2.0 kg is pulled up on a smooth incline of angle 30o with the horizontal. If the block
moves with an acceleration of 1.0 m/s2, find the power delivered by the pulling force at a time 4.0 s
after the motion starts. What is the average power delivered during 4.0 s after the motion starts?

2m
3. A string with one end fixed on a rigid wall passing over a fixed frictionless
M
pulley at a distance of 2 m from the wall has a point mass M = 2 kg attached 1m
to it at a distance of 1 m, from the wall. A mass m = 0.5 kg attached at the
free end is held at rest so that the string is horizontal between the wall and
the pulley, and vertical beyond the pulley. What will be the speed with m

which the mass M will hit the wall when mass m is released?

m
4. One end of a spring of natural length h and spring constant k is
fixed at the ground and the other is fitted with a smooth ring of h
mass m which is allowed to slide on a horizontal rod fixed at height 370

h as shown in the figure. Initially, the spring makes an angle of 370

with the vertical when the system is released from rest. Find the
speed of the ring when the spring becomes vertical.

5. An automobile of mass ‘m’ accelerates starting from rest, while the engine supplies constant power P.
show that:
(a) the velocity is given as a function of time by v = (2Pt/m)1/2
(b) the position is given as a function of time by s = (8P/9m)1/2t3/2.

6. A stone with weight W is thrown vertically upwards into the air with initial speed v0. If a constant force
f due to air drag acts on the stone throughout its flight,
v 20
(a) show that the maximum height reached by the stone is h =
2g 1  (f / W) 
1/ 2
 W f 
(b) show that the speed of the stone upon impact with the ground is v  v0  
Wf 

7. A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as
an = at2, where ‘a’ is a constant. Find the time dependence of the power developed by all the forces
acting on the particle, and the mean value of this power averaged over the first t seconds after the
beginning of motion.

8. A rigid body of mass 400 kg is pulled vertically up by a light inextensible chain. Initially, the body is at
rest and the pull on the chain is 60 kN. The pull gets smaller at the rate of 3.60 kN each meter through
which it is raised. Find the speed of the body when it has been raised by 10 m.

9. A particle is hanging from a fixed point O by means of a string of length a. There is a small nail 'Q' in
the same horizontal line with O at a distance b (b = a/3) from O. Find the minimum velocity with
which the particle should be projected so that it may make a complete revolution around the nail
without being slackened.

10. A rod of mass m and length  is kept on a smooth wedge of mass M

as shown in the figure. If the system is released when the rod is at m

the top of the wedge, find the speed of the wedge when the rod hits H
M
the ground level. Neglect friction between all surfaces in contact. 

11. A small body starts sliding from the height ‘h’ down an inclined
groove passing into a half circle of radius h/2. Find the speed of the h/2
body when it reaches the highest point. h

12. A smooth sphere of radius R is made to translate in a straight line with constant acceleration a. A
particle kept on the top of sphere is released from there at zero velocity w.r.t. the sphere. Find the speed
of the particle with respect to sphere as a function of angle  as its slides on the spherical surface.

r
13 AOB is a smooth semicircular track of radius r. A block of A
B
mass m is given a velocity 2rg parallel to track at point A.
Calculate normal reaction between block and track when
block reaches at point O.

14. An object dropped from a height h strikes the ground with a speed of k gh where k < 2 . Calculate
the work done by airfriction.

15. A small block of mass m slides along a frictionless loop inside B

loop track as shown in figure. Find the minimum ratio R/r so
R
that the block may not lose contact at the highest point of inner
loop. r
A

Objective:

Level - I

1. Two bodies of masses m1 and m2 have equal momenta. Their kinetic energies E1 and E2 are in the ratio:
(A) m1 : m2 (B) m1 : m2
(C) m2 : m1 (D) m12 : m 22

2. A motor boat is travelling with a speed of 3.0 m/sec. If the force on it due to water flow is 500 N, the
power of the boat is:
(A) 150 kW (B) 15 kW
(C) 1.5 kW (D) 150 W

3. A chain of mass m and length  is placed on a table with onesixth of it hanging freely from the table
edge. The amount of work done to pull the chain on the table is:
(A) mg/4 (B) mg/6
(C) mg/72 (D) mg/36

4. A ball looses 15 % of its kinetic energy after it bounces back from a concrete slab. The speed with
which one must throw it vertically down from a height of 12.4 m to have it bounce back to the same
height is
(A) 2.5 m/s (B) 4.38 m/s
(C) 6.55 m/s (D) 8.25 m/s

5. A body is dropped from a certain height. When it has lost an amount of potential energy ‘U’, it
subsequently acquires a velocity ‘v’. The mass of the body is:
2U U
(A) 2 (B)
v 2v2
2v U2
(C) (D)
U 2v
6. A man of mass 80 kg runs up a staircase in 15 s. Another man of same mass runs up the staircase in 20
s. The ratio of the power developed by them will be:
(A) 1 (B) 4/3
(C) 16/9 (D) None of the above


7.  
A particle moves with a velocity v = 5iˆ  3jˆ  6kˆ m/s under the influence of a constant

 
force F  10iˆ  10jˆ  20kˆ N . The instantaneous power delivered to the particle is:

(A) 200 J/s (B) 40 J/s

8. If v, P and E denote the velocity, momentum and kinetic energy of the particle, then:
(A) P = dE/dv (B) P = dE/dt
(C) P = dv/dt (D) none of these

9. Can a body have energy without having momentum?

(A) the body which has no momentum will have no energy
(B) the body which has no momentum, will have potential energy by virtue of its position.
(C) the body which has no momentum, may have no kinetic as well as potential energy.
(D) nothing can be said

10. When a ball is thrown up, the magnitude of its momentum decreases and then increases. Does this
violate the conservation of momentum? Answer by selecting the correct statement.
(A) The conservation of momentum is violated.
(B) The conservation of momentum is not violated. Here the system includes (ball + earth + air
molecules) and the net momentum of the system remains constant.
(C) It depends upon how high the ball is thrown.
(D) Conservation of momentum is sometimes violated and sometimes not.

Multiple choice questions (More than one correct option)

11. A block of mass m is released on the frictionless incline as shown

from a height R. The incline turns into a circle of radius R/2. The
maximum height attained by the block is
R/2
(A) less than R/2 (B) greater than R/2 R
(C) equal to R (D) less than R

12. A heavy particle is projected up from a point at an angle with the horizontal. At any instant ’t’, if p =
linear momentum, y = vertical displacement, x = horizontal displacement, then the kinetic energy of
the particle plotted against these parameters can be
(A) (B)

K.E
K.E

O
t
O y
(C) (D)

K.E
K.E

O x O p2

13. The plot of velocity versus time is shown in the figure. 

A varying force acts on the body. The correct B C
statement(s) among the following is (are)
(A) In moving from A to B, work done on the body is D
A
negative. t
(B) In moving from B to C no work is done on the body. E
(C) In moving from C to D, work done by the force on the body is positive.
(D) In moving from D to E, work done by the force on the body is positive.

14. A body of mass M was slowly hauled up the rough hill by a force F which at each F
point was directed along tangent to the hill. Work done by the force: M
(A) is independent of shape of trajectory
(B) depends upon vertical component of displacement but independent of the
horizontal component
(C) depends upon both the components of displacement horizontal as well
vertical.
(D) does not depend upon coefficient of friction

15. The potential energy U (in joule) of a particle of mass 1 kg moving in xy plane obeys the law U = 3x
+ 4y, where (x, y) are the coordinates of the particle in metre. If the particle is at rest at (6, 4) at time
t = 0, then:
(A) the particle has constant acceleration
(B) the particle has zero acceleration
(C) the speed of particle when it crosses the yaxis is 10 m/s
(D) coordinates of the particle at t = 1 sec are (4.5, 2)

True or False Type Questions

1. Total energy of a system is always conserved, no matter what internal and external forces on the body
are present.

2. Work done in the motion of a body over a closed loop is zero for any force.

3. Work done in moving a body does not depend on how fast or slow the body is moved.

4. Work done by an individual force is zero if the work done by the resultant force is also zero.

5. Kinetic energy is an absolute value and it does not depend upon the frame of reference.

Fill in the Blanks

1. A ball whose K.E. is E, is thrown at an angle of 45 with the horizontal; its K.E. at the highest point of
its flight will be …………

2. A mass M lowered with the help of a string by a distance x at a constant acceleration g/2. The work
done by string is …………

3. Two bodies with masses M1 and M2 have equal kinetic energies. If P1 and P2 are their respective
momenta, then P1/P2 is equal to …………

4. Work is said to be negative when angle between force and displacement lies between ………… and
…………

5. Area under a ………… curve (taking due care of algebraic sign) gives work done by the force.

Level - II

1. How much work is done in raising a stone of mass 5 kg and relative density 3 lying at the bed of a lake
through a height of 3 metre? (Take g = 10 ms2)
(A) 25 J (B) 100 J
(C) 75 J (D) None of the above

2. A body is acted upon by a force which is proportional to the distance covered. If distance covered is
represented by s, then work done by the force will be proportional to
(A) s (B) s2
(C) s (D) None of the above

3. A car is moving along a straight horizontal road with a speed v0. If the coefficient of friction between
the tyres and the road is , the shortest distance in which the car can be stopped is
v 20 v2
(A) (B) 0
2g g
2v20
(C) (D) none of these
g

4. A block of mass m is released on top of a wedge, which is free to move on a horizontal surface.
Neglecting friction between the surfaces in contact, which of the following statement is true ?
(A) The kinetic energy of the block when it reaches the bottom of the wedge is mgH.
(B) The kinetic energy of the wedge when the block reaches the bottom is mgH.
(C) The work of normal reaction on the block in the ground reference is not zero.
(D) The potential energy of the wedge continuously changes.

5. A rail road car is moving with a constant acceleration of 1 m/s2.

A block of 5 kg is put on a horizontal rough floor in the car. At
time t = 0, velocity of the car is 5 m/s. Considering friction is a = 1 m/s
2

sufficient and block is not slipping on the floor, the work done on
the block by friction force during t = 0 to t = 2 sec will be
(coefficient of friction is s)
(A) 60 J (B) –60J
(C) 10 J (D) 600 sJ

6. A particle of mass m moves in a conservative force field where the potential energy U varies with
position coordinate x as U = U0(1 – cos ax), U0 and a are positive constants. Which of the following
statement is true regarding its motion (  x  ) ?
(A) The acceleration is constant.
(B) The kinetic energy is constant.
(C) The acceleration is directed along the position vector.
(D) The acceleration is directed opposite to the position vector.

FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Terminal), New Delhi -16, Ph 26515949, 26854102, Fax 26513942
FIITJEE Ltd. Material Provided by - Material Point Available on - Learnaf.com
Pinnacle Study Package-68 PH-WEP-55

7. A particle of mass m moves under the influence of the force F = a (sin t ˆi  cos tjˆ ), where a,  are
constants and t is time. The particle is initially at rest at the origin. The instantaneous power given to
the particle is
(A) zero (B) a2sin t/ m
(C) a2cos t/ m (D) a2(sin t+ cost)/m

8. A horizontal massive platform is moving with a constant velocity v0. At time t = 0, a small block of
mass m is gently placed on the platform. If the coefficient of friction between the block and the
platform is , the work done by the force of friction on the block in the fixed ground reference frame
(from t = 0 to a sufficiently long time) is
1 1
(A) + mv 20 (B)  mv20
2 2
1
(C) + mv20 (D) zero
2

9. A block of mass m is moving with a constant acceleration ‘a’ on a rough horizontal plane. If the
coefficient of friction between the block and ground is , the power delivered by the external agent
after a time t from the beginning is equal to:
(A) 2mat (B) mgat
(C) m(a+g)gt (D) m(a+g)at

10. A body is moved along a straight line by a machine delivering constant power. The distance moved by
the body in time t is proportional to
(A) t1/ 2 (B) t 3 4
(C) t 32
(D) t2

More than one correct option.

11. Select the correct alternative(s).

(A) Work done by the static friction is always zero.
(B) Work done by kinetic friction can be positive also.
(C) Kinetic energy of a system is always constant by applying external force on the system.
(D) Work–energy theorem is valid in a noninertial frame also.

12. A particle moves in a straight line with constant acceleration under a constant force F. Select the
correct alternative (s).
(A) Power developed by this force varies linearly with time.
(B) Power developed by this force varies parabolically with time.
(C) Power developed by this force varies linearly with displacement.
(D) Power developed by this force varies parabolically with displacement.

13. In projectile motion, power of the gravitational force

(A) is constant throughout
(B) is negative for first half, zero at topmost point and positive for rest half
(C) varies linearly with time
(D) is positive for complete path

14. Kinetic energy of a particle moving in a straight line is proportional to the time t. The magnitude of the
force acting on the particle is
(A) directly proportional to the speed of the particle
(B) inversely proportional to t
(C) inversely proportional to the speed of the particle
(D) directly proportional to t

15. A particle is taken from point A to point B under the influence of a force field. Now, it is taken back
from B to A and it is observed that the work done in taking the particle from A to B is not equal to the
work done in taking it from B to A. If Wnc and Wc are the work done by non–conservative and
conservative forces, respectively, present in the system, U is the change in potential energy and k is
the change in kinetic energy, then
(A) Wnc  U = k (B) Wc = U
(C) Wnc + Wc = k (D) Wnc U = k

True or False Type Questions

1. In circular motion, work done on a body is always zero.

2. Work done by frictional force may be positive, zero or negative.

3. Potential energy can be positive on negative but kinetic energy is never negative.

1 2
4. Potential energy of a spring given by relation kx is positive.
2

5. Area under the powertime graph gives the work done and the slope of the worktime curve represents
the power at that instant.

Fill in the blanks

1. Kinetic energy ………… if work is negative and ………… if work is positive.

2. Work done by a constant force is ………… of the actual path followed.

3. Potential energy exist for………… forces and not for………… forces.

4. If different bodies have same linear momentum, the ………… body will have the maximum kinetic
energy.

5. If the particle moves opposite to the conservative field, work done by the field will be …………and so
change in potential energy will be …………, i.e. potential energy will…………

ANSWERS TO ASSIGNMENT PROBLEMS

Subjective:

Level - O

1. Zero, because the centripetal force needed to revolve the body is always perpendicular to the
circular path.
2. Both will travel the same distance.

mv12
3. (R + ma) v 4. t
t12
Mgd cos 
5. 6. mgD
(cos    sin )
1
7. K 8. 50 m
4
9. Work done against friction along any closed path is nonzero.

3
10. WAB  K B  K A 12.  mg
4

13. No, we can associate potential energy only with a conservative force.

14. The winning team is performing over loosing team.

15. 3.83 m/s 16. 0.025 J

17. The momentum of the system remains constant. Here, the system includes ball, earth and air
molecules.
18. (a) 51.5 J, 29.7 J ; (b) 81.2 J 19. 3000 J

20. 32 watt

Level - I

1. Zero 2. (a + bd/2)d

3. mgvt sin2 4. (a) FR (b) FR
2
1 1 A
5. (a)  Mg a t 20 (b) M(g + a)a t 20 6. (h1 – h2)2g
2 2 4
2  m 2  m1  g
7. 0, mg(gt  usin) 8. v
 m1  m2 
9. (a) 304 mJ, (b) 1.75 J, (c) 3.32 m/s, (d) 22.5 cm
10. (a) zero (b) zero 11. vB = 2gh cos , vA = 2gh sin 

Level – II

1. 75 J 2. 48 W, 24 W
h k
3. v = 1.84 m/s. 4.
4 m
7. P = mRat , <P> = mRat/2 8. 43.58 m/s
2mgH
9. 2 ga 10.
M  m tan 2 
2 gh
11. 12. [2R (a sin + g – g cos)]1/2
3 3
 k2 
13. 5mg 14.  1   mgh
 2
R 5
15. 
r 2

Objective:

Level - I

1. C 2. C
3. C 4. C
5. A 6. B
7. C 8. A
9. B 10 B
11. B, D 12. B, C
13. B, D 14. A, C
15. A, D

True or False Type Questions

1. False 2. False
3. True 4. False
5. False

Fill in the blanks

Mgx
1. E/2 2
2
3. M1 ; M 2 4. /2, 
5. force displacement

Level - II

1. B 2. B
3. A 4. C
5. A 6. D
7. A 8. A
9. D 10. C
11. B, D 12. A, D
13. B, C 14. B, C
15. A, B, C

True or False Type Questions

1. False 2. True
3. True 4. True
5. True

Fill in the blanks

1. decreases, increases 2. independent
3. conservative, non conservative 4. lightest.
5. negative, positive, increase.

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