Dowork Energy
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Eswar AnandDowork Energy
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Eswar AnandWork Power Energy 1
WORK , POWER & ENERGY
INTRODUCTION
The terms work, energy and power are frequently used in everyday language. A farmer ploughing
the field, a construction worker carrying bricks, a student studying for competitive examination, all are
said to be working. Somebody who has the capacity to do work for 14–16 hours a day is said to have a
large stamina or energy. Thus work is related to energy. The word power is used in everyday life with
different shades of meaning, Physical power, economical power and so many other examples. We shall
find that these quantities namely work, energy and power in physics have a very loose correlation with the
meaning they have in day to day life.
ENERGY
The Newton’s laws of motion allow us to analyse many kinds of motion. However, the analysis is
often complicated, requiring details about the motion that we simply do not know. For example, a block
sliding along an inclined frictionless track that includes several ups and downs of various shapes, With an
initial speed from a height. Using Newton’s second law can you calculate the block speed when it
reaches the end of the track at zero height? No, not without the details of how the inclination varies all
along the track and the calculation can be very complicated.
There is a more powerful technique for analyzing motion. This other technique involves energy,
which comes in a great many forms. One can define energy as a scalar quantity whose magnitude is
associated with a system of one or more objects. If a force by say, making it move, then the energy
changes. If the magnitude of energy is assigned carefully, then energy can be used to predict the out-
come of experiments.
First we define the form of energy associated with the state of motion of an object called the “kinetic
energy”.
Kinetic energy is greater than zero as the object moves and is zero, when the object is stationary. For an
object of mass m and whose speed is v ( c , the speed of light)
1
Kinetic energy is K mv 2 _______________ (1)
2
for example a stone of mass 1 kg moves past us at 2 m/s has a kinetic energy of 2 kgm2/s2 i.e we
associate that energy with stone’s motion.
The SI unit of kinetic energy (and every other type of energy) is the joule (J)
1 Joule = 1 kgm2/s2.
Illustration 1: A locomotive weighing 1000 tons starts from rest and moves with a constant accel-
eration of 0.5 m/s2. After travelling a distance of 900 m on the track, what is the kinetic energy of
the locomotive?
Sol. v 2aS 2 0.5 900m 30 m / s .
1 1
k m v 2 106 kg (30m / s )2 4.5 108 J
2 2
Work
If an object is accelerated to greater speed by applying a force to the object, its kinetic energy
increases. Similarily, if you deaccelerate the object by applying force its kinetic energy decreases. These
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2 Work Power Energy
changes in kinetic energy can be accounted by saying that force has transferred energy to the object from
you or from the object to yourself.
In such a transfer of energy via a force, work is said to be done on the object by the force.
So, work can be defined as follows:
Work is energy transfered to or from an object by means of a force acting on the object. Energy transferred
to the object is positive work and energy transferred from the object is negative work.
In that sense, if one pushes hard against a wall, he tires, and in common sense, such effort does not
cause an energy transfer to or from the wall and thus no work is done on the wall.
Work & Kinetic Energy
Let us find an expression for work on a block that can slide along a smooth surface (horizontal) by
Newton’s second law
F
Fx
Fx = max.
a x
d
As the block moves through a displacement d the force changes the block’s velocity from initial value V0
to another value V .
Because the force is constant, the acceleration is also constant. So
v 2 v 02 2ax d
2Fx d
v v0
2 2
1 1
mv 2 mv 02 Fx d
2 2
kf k i Fx d . ______________ (2)
Thus the above equation tells us the kinetic energy has been changed by the force and the right side tells
the change is equal to Fxd.
Therefore work done on the block by the force is
W Fx d .
W Fd cos
W F .d (Work done by a constant force) _____________ (3)
To calculate work done by a force using this equation, the force must be a constant force, i.e it must not
change in magnitude or direction as the object moves and object must be rigid i.e all parts of the object
must move together in same direction.
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Sign for Work : The work done on an object by a force is positive if the angle is less than 90° and negative if
the angle is greater than 90° (up to 180°).
Unit of work is same as the unit of kinetic energy ie kg m2/s2 or joule (J).
Net Work done by Several Forces : When two or more forces act on an object, the net work done by the
individual forces. can be calculated in two ways (1) we can find the work done by each force and then sum
those works. (2) we can first find the net force Fnet of those forces and the angle between the directions
of Fnet and the displacement for and write
W Fnet d cos .
Conservative & nonconservative Forces : A force is called conservative when work done by the force be-
tween two points on the path is independent of shape of path traced, depends only on initial and final
position. C B
Net work done by the force in round trip is zero.
D
w ACB BDA 0
A
A force is called convservative when F 0
d ˆ d ˆ d ˆ ˆ ˆ ˆ
dx i dy j dz k Fx i Fy j Fz k 0
iˆ ˆj kˆ
d d d
dx dy dz = 0
Fx Fy Fz
Example : electrostatical force, gravitational force, spring force etc.
A force is called non- conservative when work done by the force depends on shape of path. independent of
initial and final position
Example : Frictional force, Viscous force.
Work Energy theorem :
From. eq (2) & eq (3) we can say that
kf ki W _____________ (4)
(Change in kinetic energy of a particle) = (net work done on the particle)
General Proof of Work Energy Theorem :
Statement : Work done by all the forces (conservative or non-conservative) acting on a particle is equal to
change in its kinetic energy.
Proof: Let us consider that the particle of mass ‘m’ is moving along the particular path w.r.t. a particular
frame of reference. We are considering the motion of the particle from point 1 to point 2. At point 1 its
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4 Work Power Energy
velocity magnitude is | v 1 | and at point 2 its velocity magnitude is | v 2 | . At point ‘p’ of the path velocity
magnitude is | v |.
Now calculating the work done by the net force on the particle,
Y
2
2
W Fnet .dr | Fnet || dr | cos v1 (Fnet)ta ngential= | Fnet| cos
1 1 v
1
Path of the
2
r1 particle
= F
1
net tan
dl ,
Fnet
2
v2
r2
O X
where |Fnet | cos = Fnet tan and | dr | = dl, elementary path length.
2
d |v |
= ma
1
tan dl , where atan =
dt
2
d |v | dl
= m dl where = |v |
1
dt dt
|v 2 |
1 1
= m | v | d | v | 2 m | v
|v1|
2 |2 m | v1 |2 k .
2
Illustration 2: A cyclist comes to a skidding stop in 10m. During this process, the force on the cycle
due to the road is 200N and is directly opposed to the motion. (a) How much work does the road
do on the cycle? (b) How much work does the cycle do on the road?
Solution: Work done on the cycle by the road is the work done by the stopping (frictional) force on the cycle due
to the road.
(a) The stopping force and the displacement make and angle of 180° ( rad) with each other..
Thus, work done by the road,
W r = Fd cos
= 200 × 10 × cos
= – 2000 J
It is negative work that brings the cycle to a halt in accordance with WE theorem.
(b) From Newton’s Third Law an equal and opposite force acts on the road due to the cycle. Its magnitude
is 200N. However, the road undergoes no displacement. Thus, work done by cycle on the road is zero.
The lesson of Example is that though the force on a body A exerted by the body B is always equal and
opposite to that on B by A (Newton’s Third Law); the work done on A by B is not necessarily equal and
opposite to the work done on B by A.
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Illustration 3: The displacement x of a body of mass 1 kg on horizontal smooth surface as a func-
t4
tion of time t is given by x. = . The work done in the first one second is
4
1 1 3 5
(A) J (B) J (C) J (D) J
4 2 4 4
dx 1
Solution: V 4t 3 t 3 1m / s
dt 4
1 1 1
Work done = change in K.E. = 1 (1)2 1 (0)2 J
2 2 2
Note: - W= F.d is not applicable to solve the above problem. This formula is applicable when force is not
dependent on time.
Work done by Variable Force : Let us now consider a force which acts always in one direction but whose
magnitude may keep on varying. We can choose the direction of the force as x-axis. Further, let us
assume that the magnitude of the force is also a function of x or say F(x) is known to us. Now, we are
interseted in finding the work done by this force is moving a body from x1 to x2.
F
F
x1 X
x2 X
dx
Work done in a small displacement from x to x + dx will be
dW = F.dx
Work can be obtained by integration of the above elemental work from x1 to x2 or
x2 x2
W= dW
x1
F .dx
x1
x2
It is important to note that Fdx
x1
is also the area under F-xgraph between x = x1 to x = x2.
Illustration 4: Woman pushes a trunk on a railway platform which has a rough surface. She ap-
plies a force of 100N over a distance of 10m. Thereafter, she gets progressively tired and her
applied force reduces linearly with distance to 50N. The total distance through which the trunk
has been moved is 20m. Plot the force applied by the woman and the frictional force, which is 50
N versus displacement. Calculate the work done by two forces over 20m.
Solution :
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B F C
100
Force (N)
50 E
A D I
0 20m
10m x
f= –50
G H
Plot of the force F applied by the woman and the opposing frictional force f versus displacement.
The plot of the applied force is shown in Fig. 6.4. At x = 20 m, F = 50 N 0 . We are given that the
frictional force f is |f| = 50 N. It opposes motion and acts in a direction opposite of F. It is therefore, shown
on the negative side of the forces axis.
the work done by the woman is
WF area of the rectangle ABCD + area of the trapezium CEID
1
WF = 100 × 10 + 100 50 10
2
= 1000 + 750
= 1750 J
The work done by the frictional force is WF area of the rectangle AGHI
Wf = 50 20
= – 1000 J
The area on the negative side of the force axis has a negative sign.
The Work Energy Theorem for Variable Force: We are now familiar with the concepts of work and kinetic
energy to prove the work-energy theorem for a variable force. We confine ourselves to one dimension. The
time rate of change of kinetic energy is
dk d 1 dv
mv 2 = m v
dt dt 2 dt
= Fv (from Newton’s second Law)
dx
= F Thus dK = Fdx
dt
Integrating from the initial position (xi) to final position (xf), we have
Kf xf
Ki
dK Fdx xi
where, Ki and Kf are the initial and final kinetic energies corresponding to x i and xf
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Work Power Energy 7
xf
or K f – K i Fdx From eq., it follows that Kf – Ki W
xi
Thus, the WE theorem is proved for a variable force.
Illustration 5: A block of mass m = 1 kg, moving on a horizontal surface with speed i 2ms –1
enters a rough patch ranging from x = 0. 10m to x = 2.01 m. The retarding force Fr on the block
in this range is inversely proportional to x over this range,
k
Fr for 0.1 < x < 2.01 m
x
= 0 for x < 0.1 and x > 2.01 m
where k = 0.5J. What is the final kinetic energy and speed f of the block as it crosses this
patch?
Solution : From Work Energy Theorem,
2.01
k dx 1 1
Kf Ki = m2i k ln( x ) |0.1
2.01
= m2i k ln 2.01/ 0.1
0.1 x 2 2
= 2 – 0.5 ln (20.1) = 2 –1.5 = 0.5 J
f 2K f / m 1ms 1
Here, note that ln is a symbol for the natural logarithm to the base e and not the logarithm to the base 10
[ln X = logeX = 2.303 log10X].
Potential Energy :The word potential suggests the possibility or capacity for action. The term potential energy
bring to one’s mind, ‘stored’ energy.
We can define potential energy as the energy associated with the configuration or arrangement of a
system of objects that exert forces on one another. If the configuration changes, then the potential energy
also changes. One type of potential energy is the gravitation potential energy that is associated with the
state of separation between objects. For example let us throw a stone upward. When it rises work done
by gravitational force is negative because the force transfers energy from kinetic energy of the stone. The
stone stops and then begin to fall back down because of gravitational force. During the fall the transfer is
reversed. The work done on the stone by gravitational force is now positive. The force transfers energy
from gravitational potential energy of the stone earth system to the kinetic energy of the stone.
Thus for either rise or fall the change U of potential energy is difined to equal the negative of work done
on the stone by gravitational force
U –W
Another type of potential energy is elastic potential energy. Which is associated with the state of com-
pression or extension of an elastic object. If we push a block towards a spring on a horizontal smooth
surface, the spring force does negative work on the block, transfering energy from kinetic energy of block
to elastic
Potential energy of spring. The block slows and eventually stops and then begins to move in opposite
direction because of spring force still in that direction. The transfer of energy is reversed . It is from
potential energy of spring to kinetic energy of the block.
Potential Energy & Conservation of Mechanical Energy
We define the change in potential energy of a system corresponding to a conservative internal force as
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8 Work Power Energy
f
U f U i F.d r
i
where W is the work done by the internal conservative force on the system as the system passes from the
initial configuration i to the final configuration f.
Type of Potential Energy
1. Elastic : We have seen that the work done by the spring force (of course conservative for an ideal spring) is -
1 2
kx when the spring is stretched or compressed by an amount x from its unstretched or natural posi-
2
tion. Thus,
1 2 1 2
U = -W = - kx or U= kx (k = spring constant)
2 2
Note that elastic potential energy is always positive.
2. Gravitational
The gravitational potential energy of two particles of masses m1 and m2 separated by a distance r is given
by
m1m2
U= G
r
Here, G = universal gravitation constant
Nm 2
= 6.67 × 10-11 kg 2
If a body of mass m is raised to a height 'h' from the surface of earth, the change in potential energy of the
system (earth + body) comes out to be:
mgh
U = 1 (R = radius of earth) or
h
U = mgh if h << R
R
Thus, the potential energy of a body at height h, i.e., mgh is really the change in potential energy of the
system for h << R. So be careful while using U = mgh, that h should not be too large. This we will discuss
in detail in the chapter of Gravitation.
CONSERVATION OF MECHANICAL ENERGY :
Suppose only conservative internal forces operate between the parts of the system and the potential
energy U is defined corresponding to these forces. There are either no external forces or the work done by
them is zero. We have
Uf - Ui = -W = -(Kf - Ki) or Uf + Kf = Ui + Ki . . . . . (i)
The sum of the kinetic energy and the potential energy is called the total mechanical energy. We see
from equation (i) that the total mechanical energy of a system remains constant if the internal
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Work Power Energy 9
forces are conservative and the external forces do not work. This is called the principle of conser-
vation of mechanical energy.
The total mechanical energy K + U is not constant if nonconservative forces, such as friction, act
between the parts of the system. We can't apply the principle of conservation of mechanical
energy in presence of nonconservative forces. The work-energy theorem is still valid even in the
presence of nonconservative forces.
Note: that only a change in potential energy is defined above. We are free to choose the zero potential energy
in any configuration just as we are free to choose the origin in space anywhere we like.
If nonconservative internal forces operate within the system, or external forces do work on the system, the
mechanical energy changes as the configuration changes. According to the work-energy theorem,
the work done by all the forces equals the change in the kinetic energy. Thus,
W c + W nc + W ext = Kf - Ki
where the three terms on the left denote the work done by the conservative internal forces, nonconservative
internal forces and the external forces.
As W c = -(Uf - Ui),
We get W nc + W ext = (Kf + Uf) - (Ki + Ui)
= Ef - Ei . . . . . (ii)
Where E = K + U is the total mechanical energy.
If the internal forces are conservative but external forces also act on the system and they do work,
W nc = 0 and from (ii)
W ext = Ef - Ei . . . . .(iii)
The work done by the external forces equals the change in the mechanical energy of the system.
Illustration 6: In the figure shown stiffness 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 the maximum elongation in the spring
Neglect the mass of the spring, pulley and that of the string.
m
Solution:
Let the maximum elongation in the spring be x, when the block is at position 2.
The displacement of the block m is also x. If E1 and E2 are the energies of the system when the block is
at position 1 and 2 respectively. Then
E1 = U1g + U1s + T1
where U1g = gravitational P.E. with respect to surface S.
m
U1S = P.E. stored in the spring. 1
h1
m 2
h2
S
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T1 = initial K.E. of the block.
E1 = mgh1 + 0 + 0 = mgh1 . . . . (i)
and E2 = U2g + U2s + T2
1 2
= mgh2 + kx 0 . . . . . (ii)
2
From conservation of energy E1 = E2
1 2
mgh1 = mgh2 + kx
2
1 2
kx mg ( h 1 h 2 ) mgx x = 2mg/k
2
Potential Energy Curves
For a conservative force is one dimension,
dU
dU = F . ds F dx Fx
x dx
Graphically the force is negative of the slope of the tangent to potential energy curve.
1 2
For example, the potential energy function of a spring block system is U = kx
2
dU d 1 2
kx kx
dx 2
=
dx
U(x)
Total energy
KE
Slope of
slope of tangent positive
tangent
negative PE
x
In a U(x) graph the points where slope of U(x) graph is zero. are the points where force is zero. i.e. the
points of equilibrium.
U(x) U(x) U(x)
x0 x x0 x x0 x
(A) (B) (C)
The points x = x0 in (A), (B) & (C) are having slope zero i,e point of equilibrium. But in A if the equillibrium
is disturbed. i.e x is made greater or smaller than x0 the slope for x > x0 becomes negative ie the force is
positive ie takes the particle away from x = x0 and for x <x0 slope is positive ie force is negative. Thus this
equilibrium is unstable. Here potential energy is maximum .Similarily for (B) the force for neighbourhood of
x = x0 are that is the equilibrium is stable. Here potential energy is minimum.
And in (C) all the points have zero slope so equilbrium is there for all points
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Work Power Energy 11
Hence equilibrium is neutral.
d 2U
If is calculated at the point of equilibrium one can determine the nature of the equilibrium .
dx 2
d 2U
If 0 , U is maximum and hence equilibrium is unstable.
dx 2
d 2U
If 0 , U is minimum and hence equilibrium is stable.
dx 2
d 2U
If 0 , then U is the same at every point. Hence the equilibrium is neutral.
dx 2
Illustration 7: In a region x 0 and x 2 a particle has potential energy function.
U (x) = A sinx. Find the conservative force F(x) and the locations where a particle under the force
is in equilibrium.
Also state the nature.
dU
SOLUTION: F ( x ) A cos x
dx
for equilibrium
F ( x ) 0 A cos x
3
x , .
2 2
d 2U
At x U ( x ) is max. 2 0 Hence.at x is unstable equilibrium.
2 dx 2
3 d 2U 3
at x U(x) is min 2
0 . Hence.at x is stable equilibrium.
2 dx 2
Work done on a system by an external force: In case of a particle a force only transfers Kinetic energy into
or out of the particle. But in case of a system more complicated an external force can change both kinetic
and potential energies of the system.
(1) No. friction involved.
The work done by a force on a system is say W. Then W = U k for example consider two blocks.
connected by a spring be the system. The spring is in its natural state. A force F displaces the first block
by say
m2 m1 F
a distance x1 and the second block moves by x2 (x1 > x2) then we can write
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1 1 1
Fx1 m1v12 m2v 22 k ( x1 x2 )2
2 2 2
where v1 and v2 are the velocities aquired by m1 and m2 till that instant.
(2) Friction involved: - In the same situation. If we have friction between block and surface, then
Fx1 f1x1 f2 x2 U k .
1 2 1 2 1
Fx1 f1x1 f2 x2 m2v 2 m1v1 k ( x1 x2 )2
2 2 2
Fx1 U k W (against friction) = U K Eth
= E mech Eth where E mech U k .
Eth = Heat generated working against friction.
Illustration 8: A small object of weight w hangs from a string of length R, as shown in figure. A
variable horizontal force P that starts at zero and gradually increases is used to pull the object
very slowly until the string makes an angle with the vertical. Calculate the work of the force P..
Solution: The sum of the works W of all the foces other than the gravitational force must equal the change of
kinetic energy plus the change of gravitational potential energy. Hence
W WP WT K U Rcos R
R
T
Since T is perpendicular to the path of its point of application, W T = 0 ;
and since the body was pulled very slowly at all times, the change of P
y
kinetic energy is also zero. Hence
w
WP U Wmg .y
where y is the distance that the object has been raised. From figure y is seen to be R 1 cos .
Therefore,
WP Wmg R 1 cos
Internal energy sources and work : Consider a person walking & a car moving on road or a man climbing up
stairs. In all these cases the force which is accelerating the body does not perform any work (as the point
of application does not move).
The energy required to work & climb by a man and to move by a car comes from the internal energy stored
in the man or in the fuel of the car.
So even when the force is responsible for the motion of the body it supplies no energy to the body.
Motion of a Particle in a Vertical Circle attached to an In-Extensible light String: Suppose a particle of
mass m is attached to an inextensible light string of length R. The particle is moving in a vertical circle of
radius R about a fixed point O. It is imparted a velocity u in horizontal direction at lowest point A. Let v be
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Work Power Energy 13
its velocity at point B of the circle as shown in figure. Here
h = R(1 - cos ) . . . . .(i)
From conservation of mechanical energy
1
m(u 2 v 2 ) mgh or v2 = u2 - 2gh . . . . . (ii)
2
The necessary centripetal force is provided by the resultant of tension T and mg cos
O
T v
mv 2 B
T - mg cos = . . . . .(iii)
R A u mg
Now, following three conditions arise depending on the value of u.
Case - 1 : (u 5 gR )
The particle will complete the circle if the string does not slack even at the highest point ( = ). Thus,
tension in the string should be greater than or equal to zero (T 0) at = . In critical case, substituting
T = 0 and = in equation (iii), we get P
T=0
vmin = gR
mv 2 min
mg =
R O
or 2
v min gR A u
umin = 5 gR
T = 6mg
or vmin = gR (at highest point)
Substituting = in equation (i) h = 2R
Therefore, from equation (ii) 2
u min v min
2
2 gh
or 2
u min gR 2 g (2 R) 5 gR or umin = 5 gR
Thus, if u 5gR , the particle will complete the circle.
At u = 5gR , velocity at highest point is v = gR , and tension in the string is zero.
Substituting = 00 and v = u = 5 gR in equation (iii),
we get T = 6 mg or in the critical condition tension in the string at lowest position is 6 mg. This is
shown in figure.
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If u < 5gR , following two cases are possible.
Case - 2: 2 gR u 5 gR
If u < 5 gR , the tension in the string will become zero before reaching the highest point.
From equation (iii), tension in the string becomes zero (T = 0)
v2 2 gh u 2
where cos = or cos =
Rg Rg
Substituting, this value of cos in equation (i), we get
2 gh u 2 h 1 u 2 Rg
1 h h 1 . . . .(iv)
Rg R 3 g
or we can say that at height h 1 tension in the string becomes zero. Further, if u < 5gR ,
velocity of the particle becomes zero when
u2
0 = u - 2gh
2
or h= h2 (say) . . . .(v) v
2g
T=0
v=0
u
i.e. at height h2 velocity of particle becomes zero.
Now, the particle will leave the circle if tension in the string becomes zero but velocity is not
zero or
T = 0 but v 0. This is possible only when h1 < h2
u 2 Rg u 2
or or 2u2 + 2Rg < 3u2
3g 2g
or u2 > 2Rg or u > 2 Rg
Therefore, if 2gR u 5gR , the particle leaves the circle.
From equation (v), we can see that h > R if u2 > 2gR. Thus, the particle, will leave the circle when h > R or
900 < < 1800. This situation is shown in the figure.
2gR u 5gR or 900 < < 1800
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Case - 3: (0 < u < 2 gR )
The particle will oscillate if velocity of the particle becomes zero but tension in the string is not zero.
or v = 0, but T 0. This is possible when
h2 < h1 O
v=0
T0
u 2 u 2 Rg P
or or 3u2 < 2u2 + 2Rg hR
2g 3g
u
A
or u2 < 2Rg or u < 2 Rg
Further, if h1 = h2, u = 2 Rg and tension and velocity both becomes zero simultaneously..
Further, from equation (v), we can say that h < R if u < 2 Rg .
Thus, for 0 < u < 2gR , particle oscillates in lower half of the circle (00 < < 900). This situation
is shown in the figure,
0<u< 2 gR or 00 < < 900 .
A heavy particle hanging from a fixed point by a light inextensible string of length l
Illustration 9:
is projected horizontally with speed (gl ). 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 the string from vertical is . Resolving mg along the string and perpendicular to the string, we
get net radial force on the particle at B i.e.
FR = T - mg cos . . . . (i)
If v be the speed of the particle at B, then
mv 2 O
FR = . . . . (ii)
l T
B
From (i) and (ii), we get
mg cos
mv 2 A
T - mg cos = . . . . (iii) mg sin mg
l
Since at B, T = mg
mv 2
mg(1 - cos ) =
l
NARAYANA IIT ACADEMY – INDIA
16 Work Power Energy
v2 = gl(1 - cos ) . . . . .(iv)
Conserving the energy of the particle at point A and B, we have
1 2 1
mv 0 mgl (1 cos ) mv 2
2 2
where v0 = gl and v = gl (1 cos )
gl = 2gl(1 - cos ) + gl(1 - cos )
cos = 2/3 . . . . .(v)
gl
Putting the value of cos in equation (iv) we get v = .
3
A Body Moving Inside a Hollow Tube:
The same discussion holds good for this case, but instead of tension in the string we have the normal
reaction of the surface. If N is the normal reaction at the lowest point, then
mv12 v12 N mg
N - mg = ; N = m g
r r O
At the highest point of the circle,
N
mv 22 v1
N + mg =
r
mg
v 22
N = m g , If the circle has to be completed, then the condition is v 5rg
r 1
Body slides down on a Spherical Surface and flies off at a point
Consider the point C where the mass is, at a certain instant. The forces are the normal reaction R and the
weight mg. The radial component of the weight is mg cos acting towards the center. The centripetal
force is
B
2 R
mv C
mg cos - R = N
r
v
mg
where v is the velocity of the body at C.
O
v2
R = m g cos
r
. . . . . (i)
A
The body flies off the surface at the point where R becomes zero.
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 17
v2 v2
i.e., g cos = ; cos = . . . . (ii)
r rg
To find v, we use conservation of energy
1 2
i.e., mv mg ( BN ) = mg(OB - ON) = mgr(1 - cos )
2
v2 = 2rg(1 - cos )
v2
2(1 - cos ) = . . . .(iii)
rg
From equation (ii) and (iii) we get
cos = 2 - 2 cos ; 3 cos = 2
2 2
cos = ; = cos-1 . . . . (iv)
3 3
This gives the angle at which the body goes off the surface. The height from the ground of that point = AN
= r(1 + cos )
2 5
= r 1 r
3 3
Illustration-10: A point mass ‘m’ starts from rest and slides down the surface of a frictionless solid
sphere of radius ‘R’ as shown in the figure. At what angle will this body break off the surface of
sphere? Find the velocity with which it will break off.
Solution: m
R
Applying COE, we have at the point A and B, O
R
1
we have mgr (1 – cos ) = mv 2 . . …(i)
2
Force equation gives, mg cos - N = mv2/r . . …(ii)
N = 0 for break off.
v= gr cos . . . . (iii)
2
Putting it in (i) We get cos θ = 2/3 Putting this in (iii) we get v = gr .
3
NARAYANA IIT ACADEMY – INDIA
18 Work Power Energy
Power
From a particle viewpoint, it is interesting to know not only the work done on an object but also the rate at
which the work is being done. The time rate of doing work is called power.
If an external force is applied to an object (Which we assume as a particle), and if the work done by this
force is W in the time interval t, then the average power during this interval is defined as
W
P
t
The work done on the object contributes to increasing the energy of the object. A more general definition
of power is the time rate of energy transfer. The instantaneous power is the limiting value of the average
power as t approaches zero.
W dW
i.e., P = lim
t 0 t dt
where we have represented the infinitesimal value of the work done by dW (even though it is not a change
and therefore not a differential).
dW d r
P= F. F.v
dt dt
dr
where we have used the fact that v .
dt
This SI unit of power is joule per second (J/s), also called watt(W)' (after James Watt);
1W = 1 J/s = 1 kg.m2/s3.
Illustration 11: An elevator can carry a maximum load of 1800 kg (elevator + passengers) is moving
up with a constant speed of 2 ms–1. The frictional force opposing the motion is 4000 N. Determine
the minimum power delivered by the motor to the elevator in watts as well as in horse power.
Solution : The downword force on the elevator is
F = mg + Ff = (1800 × 10 ) + 4000 = 22000 N The motor must supply enough power to balance this force.
Hence,
P = F.v = 22000 × 2 = 44000W = 59 hp
Illustration 12: Sand drops vertically at the rate of 2 kg s–1 on to a conveyor belt moving horizon-
tally with a velocity of 0.1 ms–1. Calculate (i) the extra power needed to keep the belt moving. (ii)
the rate of change of kinetic energy of the sand. Why is the power twice as great as the rate of
change of kinetic energy ?
Solution : (i) Force required to keep belt moving = rate of increase of horizontal momentum of sand = mass per
second (dm/dt) × velocity change = 2 × 0.1 = 0.2 newton.
Therefore, power = work done per second = force × rate of displacement
= force × velocity = 0.2 ×0.1 = 0.02 W
1
(ii) Kinetic energy of sand = mv 2
2
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 19
1 2 dm 1
Therefore, rate of change of energy v , since v is constant 0.12 2 0.01 W
2 dt 2
Thus the power supplied is twice as great as the rate of change of kinetic energy. The extra power is due
to the fact that the sand does not immediately assume the velocity of the belt, so that the belt at first
moves relative to the sand. The extra power is needed to overcome the friction between the sand and belt.
Illustration 13: A pump is required to lift 1000 kg of water per minutes from a well 20 m deep and
eject it at a rate of 20 m/s. (a) How much work is done in lifting water ? (b) How much work is
done in giving it a KE ?
Solution: (a) Work done in lifting water = gain in PE (potential energy)
work = 1000 × g × 20 = 1.96 × 105 J per minute
(b) Work done (per minute) in giving it KE = 1/2 mv2
= 1/2 (1000) (20)2 = 2 × 105 J per minute
KEY CONCEPTS
Work done on a particle by all the forces acting on it is equal to the change in its kinetic energy.
Work done on a system by all the (external and internal) forces is equal to the change in its kinetic
energy.
A force is called conservative if the work done by it in closed path is always zero. The force of gravitation,
Coulomb force, force by a spring etc. are conservative. If the work done by it during a round trip is not zero,
the force is nonconservative. Friction is an example of nonconservative force.
The change in the potential energy of a system corresponding to conservative internal forces is equal to
negative of the work done by these forces.
If no external forces act (or the work done by them is zero) and the internal forces are conservative, the
mechanical energy of the system remains constant. This is known as the principle of conservation of
mechanical energy.
If some of the internal forces are nonconservative, the mechanical energy of the system is not constant.
If the internal forces are conservative, the work done by the external forces is equal to the change in
mechanical energy.
A particle has to be given 5gR velocity at the bottom to complete a vertical circle tied to string.
If 2gR v 5gR the particle leaves the circular path.
W
Pav
t
dw
P instantaneous = F .v dt
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20 Work Power Energy
SINGLE CORRECT ANSWERS
LEVEL-1
Work - Energy Theorem
1. An object of mass m is tied to a string of length l and a variable horizontal
force is applied on it which is initially is zero and gradually increases L
until the string makes an angle with the vertical. Workdone by the F
force F is
(A) mgl(1 – sin) (B) mgl
(C) mgl(1 – cos) (D) mgl(1 + cos )
2. A block of mass m moving with speed v compresses a spring through distance x before its speed is
halved. The value of spring constant is
3mv 2 mv 2 mv 2 2mv 2
(A) (B) (C) (D)
4x 2 4x 2 2x 2 x2
3. A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging
vertically down over the edge of the table. If g is acceleration due to gravity, then the work required to pull
the hanging part onto the table is
MgL MgL MgL
(A) MgL (B) (C) (D)
3 9 18
4. A ball falls under gravity from a height 10 m with an initial velocity v0. It hits the ground, losses 50% of its
energy in collision and it rises to the same height. The value of v0 is
(A) 14 m/s (B) 7 m/s (C) 28 m/s (D) 9.8 m/s.
5. A particle falls from rest under gravity. Its potential energy with respect to the ground (PE) and its kinetic
energy (KE) are plotted against time (t). Choose the correct graph
PE KE KE
PE KE
(A) (B) (C) PE (D) PE
KE
t t t t
6. A particle of mass m is fixed to one end of a light rigid rod of length l and rotated in a vertical circular path
about its other end. The minimum speed of the particle at its highest point must be
(A) zero (B) g (C) 1.5 g (D) 2 g
7. A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a stunt film
shooting. The car moves up along an inclined cliff at a speed v and makes a jump to leave the cliff and
lands at some distance. Let R be the maximum height of the car from the top of the cliff. The tension in the
string when the car is in air is–
mv 2 mv 2
(A) mg (B) mg – (C) mg + (D) zero.
R R
8. A simple pendulum has a string of length l and bob of mass m. When the bob is at its lowest position, it
is given the minimum horizontal speed necessary for it to move in a circular path about the point of
suspension. The tension in the string at the lowest position of the bob is –
(A) 3mg (B) 4mg (C) 5 mg (D) 6 mg
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Work Power Energy 21
9. An electric pump on ground floor of a building takes 10 minutes to fill a tank of volume 30 m3 with water. If
the tank is 60 m above the ground and efficiency of pump is 30%, how much electric power is consumed
by the pump in filling the tank (g = 10 m/s2)
(A) 100 KW (B) 150 KW (C) 200 KW (D) 250 KW
10. An object of mass (m) is located on the horizontal plane at the origin O. The body acquires horizontal
velocity V. the mean power developed by the frictional force during the whole time of motion is ( =
frictional coefficient)
1 V 3
(A) mgV (B) mgV (C) mg (D) mgV
2 4 2
11. The distance x moved by a body of mass 0.5 kg by a force varies with time t as x = 3t2 + 4t + 5 where x
is expressed in metres and t in seconds. The work done by force in first 2 seconds.
(A) 25 J (B) 50 J (C) 60 J (D) 100 J
12. Work done in time t on a body of mass m which is accelerated from rest to a speed v in time t1 as a
function of time t is given by
2
1 v 2 v 2 1 mv 2 1 v2 2
(A) m t (B) m t (C) t t (D) m 2 t
2 t1 t1 2 t 1 2 t1
13. A particle moves under the effect of a force F = kx2 from x = 0 to x = 4 the work done by force is
8k 32k 64k 128k
(A) (B) (C) (D)
3 3 3 3
14. A block of mass M is hanging over a smooth and light pulley through a light string. The other end of the
string is pulled by a constant force F. The kinetic energy of the block increases by 20 J in 1s.
(A) the tension in the string is Mg (B) the tension in the string is F
(C) the work done by the tension on the block is 20 J in the above 1s
(D) the work done by the force of gravity is – 20 J in the above 1s
15. An ideal spring with spring–constant k is hung from the ceiling and a block of mass M is attached to its
lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the
spring is
4Mg 2Mg Mg Mg
(A) (B) (C) (D)
k k k 2k
.
16. In the figure, a ball a is released from rest when the spring is at its natural
(unstretched) length. For the block B, of mass M to leave contact with
A
the ground at some stage, the minimum mass of A must be –
(A) 2 M (B) M
(C) M/2 (D) a function of M and the force constant of the spring B M
17. A sphere of mass 2 kg is moving on a frictionles horizontal table with velocity v. It strikes with a spring
(force constant = 1 N/m) and compresses is by 4m. The velocity v of the sphere is
(A) 4 m/s (B) 2 2 m/s (C) 2 m/s (D) 2 m/s
18. Two bodies of masses m1 and m2 have equal kinetic energies. If P1 and P2 are their respective momentum,
the ratio P1 : P2 is equal to
(A) m1 : m2 (B) m2 : m1 (C) m1 : m2 (D) m12 : m2 2
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22 Work Power Energy
19. The displacement x of a particle moving in one dimension, under the action of a constant force, is related
to the time t by the equation t x 3 where x is in meters and t in seconds. The displacement of the
particle when its velocity is zero is
(A) 0 (B) 6 m (C) 12 m (D) 18 m
20. If the K.E. of a body is increased by 300%, its momentum will increase by
(A) 100% (B) 150 % (C) 300 % (D) 175 %
21. A cord is used to lower vertically a block of mass M a distance d at a constant downward accleration of
g
. Then the work done by the cord on the block is
4
d d d
(A) Mg (B) 3 Mg (C) – 3 Mg (D) Mgd
4 4 4
22. A rope ladder with a length l carrying a man with a mass m at its end is attached to the basket of ballon
with a mass M. The entire system is in equilibrium in the air. As the man climbs up the ladder into the
balloon, the baloon descends by a height h. Then the potential energy of the man
(A) increases by mg l h (B) increases by mgl
(C) increases by mgh (D) increases by mg (2l –h)
23. When a long spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 10 cm, the
potential energy stored in it will be
(A) 25 U (B) U/5 (C) 5 U (D) 10 U
24. Two springs have their force constants K1 and K2. These are extended through the same distance x. If their
elastic energies are E1 and E2, then E1 : E2 is equal to
(A) K1 : K2 (B) K2 : K1 (C) K 2 : K1 (D) K12 : K 2 2
25. An engine pumps up 100 kg of water through a height of 10 m in 5 seconds. Given that the efficiency of the
engine is 60%. If g = 10 m/s2, the power of the engine is
(A) 3.3 kW (B) 0.33 kW (C) 0.033 kW (D) 33 kW
26. A particle of mass M moves in a circle of radius R with a constant speed V. the work done when it
completes one circle is
MV 2 1 MV 3
(A) (2R ) (B) MV
2
(C) (2R ) (D) zero
R 2 R
27. The power of pump, which can pump 200 kg of water to a height of 50 m in 10 sec, will be
(A) 10 103 watt (B) 20 × 103 watt (C) 4 ×103 watt (D) 60 × 103 watt
28. Power applied to a particle varies with time as P = (3t 2 2t 1) watt, where t is in second. Find the
change in its kinetic energy between time t = 2 s and t = 4s
(A) 32J (B) 46J (C) 61 J (D) 102 J
29. A simple pendulum consists of light string from which a spherical bob of mass M is suspended. The
distance between the point of suspension and the centre of the bob is l. The bob is given a tangential
velocity v at the position of equilibrium (bottom). What can be maximum value of velocity v, so that the
pendulum oscillates without the string becoming slack?
(A) gl (B) 2gl (C) 4gl (D) 5gl
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Work Power Energy 23
LEVEL-1
KEY
1. C 2. A 3. D 4. A 5. A 6. A 7. D
8. D 9. A 10. B 11. C 12. D 13. C 14. B
15. B 16. C 17. C 18. B 19. C 20. A 21. C
22. A 23. C 24. A 25. A 26. A 27. D 28. A
29. B
HINTS & SOLUTIONS
1. By W-E theorem
W mg W F W t 0 mgl 1 cos 0 WF WF mgl 1 cos
2. KE Wkx
1 1 v2 1 2 1 3 1 3mv 2
mv 2 m Kx mv 2 Kx 2 K
2 2 4 2 2 4 2 4x 2
}
1
3
3.
Its centre of gravity will be at a distance of L/6.
M 1 M L
Mass M L Units 1 Unit. Mass of of length is
L 3 L 3
M L L MgL
g .
L 3 6 18
1 1 2
mv 0 mgh mgh v 0 14m / s
2 2
4.
1
5. P.E = mgh K.E = mv 2 As the ball falls potential energy decreases and kinetic energy increases.
2
mv 2
6. T mg if T 0 , v gl
l
7. Acceleration of car is “g” so g eff g a
mv 2
8. Velocity required 5gr T mg 6mg
r
w mgh 30 30 103 kg 10 60
9. PO / P P
t t 100 10 60
NARAYANA IIT ACADEMY – INDIA
24 Work Power Energy
V V
10. Time taken by the object to come to rest is t
a g
Total change in kinetic energy K
1
mV2 0
2
K 1 / 2mV 2 1
Mean power mgV Hence choice (A), (C) and (D) are wrong.
t V / g 2
dx 1
11. v
dt
6t 4 W KE
2
0.5 162 42 60J
1 v 1 v2 2
12. W KE mv 2 v t W m t
2 t1 2 t12
4
k
13. W Fdx Kx 2dx = 64
0 3
14. By action-reaction T = F Wt Wmg 20J
1 2 2mg
15. mgh = Kx x
2 k
Mg
16. To make B leave contact Kx = Mg x
k
Mg
Before comming to rest A has to fall x
k
2
Mg 1 Mg M
mg k m
k 2 k 2
1 1
17. (b) Loss in K. E = gain in K.E. mv 2 kx 2
2 2
or, 2mv 2 1 (4)2 v2 = 8 v = 2 2 m/s
2 2
1 1 P1 P P1 m1
18.
2
KE1 KE 2 m1v1 m 2 v 2
2
2
2 2 2m1 2m 2 P2 m2
19. (a) t x 3
x (t 3)2 t 2 6t 9
dx
velocity v 2t 6 0 (given) or, t = 3secs x = (3 –3)2 = 0
dt
2
P2 300 P1
20. KE KE KE 4KE KE
1
2m 100 2m
P1 2mKE1 2m4KE 2 2mKE
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 25
P1 2 P
x x x
PP P1 P 1 2P 1
100 100 100
x 100%
21. (c) The block is coming down with acceleration g/4.
Thus,
Mg 3
Mg – F = Ma = F= Mg The work done by the cord on the block is
4 4
3 3
W = Fd cos Mg d cos180 =– Mgd
4 4
22. (a) The increase in P.E. of the man will be mg (l –h) because the man rises up by a height (l –h).
1 2
23. (a) P.E. of spring, U = kx
2
2 2
U1 x1 U 2cm
or,
U2 x2 U2 10cm
or U2 = 25 U
E1 1/ 2 K1x 2 E1 K 1
24.
E 2 1/ 2 K 2 x 2 E2 K2
25. m 100, h 10, t 5, 60%
W mgh 60 100 10 10 60
P P P 2000
t t 100 5 100
2 104 1
P 104 0.3 104
6 3
26. No work is done by the centripetal force on the particle in the circular path because the direction of motion
of the particle is perpendicular to the direction of force.
W = Fd cos = Fd cos 90 = 0
W 200 10 50
27. (a) P = 10 103 watt
T 10
4
Pdt | t
3
28. (b) W = t2 t | 46J
2
1 2
29. For oscillation, string can become horizontal at the extreme position. PE mv mgh v 2 gl
2
NARAYANA IIT ACADEMY – INDIA
26 Work Power Energy
SINGLE CORRECT ANSWERS
LEVEL-2
1. A body of mass m is slowly pulled up the hill by a force F which at each point F h
was directed along the tangent of the trajectory as shown in figure. All surfaces
are smooth. Find the work performed by this force :
l
(A) mgl (B) – mgl (C) mgh (D) zero
2. A mass of 1 kg is acted upon by a single force F (4iˆ 4ˆj)N. Due to force, mass is displaced from (0,
0) to (1m, 1m). If initially the speed of the particle was 2 m/s, its final speed should approximately be :
(A) 6 m/s (B) 4.5 m/s (C) 8 m/s (D) 7.2 m/s
3. Work required to raise a stone of mass 5 kg and relative density 3, lying at the bed of a lake, through a
height of 3 m is
(A) 25 J (B) 40 J (C) 75 J (D) 100 J
4. A block of mass m is allowed to slide down a fixed smooth inclined plane of angle and length . The
magnitude of power developed by the gravitational force when the block reaches the bottom is
(A) 2m 2 (g sin )3 (B) (2 / 3)m3g 2 sin (C) (2 / 3)m 2 2g cos (D) (1/ 3)m3g 2 sin
5. A body of mass m, having momentum p, is moving on a rough horizontal surface. If it is stopped in a
distance x, the coefficient of friction between the body and the surface is given by
p2 p2 p p
(A) (B) (C) (D)
2gm 2 x 2gmx 2gmx 2gm 2 x
6. A body of mass 5 kg rests on a rough horizontal surface of coefficient of friction 0.2. The body is pulled
through a distance of 10 m by a horizontal force of 25 N. The kinetic energy acquired by it is (take g = 10
ms–2)
(A) 200 J (B) 150 J (C) 100 J (D) 50 J
7. A ball is thrown up with a certain velocity at angle to the horizontal. The kinetic energy KE of the ball
varies with horizontal displacement x as:
(a) (b)
KE KE
O O
x x
(c) KE (d) KE
O O
x x
8. A particle is acted by a force F = kx, where k is a positive constant. Its potential energy at x=0 is zero.
Which curve correctly represents the variation of potential energy of the block with respect to x?
U U U U
x x x x
(A) (B) (C) (D)
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Work Power Energy 27
9. Two equal masses are attached to the two ends of a spring of spring constant k. The masses are pulled
a part symmetrically to stretch the spring by a length x over its natural length. The work done by the
spring on each mass is –
1 2 1 2 1 2 1 2
(A) kx (B) – kx (C) kx (D) – kx
2 2 4 4
10. A simple pendulum is oscillating without damping. When the displacement of the bob is less than
maximum, its acceleration vector a is correctly shown in [2002]
a
(A) (B) (C) (D) .
a
a a
11. A block slides down an inclined plane of inclination with constant velocity. It is then projected up the
inclined plane with an initial speed u. How far up the incline will it move before coming to rest ?
(A) u2/4 g sin (B) u2/g sin (C) u2/2 g sin (D) u2/g
12. The displacement x of a body of mass 1 kg on horizontal smooth surface as a function of time t is given
t4
by x . The work done in the first one second is
4
1 1 3 5
(A) J (B) J (C) J (D) J
4 2 4 4
13. An object of mass (m) is located at the origin of a vertical plane . The body is projected at an angle
with velocity V. The mean power developed by the gravitational force during the interval of time till it
reaches maximum height
mgu sin mgu sin mgu sin
(A) mgu sin (B) (C) (D)
2 3 4
14. A uniform rod of mas m and length l is made to stand vertically on one end. The potential energy of the rod
in this position with respect to its lower end is
mgl mgl mgl
(A) (B) (C) (D) mgl
4 3 2
15. A uniform steel rod of mass m and length l is pivoted at one end. If it is inclined with the horizontal at an
angle , its change in potential energy will be
1 1
(A) mgl cos (B) mgl sin (C) mgl cos (D) mgl sin
2 2
16. In a hydroelectric power station, the height of the dam is 10 m. How many kg of water must fall per
second on the blades of a turbine in order to generate 1 MW of electrical power ? Take g = 10 ms–2
(A) 103 kgs 1 (B) 104 kgs 1 (C) 105 kgs 1 (D) 106 kgs 1
17. A body of mass m is dropped from a certain height. It has a velocity v when it is at height h above the
ground. Which of the following will remain constant during the free fall ?
(A) v 2 2gh (B) v 2 2gh (C) v 2gh (D) v 2gh
NARAYANA IIT ACADEMY – INDIA
28 Work Power Energy
18. A bullet is fired normally on an immovable wooden plank. It loses 25% of its momentum in penetrating a
thickness of 3.5 cm. The total thickness penetrated by the bullet is
(A) 8 cm (B) 10 cm (C) 12 cm (D) 14 cm
19. A ladder 2.5 m long and of weight 150 N has its centre of gravity 1 m from its bottom. A weight of 40 N is
attached to the top end. The work required to raise the ladder from the horizontal position to the vertical
positions is
(A) 190 J (B) 250 J (C) 285 J (D) 475 J
20. A conservative force F is acting on a body and body moves from Point A to B and then B to A then work
done by the force is
(A) W 0 (B) W 0 (C) W 0 (D) W 0 .
21. The potential energy of a particle varies with position x according to the relation U( x ) 2 x 4 27 x the
3
point x is point of
2
(A) unstable equilibrium (B) stable equilibrium (C) neutral equilibrium (D) none of these.
22. A particle of mass M is moving in a horizontal circle of radius ‘R’ under the centripetal force equal to K/R2,
where K is constant. The potential energy of the particle is
(A) K/2R (B) –K/2R (C) K/R (D) –K/R
23. A pendulum of mass m and length l is suspended from the ceiling of a trolley which
a
has a constant acceleration a in the horizontal direction as shown in figure. Work
done by the tension is (in the frame of trolley)
(A) mgl 1 cos (B) mal sin (C) mal cos (D) zero
24. Velocity - time graph of a particle moving in a straight line is as shown v(m/s)
in figure. Mass of the forces is 2kg. Work done by all the forces
10
acting on the particle in time interval between t =0 toe r = 10s is :
10
(A) 300J (B) – 300J t(s)
(C) 400J (D) – 400J –20
25. An ice cube of size a = 10cm is floting in a tank (base are A = 50cm 50cm) partially filled with water..
The change in gravitational potential energy when ice melts completely is : (density of ice is 900Kg/m3)
(A) – 0.072J (B) – 0.24J (C) – 0.016J (D) 0.045J
KEY
1. C 2. B 3. D 4. C 5. A 6. B 7. C
8. B 9. D 10. C 11. A 12. B 13. B 14. C
15. B 16. B 17. A 18. A 19. B 20. C 21. B
22. C 23. D 24. A 25. D
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 29
HINTS & SOLUTIONS
1. Wnet K
as particle is moving slowly, this means K 0
WN WF Wmg K but WN 0, as N d r
O Wf mgh 0
WF mgh . Ans. (C)
1m 1m
2. Wnet F.d r 4 dx 4 dy = 8 J and W net = K
0m 0m
1 1
8 (1) v 2 (1) (2) 2
2 2
v 4.5 m/s Ans. (B)
m mg
3. Fbuoyant w g
3
mg 2mg
Fapplied = mg
3 3
2mg
work done by applied force 3 = 100 J Ans. (D)
3
4. p F.v
mg sin 2g sin
2m 2 g 3 sin 3 Ans. (A)
5. Force of friction mg . Therefore, retardation a mg / m g . Also 2ax v 2 or 2am2 x m 2 v 2 . But
p2
But a g. Therefore, 2gm x p or
2 2
p = mv. Therefore, 2am x p
2 2
2 gm 2 x
Hence the correct choice is (A)
6. Friction force mg 0.2 5 10 10 N . Effective force F = applied force – frictional force = 25 – 10 =
15 N. Kinetic energy = work done by force F in pulling the body through a distance S (= 10 m) = 15 × 10
= 150 J, which is choice (B).
dv dk
7. Velocity is min but not zero at max. height also 0 hence 0
dt dt
1 2
8. U Fdx =– Kx
2
NARAYANA IIT ACADEMY – INDIA
30 Work Power Energy
1 2
9. Work done on both block are the same and total work done by spring is kx .
2
1 2
Hence on each block kx
4
10. The net acceleration is resultant of aJ and aC
11. As acceleration is zero on the inclined plane this means force of friction = mg sin
Work Done by friction = mg l sin
1
Work Done by gravity = mg l sin K.E. mu 2
2
1 u2
Thus, 2mg l sin mu 2 l
2 4g sin
dx 1 1 1 1
4t 3 t 3 1 m / s Work done = change in K.E. = 1 1 1 0 J
2 2
12. V
dt 4 2 2 2
Note : W F d is not applicable to solve the above problem. This formula is applicable when force is not
dependent on time.
mgh mgu sin
13. avg. Power
t1/2 2
14. The potential energy in the vertical position = work done in raising it from horizontal position to vertical
position. In doing so, the mid–point of the rod is rasied through a height h = l/2. Since the entire mass of
the rod can be assumed to be concentrated at the mid–point (centre of gravity), the work done = mgh =
mgl/2. Hence the correct choice is (C)
15. The weight of the rod acts at the centre of gravity which is at distance of l/2 from the pivoted end. When the
end rod makes an angle with the horizontal, the vertical height of the centre of gravity is
l 1
h sin i.e., the centre of gravity rises by an amount h. Therefore, P.E. mgh mglsin
2 2
Hence the correct choice is (B)
16. Let M kg of water fall per second. The power is
P = rate at which work is done = mass per second ×g × h = Mgh But P = 1 M W = 106 W, h = 10 m.
P 106
Therefore M 104 kgs 1 , which is correct choice (B)
gh 10 10
17. The total energy = KE + PE remains constant during the free fall, i.e.,
1 v2
mgh mv 2 constant or gh constant Hence the correct choice is (A)
2 2
18. Let u cms1 be the speed of the bullet. Since the mass of the bullet remains unchanged, its speeds
3u
becomes v cms1 after it penetrates a distance x = 3.5 cm. The retardation a due to the resistance
4
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 31
2
3u
of the wooden is given by u v 2ax or u 2a 3. 5
2
2 2
4
u2
which gives a cms 2 . The bullet will come to rest when its velocity v' 0 . If x ' is the thickness
16
u 2 16
penetrated by the bullet, then u 2 v'2 2ax ' or x ' 2u 2 8cm
Hence the correct choice is (A)
19. Work done = increase in potential energy in (i) raising the weight 150 N of the ladder through a height 1 m
and (ii) raising a weigth 40 N through 2.5 m
= 150 N × 1 m + 40 N × 2.5 m = 250 Nm = 250 J
20. Work done for cyclic process is zero.
du 3
21. U x 2 x 4 27 x F 8 x3 27 0 for equilibrium x
dx 2
d 2u 3 3
2
24 x 2 0 at x Hence U is min at x so stable.
dx 2 2
b
22. U F .dr F k2 attractive so U K
a R R
23. Work done by tension is zero because T dS
Choice (A) represents work done the gravity T
Choice (B) represent work done by Psuedo force ma
Choice (C) is ruled out. mg
1
24. From work - energy theorem W K .E. = Kf Ki = m(f2 i2 )
2
1
= 2(400 – 100) 300J
2
25. Relative dencity of ice is 0.9, i.e., 90% valume of ice is immersed in warer. Whne ice melts completely,
level of water does not change. h=GG’ = 0.5 cm
1cm
G
G
9cm
U mgh ; (0.1)3 (900)(10)(0.5 102 ) j 0.045J
NARAYANA IIT ACADEMY – INDIA
32 Work Power Energy
SINGLE CORRECT ANSWERS
LEVEL-3
Objectives with single correct choice
1. 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) t3/4 (C) t3/2 (D) t2
2. A wind–powered generator converts wind energy into electric energy. Assume that the generator converts
a fixed fraction of the wind energy intercepted by its blades into electric energy. For wind speed V, the
electrical power output will be proportional to
(A) V (B) V2 (C) V3 (D) V4
3. A particle, which is constrained to move along the x–axis, is subjected to a force in the same direction
which varies with the distance x of the particle from the origin as F (x) = – kx + ax3. Here k and a are positive
constants. For x 0, the functional form of the potential energy U (x) of particle is
U(x) U(x) U(x) U(x)
(A) (B) (C) (D)
x x x x
4. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration
ac is varying with time as ac = k2rt2, where k is a constant. The power delivered to the particle by the forces
acting on it is –
1
(A) 2 mk2r2t (B) mk2r2t (C) mk4r2t5 (D) 0
3
5. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the
centre. At a certain instant of time, the stone is at its lowest position, and has a speed u. The magnitude
of the change in its velocity as it reaches a position where the string is horizontal is :
(A) u 2 2gL (B) 2gL (C) u 2 gL (D) 2(u 2 gL) .
6. A force F K(yiˆ xj)
ˆ , where k is a positive constant, acts on a particle moving in the xy plane.
Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0), and the parallel
to the y-axis to the point (a, a). The total work done by the force on the particle is
(A) – 2Ka2 (B) 2Ka2 (C) – Ka2 (D) Ka2
7. A smooth sphere of radius sphere of radius R is made to translate in a straight line with a constant accelera-
tion a = g. A particle kept on the top of the sphere is released from there at zero velocity with respect to the
sphere. The speed of the particle with respect to sphere as a function of as it slides down is
(A) Rg (sin cos ) (B) Rg (1 cos sin )
(C) 4Rg sin (D) 2Rg (1 sin cos )
8. A particle is projected with speed u in air at an angle with the horizontal. The graph showing the variation
of instantaneous power due to gravity P with time t will be
P P P
(A) (B) (C) (D)
t t
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 33
x4 x2
9. The potential energy of a 1 kg particle free to more along the x-axis is given by U(x) = J
4 2 .The
total mechanical energy of the particle is 2.J. Then, the maximum speed in (m/s) is
(A) 1/ 2 (B) 2 (C) 3 / 2 (D) 2
10. Particle moves in a straight line with retardation proportional to its displacement. The loss of kinetic
energy during a displacement x is proportional to
(A) x2 (B) ex (C) x (D) logex
11. A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horixontal force in the positive
direction of x-asix is applied to the block. The force is given by F (4 x 2 )iˆ N , where x is in meter and
the initial position of the block is x = 0. The maximum kinetic energy of the block between x = 0 and x =
2.0m is :
(A) 2.33 J (B) 8.67 J (C) 5.33 J (D) 6.67 J
KEY
1. C 2. C 3. D 4. B 5. D 6. C 7. D
8.C 9. C 10. A 11. C
SOLUTIONS :
dv dx
1. P Fv mv v 2 t or v t 1/ 2 t 1/ 2 or x t 3 / 2
dt dt
2. Pwind = F.v = Av 2 .v P v3 .
k k
3. F ( x ) – kx ax 3 ; F( x ) 0 for x 0 x ; So slope is zero at x = 0 x
a a
d |v | v2
4. P Ft .v mat .v ; at ; ac k rt
2 2 k 2 rt 2
dt r
d |v |
v krt kr at = kr P mkr .krt mk 2 r 2 t
dt
uf
5. intial velocity = u final velocity = u 2 2gl | v | ui u0
2 2
=
2 u 2 gl
ui = 4
yiˆ xyˆ . dxiˆ dyiˆ
( a,a ) ( a,a )
6. W
(0,0)
= k ydx xdy kxy (0,0)
= - ka2
1
7. (d) maR sin mgR (1 cos ) = mv 2 ma
2
mg
R
8. (c)
Pmg mgjˆ uiˆ (v gt ) jˆ
= -mg (v –gt).
NARAYANA IIT ACADEMY – INDIA
34 Work Power Energy
x 4 x 2 dU
9. (c) U , x3 x 0 x 0, x 1
4 2 dx
d 2U d 2U 1
2
3x 2 1 ve for x 1 U 1
dx dx 2 4
K max Ul min T .E 2J
1 3
9 Kmax mv 2 v
K max 2 2
4
dv dv dx dv
10. (a) a x kx Kx v Kx vdv Kxdx
dt dx dt dx
1 kx 2
2
m v 2 u2 2
x x3
11. From work energy theorem kinetic energy of block at x = x is K 0
(4 x 2 )dx or K 4 x
3
dK
For X to be maximum, 0 or 4 – x2 = 0 or x = 2 m
dx
d 2K
at x = +2 m, is negative, i.e., kinetic energy K is maximum.
dx 2
LEVEL-1
MULTIPLE CHOICE QUESTIONS
1. Select the cotrrect alternatives:
(A) Work done by static friction is always zero (B) Work done by kinetic friction can be positive also
(C) Kinetic energy of a system can not be increased without applying any external force on the system
(D) Work energy eheorem is valid in non-inerial frames also.
2. A block of mass 2kg is hanging over a smooth and light pulley through a light string. The other end of the
string is pulled by a constant force F=40N. The kinetic energy of the particle increases 40J in a given
interval of time. Then : )g = 10m/s2)
(A) tention in the string is 40N
(B) displacement of the block in the given interval of time is 2m
(C) work done by gravity is –20J (D) work done by tension is 80J.
3. Displacement time graph of a particle moving in a straight line is as shown in figure. select the current
alternative(s):
.
S C
(A) Work done by all the forces in region OA and BC is positive
(B) Work done by all the forces in region AB is zero B
(C) Work done by all the forces in region BC is negative A
(D) Work done by all the forces in region OA is negative.
O
P(watt)
4. Power of a force acting on a block varias with time t as
shown in the figure. The angle between force acting
on the block and its velocity is :
10
(A) acure at t = 1s 6 8 10
o 4
(B) 90 at t = 3s 2 t(s)
(C) obtuse at t = 7s
–10
(D) change in kinetic energy from t = 0, to t = 10s is 20J.
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 35
5. A particle is acted upon by a conservative force F (7iˆ – 6 ˆj ) N (no other force is acting on the particle).
Under the influnce of thes force particle moves from (0, 0) to (–3m, 4m) then :
(A) work done by the force is 3J (B) work done by the force is –45 J
(C) at (0, 0) speed of the particle must be zero (D) at (0, 0) speed of the particle must not be zero.
6. In which of the following is no work done by the force?
(A) A man carrying a bucket of water, walking on a level road with a uniform velocity
(B) A drop of rain falling vertically with a constant velocity
(C) A man whirling a store tied to a string in a circle with a constant speed
(D) A man walking up on a staircase.
7. The work done by a force on a body does not dependend upon
(A) The mass of the body (B) The displacement if the body
(C) the intial velocity of the body
(D) the angle between the force vector and the displacement vector.
8. A particle is acted upon by a force if constant magnitude which is always perpendicular to the velocity of
the particle. The motion of the particle takes place in a plane. It follows that
(A) The velocity of the particle is constant (B) the acceleration of the particle is constant
(C) the kinetic energy of the particle is constant (D) the particle moves in circular path.
9. Which of the following forces are conservative?
(A) Coulomb force between charged particles at rest(B) Force of compressed elastic spring
(C) Gravitational force between two masses (D) Frictional force.
10. Figure show the force F (in newton) acting on a body as a function of x. the work done in moving the body
.
(A) from x = 0 to x = 1 m is 2.5 J F
5
(B) from x = 1 m to x = 3 m is 10 J
(C) from x = to x = 4 m os 15 J
(D) from x = 0 to x = 4 m is 12.5 J.
12 3 4
x
11. A uniform rod has mass m and length l. The potential energy of the rod relative to the end in contact when
(A) it stands vertically is zero (B) it stands vertically is mgl/2
(C) it is inclined at an angle with the vertical is mgl cos /2
(D) it is inclined at an angle with the vertcal is mgl sin /2.
12. A body is subjected to a vonstant force F in Newton given by F iˆ 2 jˆ 3kˆ . Where iˆ , ĵ and k̂ are
unit vectors along x, y and z axes respectively. The work done by this force in moving the body through a
distance of
(A) 4m along the z-axis is 12J (B) 3m along y-axis is 6J
(C) 4m along the z axis and then 3m along the y-axis is 18 J
(D) 4m along the z-axis and then 3m along the y-axis is (12)2 (6)2 J .
13. A block od mass 2kg, initially at rest on a horizontal floor, moves under the action of a force of 10N. The
coefficient of friction between the block and the floor is 0.2. if g = 1ms–2.
(A) the work dine by the applied force in 4s is 240J (B) the work dine by the frictional force in 4s is 96J
(C) The work done by the net force in 4s is 336J (D)the change in kinetic energy of the block in 4s is 144J
14. In the system shown in the figure the mass m moves 4m
in a circular are of angular amplitude 60o. Mass 4m
60o
m
A
NARAYANA IIT ACADEMY – INDIA m
B
36 Work Power Energy
is stationary . Then:
(A) the minimum value of coefficient of friction between the same
of mass 4m and the surface of the table is 0.50
(B) the work dine by gravitational force in the block m is positive when it moves from A to B
(C) the power delovered by the tention when m moves from A to B is zero
(D) The kinetic energy of m in position B equals the work done by gravitational force on the block when its
moves from position A to B.
15. A strip of wood mass M and length l is placed on a smooth horizontal surface. An insect of mass m starts
at one end of the strip and walks to the other end in time t, moving with a constant speed.
l
(A) The speed of the insect as seen from the ground is
t
l M
t M m
(B) The speed of the strip as seen from the ground is
l m
t M m
(C) The speed of the strip as seen from the ground is
2
1 l
(D) The total kinetic energy of the system is ( m M )
2 t
A B
16. U (r ) : Where r is the distance from the centre of the force and A and B are positive constants.
r2 r
2A
(A) The equlibrium distance is given by
B
B2
(B) The work required to move the particle from equilibrium distance to infinity is
4A
B
(C) The work required to move the particle from equilibrium distance to infinity is
4A
(D) Only (A)
Rough
17. In the figure shown upper block is given a velocity of 6m/s and
1kg 6m/s
lower block 3 m/s. When relative motion between them is
2kg 3m/s
stopped.
Smooth
(A) Work done by friction on upper block is negative (B) Work done by friction on both blocks is positive
(C) Work done by friction on upper block is –10J (D) Net work done by friction is zero.
18. A body is moved along a stright line by a machine delivering a power proportional to time t.
(A) Velocity is prapotinal to t (B) Displacement is propotinal to t3
(C) Work done is prapotinal to t 2
(D) Velocity is prapotinal to t2.
19. A body of mass 1kg is suspended from an inextensible string of length 1 m. When the string makes an
angle 60o with vertical, speed of the bob is 4 m/s.
(A) Net acceleration of the bob at the instant is 10m/s2
(B) Net acceleration of the bob at the instant is 18.2m/s2
(C) The bob will rise to a maximum height with respect to bottom most point > 1.3m
(D) The bob will rise to a maximum height with respect to bottom most point < 1.3m
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 37
20. A pendulum is released from point A as shown in figure. At some instant
net force on the bob is making an angle with the string:
(A) for =90o particle is at A (B) for =0o particle is at B C A
o
(C) for =90 particle at the position where string is horizontal
B
(D) For = 0o particle is at A
KEY
1. B,D 2. A,B,D 3. B,D 4. A,C,D 5. B,D
6. A,B,C 7. A,C 8. C,D 9. A,B,C 10. A,B,C
11. B,C 12. A,B,C 13. A,B,D 14. A,B,C,D 15. A,C
16. A,B 17. A,C 18. A,C 19. B,D 20. A,B,C
SOLUTIONS
1. [B,D] T = F = 40N
2. [A,B,D]
W net = KE or (40–20) s = 40 s = 2m 2kg
work done by gravity is –20 x 2 = –40J
and work done by tension is 40 x 2 = 80J
mg = 20N
3. [B,D] In region OA particle is acccelerated, in region AB particle has uniform velocity while in
region BD particle is deceleration., Therefore, work done is positive in region OA , zero in region AB
and negative in region BC.
4. [A,C,D] Power is the dot product of velocity. At 1s and 3s, power is positive. Hence, angle is
acute. At 7s power is negative. Hence angle is obtuse.
5. [B,D]
6. [A,B,C]
7. [A,C]
8. [C,D]
9. [A,B,C]
10. [A,B,C] Area of the graph drawn between force and position = work done
11. [B,C]
12. [A,B,C]
13. [A,B,D]
14. [A,B,C,D]
15. [A,C] mv1 = Mv2, where v1 and v2 are speeds of mass m and M, as seen from ground. The
velocity of m relative to M is v12 = v1 – (– v2).
m v1 M
v2
1 1
Hence, t l
v12 v1 v 2 or v1 + v2 = l/t.
dU 2 A B 2A
16. [A,B] F 3 2 at eqyilibrium F= or,, r
dr r r B
2A B2 B2
At infinity U = 0 r ,U U .
B 4A 4A
17. [A,C] From conservation of linear momentum (1 + 2)v = (6 x 1) + (2 – 3) v = 4m/s (of both the
blocks)
1
From work energy therom i.e., W total = KE on 1kg block, Wf 1 (42 62 ) 10J
2
NARAYANA IIT ACADEMY – INDIA
38 Work Power Energy
1
on 2kg block Wf 2(42 32 ) 7J Net work done by friction is –3J.
2
18. [A,C] P t W Pdt t dt or W t 2
ds ds
Hence, v 2 t 2 or v t further, v t
dt dt
or ds t dt or s t 2 (by intigration)
19. [B,D] at = g sin 60o = 5 3 m/s2
v2
an = 16m / s 2 a = at2 an2 18.2m / s 2
R
R
Height from bottom at 60o : h1 = R(1– cos60o)= =0.5m
2
v 2 16
Further height reaised, h2 0.8m , Now since h + h = 1.3m > R
2g 20 1 2
The bob will rise to = 90o, but there string will slack and total height will be < 1.3m.
20. [A,B]
LEVEL-2
MULTIPLE CHOICE QUESTIONS
1. One end of a light spring of force constant k is fixed to a wall and the other end is tied to a block placed on
1 2
asmooth horizontal surface. In a displacement, the work done by the spring is kx . The possible
2
case(s) may be:
(A) the spring was initially streched by a distance x and finally was in its natural length
(B) the spring was initially in its natural length and finally it was compressed by a distance x
(C) the spring was initially conpressed by a distance x and finally was in its natural length
(D) the spring was initially in its natural length and finally stretched by a distance x.
2. 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 mean position of rest is x0 then :
3F
(A) increase in energy stored in spring is kx02 (B) x0
2k
2F
(C) x0 (D) work done by applied force F is Fx0.
k
3. The potential energy U in joule of a particle of mass 1kg 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 10m/s
(D) co-ordinates of particle at t = 1s are (4.5, 2)
4. A small block of mass m is released from rest from position A inside a smooth A R
hemispherical bowl of radius R as shown in figure. Choose the wrong option :
(A) acceleration of block is constant throughout (B) acceleration of block is g at A
B
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 39
(C) acceleration of block is 3g at B (D) acceleration of block is 2g at B.
5. A force F = – kx3 is acting on a block moving along x-axis. Here, k is a positive constant. Work done by
this force is :
(A) positive in displacing block from x = 3 to x = 1 (B) positive in displacing block from x = –1 to x =3
(C) negative in displacing block from x = 3 to x = 1 (D) negative in displacing block from x = –1 to x =3.
s
6. Displacement time graph of particle moving in a straight line is as shown in
figure. From the graph we can conclude that work done on the block is :
(A) positive from 0 to t1 (B) negative from t1 to t2
(C) zero from t2 to t3 (D) negative from t3 to t4. t1 t2 t3 t4 t
7. A simple pendulum of length L and having a bob of mass m is oscillation in a plane about a vertical line the
angular limits – and . At a time when the angular disp;acement is , the tension in the string
is T and the velocity of the bob is . Which of the following relations will hold?
(A) T cos =Mg (B) T–Mg cos =M 2/L
(C) The magnitude of tangential acceleration of the bob is |aT| = g sin
(D) tention is minimum at = 0o.
8. Two inclined frictionless tracks of different inclinations meet at A form
where two block P and Q of different masses are allowed to slide down
from rest at the same time, one on each track, as shown in figure.
(A) Both blocks will reach the bottom at the same time
(B) Block Q will reach the bottom earlier than block P
(C) Both blocks reach the bottom with the same speed
(D) Block Q will reach the bottom with a higher speed than block P
9. Choose the correct statements from the following :
(A) When a conservative force does positive work on a body, its potential energy increases
(B) When a body does work against friction, its kinetic energy decreases
(C) The rate of change of total momentum of a many-particle system is proportional to the net external
force acting on the system
(D) The rate of change of total momentum of a many-particle system is proportional to the net internal
force acting on the system.
10. A body of m is moving in a straight line at a constant speed . Its kinetic energy is k and the ,magnitude
of its momentum is p. which of the following relations is /are correct?
2k 2k
(A) p 2mk (B) p (C) 2k p (D)
m p
11. A block of mass m is taken from the botom of an inclined plane to its top and then allowed to slow down
to the bottom again. The length of the inclined plane is L and the coefficient of friction between the block
and the plane is . The inclination if the plane is .
(A) The work done by the gravitational force over the round trip is zero.
(B) The work done by the applied force over the upward journey is mgL(sin + cos ).
(C) The work done by the frictional force over the round trip is zero
(D) The kinetic energy of the block when it reaches the bottom is mgL(sin – cos ).
NARAYANA IIT ACADEMY – INDIA
40 Work Power Energy
KEY
1. A,C 2. C,D 3. A,C,D 4. A,C 5. A,D
6. A,B,C 7. B,C 8. B,C 9. B,C 10. A,C
11.A,C,D
SOLUTIONS
1. [A,C] Work done by spring force = 1/2k x12 – 1/2k x22 where
x1 initial change in length,
x2 final change in length
2. [C,D] in fig.1 mg = kx1
1 1
k x0 x1 kx12
2
in fig.2 Fx0 + mg x0 =
2 2
m x1
kx0
F= ; m x0
2
fig.1
m
F
x0 = 2 fig.2
k
du ˆ du ˆ
3. [A,C,D] F i j ; 3iˆ 4 jˆ
dx dy
a 3iˆ 4 ˆj ; a 5
on y axis x = 0, after time t sec. the particle crosses y axis.
0 – 6 = 1/2 x (– 3)t2, t = 2. vx = (– 3) x 2 = – 6
vy = (– 4) x 2 = – 8. v 10m / s.
t = 1sec., x = 6 – 1/2 x 3 = 4.5
y = 4 – 1/2 x 4 = 2
v 2 2gR
4. [A,C] at B acceleration of block = 2g
R R
x2
5. [A,D] w Fdx
x1
6. [A,B,C] Slope of the graph at a point = velocity, t1 = 0 to t1 velocity increases.
t1 to t2 velocity dicreases, t2 to t3 velocity zero and t3 to t4 velocity is constant.
7. [B,C]
8. [B,C] Velocity v = 2gh
2
t= s
gh
for A B s > S for AC.
9. [B,C]
10. [A,C] kinetic energy k = 1/2mv2 , p = mv
11. [A,C,D]
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 41
LEVEL-3
MULTIPLE CHOICE QUESTIONS
1. A smooth track in the form of a quarter circle of radius 6m lies in the vertical o 6m P2
plane. A particle moves from P1 to P2 under the action of forces F1, F2 and F3.
F1
Force F1 is always toward P2 and always 20N in magnitude. Force F2 is 6m F2
always acts horizontally and is always 30N in magnitude. Force F3 always F3
acts tangentially to the track and is of magnetude 15N. Select the correct P1
alternative(s):
(A) Work done by F1 is 120J (B) Work done by F2 is 180J
(C) Work done by F3 45 (D) F1 is conservative in nature.
2. A block of mass M is attached with a spring constant k. The whole k
a
arrangement is placed on a vechile as shown in the figure. If the M
vehicle starts moving towards right with an acceleration a (therer is
no friction anywhere), then :
Ma 2Ma
(A) maximum elongation in the spring is (B) maximum elongation in the spring is
k k
2ma
(C) maximum compression in the spring (D) maximum compression in the spring is zero.
k
3. A long block A is at rest on a smooth horizontal surface. A small block v
B
B, whose mass is half of A, is placed on A at one end and projected A
along A with some velocity u. The coefficient of friction between the
blocks is .
u
(A) The blocks will reach a final common velocity
3
(B) The work done against friction is two-thirds of the initial kinetic energy of B.
2
(C) Before the blocks reach a common velocity, the acceleration of A relative to B is g
3
3
(D) Before the blocks reach a common velocity the acceleration of A relative to B is g
2
4. A ball of mass m is attached to the lower end of a light vertical spring of force cinstant k. The upper end of
the spring is fixed. The ball is released from rest with the spring at its normal(unstreched) length, and
comes to rest again after descending through a distance x.
(A) x = mg/k (B) x = 2mg/k
(C) The ball will have no acceletation at the position where it has descended through x/2.
(D) The ball will have an upward acceleration equal to g at its lowermost position.
5. A small ball is connected to a block by light string of length l. Both are initially on u
the ground. There is sufficient friction on the ground. There is sufficient friction on
the ground to prevent the block from slipping. The ball is projected vertically up
with a velocity u, whrer 2gl < u2 < 3gl. The center of mass of the ‘block + ball’ system is C
(A) C will move along a circle (B) C will move along a parabola
(C) C will move along a straight line
NARAYANA IIT ACADEMY – INDIA
42 Work Power Energy
(D) The horizontal component of the velocity of the ball will first increase and then decrease.
KEY
1. B,C,D 2. B,D 3. A,B,D 4. B,C,D 5. A,D
SOLUTIONS
o 6m P2
P2
1. [B,C,D] Work done by F1 is W 1 = P F1 cos ds
1
Here, ds = (6)d (–2 ) = – 12d and F1 = 20N 6m F1 F2
0 F3
W 1 = – 240 x / 4 cos d 240 sin 120 2J
4 P1
F1 is conservative because it is always directed towards a fixed point P2.
6( / 2) 3
W 2 = F2 (OP2) = (30)(6) =180J and W 3 = F1 F3 ds 15ds
0 0
15 x 0 45
3
1 2 2Ma
2. [B,D] Max = Kx X=
2 K
3. [A,B,D] mu = (m + 2m) v, v is the final common velocity
1 1 1 u2 1 1 2 1
mu 2 (3m ) (3m ) mu 2 1 mu 2 .
2 2 2 9 2 3 3 2
The force of friction between the blocks is mg .
mg g
Acceleration of A (to the right) = a1 =
2m 2
mg
Acceleration of B (to the left) = a2 = g
m
3
Acceleration of A relative to B = a1 –(– a2) = g .
2
1 2 2mg
4. [B,C,D] mgx kx or x When the ball is at its lowest position, string force = kx = 2mg
2 k
5. [A,D] As the block does not move, the ball moves along a circular path of radius l. the center of
mass of the system always lies some where on the string.
v
Let v = speed of the ball when the string makes an angle with the horizontal.
V
1 1
mv 2 mu 2 mg / sin l
2 2 u
The horizontal component of v = V = v sin = sin u 2 2gl sin
dV u2
For V to be maximum, 0 , which gives sin .
d 3gl
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 43
PASSAGE
(Linked comprehension type)
This section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questions have to
be answered. Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
PASSAGE: 1
The work done by all forces on a body is equal to change of kinetic energy of the body. This is true for both
constant and variable force (variable in both magnitude and direction). For a particle W = k. For a
system, W int + W ext = Kcm or WExt + Wnonconservative = K+ U
In the absence of external and nonconservative forces, total mechanical energy of system remains
conserved.
1. I - Work done in raising a box on a platform depends on how fast it is raised
II - Work done by force depends on the frame of reference
(A) I - False II - True (B) I - False II - False
(C) I - True II - False (D) I - True II - True.
2. The total energy of a system is
(A) always conserved in presence of external forces.
(B) always conserved in absence of internal forces.
(C) always conserved in presence of no external force and no internal non conservative force
(D) none of the above.
3. Work done in motion of a body over a closed loop is
(A) always zero for any force (B) always zero for conservative forces
(C) only (A) (D) None of the above.
PASSAGE : 2
I. A small block of mass 200 gram is placed at the bottom of an inclined plane
which is 10 m, long and 3.2 m high. Coefficient of friction between the block and
inclined plane is 0.1 [g = 10 m/s2].
4. Work required to lift the block from ground and put it at the top
(A) 3.2 J (B) 6.4 J (C) 9.6 J (D) 1.6 J.
5. Work required to lift the block up the incline and taking it to top is
(A) 3.2 J (B) 6.4 J (C) 8.3 J (D) 10 J.
6. If from the top it falls off the incline and drops vertically with what speed it will hit the ground
(A) 2 m/s (B) 4 m/s (C) 8 m/s (D) 10 m/s.
PASSAGE : 3
Work done by a constant force acting on a particle is defined as W F.r where r rf ri , rf is the
final position of particle and ri is the initial position of particle. If the force is variable work done is defined
rf
as W , where F is the force at any general position r between ri and rf , and dr is representing
F.dr
ri
infinitely small displacement of particle. dr dxiˆ dyjˆ dzkˆ .
7. A particle is thrown from a tower of height h at an angle above horizontal. Work done by gravity during
its time of flight is (mass of particle is m)
(A) 0 (B) mgh cos (C) mgh sin (D) mgh.
NARAYANA IIT ACADEMY – INDIA
44 Work Power Energy
8. Given: F 2iˆ 4ˆj, ri 2iˆ 3jˆ and rf 3iˆ 4ˆj Work done is (use SI units)
(A) 4 J (B) 6 J (C) 10 J (D) zero.
9. A particle thrown from ground with a velocity v at an angle above horizontal. Work done by gravity in
2v sin
t equals
g
mgv2 sin 2 mgv2 sin 2 2mgv 2 sin 2
(A) (B) (C) zero (D) .
2g g g
PASSAGE – 4
A
A block of mass m = 1kg is released from point A along a smooth
track as shown. Part AB is circular with radius r1 = 4m and C
circular at C with radious r2. Height of point A is h1 = 2m and of h1
c is h2 = 1m. h2
(g = 10 m/s2).
B
10. The force exerted by block on the track at B is
(A) 10 N (B) 20 N (C) 30 N (D) 40 N.
11. The minimum safe value of r2 so that the block does not fly off the track at C is
(A) 1 m (B) 2 m (C) 1.5 m (D) 3 m.
12. The work done by gravitational force from A to C is
(A) 10 J (B) 20 J (C) 30 J (D) 40 J.
PASSAGE : 5
A chain of length l = R / 4 is placed. On a smooth hemispherical surface of radius R with one of its ends
fixed at the top of the sphere. Mass of chain is 2kg and R = 1m. (g = 10 m/s2).
13. The gravitational potential energy of the chain considering reference level at the base of hemisphere is
(A) 20J (B) 20 2 J (C) 40 J (D) 40 2 J.
14. If the chain sliped down the sphere, kinetic energy of the chain when it has sliped through an angle
4
(A) 23.4 J (B) 63.44 J (C) 80 J (D) 97.4J.
15. The tangential acceleration of the chain when it starts sliding down.
40 1 20 1 1
(A) 1 (B) 1 (C) 10 1 (D) zero.
2 2 2
KEY
1.A 2. C 3. B 4.B 5. C 6. B 7. D
8. B 9. C 10. B 11. B 12. A 13. C 14. A
15. A
HINTS & SOLUTIONS
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 45
1. Wmg is path and rate independent
2. Total energy is conserved when there is no external and no internal nonconservative force.
3. Work done for conservative forces are path independent.
4. WF mgh
5. WF Wmg W fr K .E 0
6. v 2 gh
7. Wmg mgh
8. F .r F . r2 r1
2v sin
9. t is time of flight and vertical displacement is zero.
g
2
mv 2
10. v 2 2gh1 N mg
r1
2
mv c 1
11. At C N 0 mg mgh1
2
mv c mgh2 r2 2( h1 h2 )
r2 2
12. Wmg mg ( h1 h2 ) U
4
m M 4
13. dm Rd
l
PE
0
l
Rd gR cos R
d
M
= gR 2 sin 45 = 40J.
l
/2 M 2 m
14. Uf R g cos .d Ui gR 2 sin 45 U KE
/ 4 l l
15. Tangential force on differential element dFt dmg sin
/4 m gR
Ft l Rd g sin [1 cos 45)
0
l
MATRIX MATCH TYPE QUESTIONS
p q r s
This section contains 1 question. Each question contains statements given in two A p q r s
column which have to be matched. Statements (A, B, C, D) in Column I have to
be matched with statements (p, q, r, s) in Column II. The answers to these B p q r s
questions have to be appropriately bubbles as illustrated in the following example.
p q r s
If the correct matches are A–p, A–s, B–q, B–r, C–p, C–q and D–s, then the correctly C
bubbled 4 × 4 matrix should be as follows : p q r s
D
1. Column I Column II
NARAYANA IIT ACADEMY – INDIA
46 Work Power Energy
(A) Work done by conservative force is (p) Is dependent upon path.
(B) Work done by non-conservative force is (q) Independent of path.
(C) Work done by friction force (r) Kf – Ki = work done by all the forces.
(D) Work energy theorem is (s) Can be +ve, –ve and zero.
2. Column I Column II
(A) Work (p) Slope of momentum P-t graph
(B) Power (q) Slope of W-t graph
(C) Force (r) Area under F-s graph
(D) Impulse (s) Area under F-t graph
(t) Unit in SI system Joule
3. Column I Column II
(A) Force is equal to (p) Work is path independent
(B) For the conservative force (q) Rate of change of linear momentum
(C) Power is equal to (r) Rate of work done
(D) Area of Power-time curve gives (s) Product of force to the velocity
(t) Work done
(u) Negative of the potential energy gradiant
4. Column I Column II
(A) Example of conservative force (p) Work done by all forces
(B) Potential energy defined for only (q) Force constant (spring constant)
(C) If two spring are stretched by same (r) Gravitational force
force, then ratio of their potential energy
is in the inverse ratio of their
(D) Change in kinetic energy is equal (s) Viscous force
(t) Conservative force
5. A block of mass m hanging by an string of length l. The block is given a velocity v horizontally. The One-
point of the string is fixed. The velocity when
Column I Column II
(A) Block rises to an angle θ with the (p) Zero
vertical (Tension is zero on that point)
(B) Block rises above the level of O (q) > 2gl
(C) Block complete full circle (r) gl 2 3cos
(D) When block just complete the circle, (s) Zero
tension at highest point is
(t) 5gL
KEY
7. A (Q,S); B (P,S); C (P,S); D (R) 9. A r,t B q C p D s
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 47
10. A q,u B p,u C r,s D t 20. A r B t C q D p
21. A r B q C t D p
ASSERTION AND REASONING TYPE QUESTIONS
This section contains 2 questions. Each question contains STATEMENT–1 (Assertion) and STATEMENT–
2 (Reason). Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
(B) Statement-1 is True, Statement-2 is True; Statement–2 NOT a correct explanation for Statement-1.
(C) Statement–1 is True, Statement–2 is False
(D) Statement –1 is False, Statement–2 is True.
1. STATEMENT – 1 When a body moves vertically upwards, then work done by force of gravity is negative.
because
STATEMENT – 2 When a body moves upwards its kinetic energy decreases and potential energy
increases.
2. STATEMENT – 1 Work done in gravitational field in cyclic process is zero.
because
STATEMENT – 2 Work done in conservative field does not depend upon path.
3. STATEMENT – 1 Work done by friction on a body sliding down an inclined plane is positive.
because
STATEMENT – 2 Work done is greater than zero, if angle between force and displacement is acute.
4. STATEMENT – 1 Work done in gravitational field in cyclic process is time dependent.
because
STATEMENT – 2 Work done in conservative field does not depend upon path.
5. STATEMENT – 1 The rate of change of total momentum of a many particle system is proportional to
sum of the internal force of a system.
because
STATEMENT – 2 Internal force can change the kinetic energy but not the momentum of the system.
KEY
1. B 2. A 3. D 4. D 5. C
HINTS & SOLUTIONS
1. By W–E theoroem
2. Work done is path independent for gravitational force
Wext Wmg 0 W F –W mg .
NARAYANA IIT ACADEMY – INDIA
48 Work Power Energy
3. Work done by friction is negative..
4. Work done is not time dependent.
dPsys
5. Fext
dt
INTEGER TYPE QUESTIONS
X Y Z W
This section contains 8 questions. The answer to each of the questions is a single
0 0 0 0
1 1 1 1
digit integer, ranging from 0 to 9. The appropriate bubbles below the respective 2
3
2
3
2
3
2
question numbers in the ORS have to be darkened. For example, if the correct 4 4 4 4
answers to questions numbers 1, 2, 3, 4 and 5 (say) are 2, 1, 4, 2 and 3,
5 5 5 5
6 6 6 6
respectively then the correct darkening of bubbles will look like the following. 7
8
7
8
7
8
7
9 9 9 9
1. An object of mass 1 kg moves under the action of a force F in a straight line with its velocity V, chang-
ing with displacement x as v 2 x . Find the work done in a displacement from x = 0 to x = 2. [ in
Joule
2. A block of mass 2kg is pulled by a constant power 100W is placed on a rough horizontal plane. The
frictional coefficient between block and surface is 1. Find the maximum velocity of block.
3. Block A has a weight of 300 N and block B has weight 50N. Calculate the
distance A must descend from rest before it obtains a speed of 4 m/s (Ne-
glect the mass of cord and pulleys). (Take g = 10 m/s2)
A
4. A particle of mass m moves along a circle of radius R with a normal acceleration varying with time as
a n bt 2 , where b 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 2 seconds after the beginning of
motion.(m = 1, v = 2, r = 1)
5. Two blocks A and B are connected to each other by a string and a spring; the B
string passes over a frictionless pulley as shown in the figure. Block B slides over
the horizontal top surface of a stationary block C and the block A slides along the
C
vertical side of C, both with the same uniform speed.The coefficient of friction
A
between the surface and blocks is 0.5. K= 2000N/m. If mass of A is 2 kg calculate
mass of B.
KEY
1. 4 2. 5 3. 2 4. 2 5. 4
HINTS & SOLUTIONS
1. From the defination of acceleration
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 49
a .
dV d 2 x dx
dt dx dt
2
= .2 x a 2m / s 2
2 x
F= ma 1 2 2 N
2
dx Fdx
0
W 50 x 0 100 J
2
2. P F .V constant
P
F
V
1
F As V , F
V
When
Net force on block becomes zero i.e. its maximum velocity
100
P mg Vmax ; Vmax 5m / s
1 2 10
3. SA h
and S B 2h
VA 4m / s VB 8m / s
From energy conservation decreases in P.E. of A
= Increase in P.E. B + Increase in K.E. of both A and B
1 1
300h 50 2h 30 4 2 5 82
2 2
300h 100h 240 160
200h 400
h 2m
dv
4. v bR t bR
dt
For circular motion work done by normal force is zero. For tangential forces.
dv
Ft m m bR P Ft .v Ft v cos as 00
dt
NARAYANA IIT ACADEMY – INDIA
50 Work Power Energy
P t dt T
mbRT dt
0
P Ft v mbRt Average power =
T
T
dt
0
0
T
mbR t 2 / 2mbRt
0
T 2
5. Normal reaction between blocks A and C will be zero.
Therfore, there will be no friction between them.
Both A and B are moving with uniform speed. Therefore,
net force on them should be zero
For equilibrium of A:
mA g kx
T =kx kx
m A g (2)(10)
x 0.01m For equilibrium of B:
k 2000
mB g = T = kx = mAg B A
T
mA 2 mBg
mB 4kg
0.5 T
mAg
SUBJECTIVE PROBLEMS
1. A body of mass m was slowly hauled up the hill. (fig) by a force E which E
at each point was directed along a tangent to the trajectory. Find the
work performed by this force, if the height of the hill is h, the length of its m
h
base l, and coefficient of kinetic friction k.
2. A small body A starts sliding from the height h down an inclined groove
passing into a half - circle of radius h/2 (see figure). Assuming the friction
to be negligible, find the velocity of the body at the highest point of its
trajectory (after breaking off the groove).
3. A small bar resting on a smooth horizontal plane is attached by threads to a
point P (figure) and by means of a weightless pulley, to a weight B possessing
the same mass as the bar itself. Besides, the bar is also attached to a point P
A
O by means of a light non-deformed spring of length l0 = 50cm and the stiffness
K = 5 mg/l0, where m is the mass of the bar. The thread PA having been burned,
the bar starts moving. Find its velocity at the moment when it is breaking off the
plane. B
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 51
4. In the figures (a) and (b) AC, DG and GF are fixed
A D
inclined planes, BC = EF = x and AB = DE = y. A small
G
block of mass M is released from the point A. It slides
down AC and reaches C with a speed VC. The small
block is released from rest from the point D. It slides
down DGF and reaches the point F with speed VF. The
coefficients of kinetic frictions between the block and B C E F
both the surfaces AC and DGF are m. Calculate VC (a) (b)
and VF.
5. A lead bullet just melts when stopped by an obstacle. Assuming that 25 % of the heat is absorbed by the
obstacle, find the velocity of the bullet if its initial temperature is 27ºC. (Melting point of lead = 327º,
specific heat of lead = 0.03 calories/gm/ºC, latent heat of fusion of
lead = 6 calories/gm, J = 4.2 J/calorie).
6. A 0.5 kg block slides from the point A (see figure) on a horizontal track
with an initial speed of 3m/s towards a weightless horizontal spring of
length 1 m and force constant 2 Newton/m.The part AB of the track is A B D C
frictionlessand the part BC has the coefficients of static and kinetic
friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2m and 2.14 m respectively find the
total distance through which the block moves before it comes to rest completely (Take g = 10 m/s2).
7. A string with one end fixed on a rigid wall, passing over a fixed frictionless
pulley at a distance of 2m from the wall, has a point mass M = 2 kg attached
M
to it at a distance of 1m from the wall. A mass m = 0.5 kg is attached to the
free end. The system is initially held at rest so that the string is horizontal
between wall and pulley and vertical beyond the pulley as shown in figure.
What will be the speed with which the mass will hit the wall when the mass m
m is released ?
8. A bullet of mass M is fired with a velocity 50 m/s at an angle with the horizontal. At the highest point of
its trajectory, it collides head–on with a bob of mass 3M suspended by a massless string of length 10/3
metres and gets embedded in the bob. After the collision the string moves through an angle of 120º. Find
(i) the angle ,
(ii) the vertical and horizontal co–ordinates of the initial position of the bob with respect to the point of
firing of the bullet. (Take g = 10 m/s2).
9. A stone with weight ‘w’ is thrown vertically upward into the air with initial speed v0. If a constant force f due
to air drag acts on the stone throughout its flight.
v 02
(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 v 0
w f
10. Show that a particle projected with velocity (2ag) from the lowest point of a vertical circle of radius a
and moving inside it will just reach the end of the horizontal diameter; while if projected with velocity
(5ag) , it will just reach the highest point. Prove that the reaction at any point in the first case is
proportional to the depth below the horizontal diameter and in the second case to the depth below the
highest point.
NARAYANA IIT ACADEMY – INDIA
52 Work Power Energy
ANSWERS
Subjective Solutions :
dl dy
1. WF Wfr Wmg KE 0 WF Wmg Wfr dWfr mg cos .dl
dx
Wfr mgdx = mgl WF mgh mgl mgh mgl .
2. v 2 (at end of track) = 2gh.
mu 2
mg cos
Let body break at angle the h ...... (1)
2
u
h 2 2
u 2 v 2 2g 1 cos solving cos = &u gh .
2 3 3
2 2 8
v at highest pt is u cos gh gh
3 3 27
3. Let L be the extension of the spring.
l 2
l0 x 2 l0 Apply C.M.E
1 1 1
mgx = mv 2 mv 2 Kl2
2 2 2
2
5g 2
v 2 gx l0 x 2 l0
2l0
l0
T cos mg cos
2
l0 x 2 T K l
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 53
K 2
2
l0 x 2 l0 l0 / l0 x 2 mg
5mg
l0 2
2
l0 x 2 l0 l0 / l0 x 2 mg Solving x
3
4
l0
2
3l g 5 g 2 9l0
2
v 2 0 l0 l0
4 2 l0 4
19gl0
v 2
32
v 1.7m / s
4. In both the cases work done by friction will be Mgx
1 1
MVC2 MVF2 Mgy Mgx
2 2
VC VF 2gy 2gx
5. Heat energy required to just melt the bullet Q = Q1 + Q2
Here, Q1 = ms
= (m×103) (0.03 × 4.2) (327 – 27)
= (3.78 × 104 m)
= Q2 = mL = (m × 103) 6 × 4.2)
= (2.52 × 104m)
Q = (6.3 × 104m)
1
If v be the speed of bullet, then 75% of mv 2 should be equal to Q. Thus,
2
1
0.75 × m v 2 6.3 104 m v= 409. 8m/s
2
6. From A to B, there will be no loss of energy. Now let block compresses the spring by an amount x and
comes momentarily to rest. Then, loss of energy will be equal to the work done against friction. There-
fore,
x
kx
A B C v=0 C f
1 1
k mg (BD x ) mv 2 kx 2
2 2
1 1
Substituting the values (0.2) (0.5) (10) (2.14 + x) = (0.5)(3)2 (2)( x )2
2 2
Solving this equation, we get x = 0.1m
NARAYANA IIT ACADEMY – INDIA
54 Work Power Energy
7.
1m
2m
M
vcos0
5
1m 1
5m
v ( 5 – 1)m
m 2
2
2v
or , (2) (9.8) = (0.5) (9.8) 1
2
1
5 1 v 2 0.5
2 5
Solving this equation, we get v = 3.29 m/s
8. (a) At the highest point, velocity of bullet is 50 cos .
So, by conservation of linear momentum
50
V cos ...... (1)
4
4mv 2 g 50
T 4mgcos1200 But T 0 v 0 or v2 l ........ (2)
l 2 3
10
( as l m and g = 10 m/s) Apply C.M.E
3
1 1 10 10
4mV 2 4mv 2 4mg sin300
2 2 3 3
1 2 1 2
V v 5g
2 2
2 2
1 50 1 5 25 35
cos g 5g cos g Cooling 300
2 4 23 2 3
R 1 u2 sin 2 10
m 108.25m or v 2 V 2 3(10) or v 2 V 2 100
2 2 g 3
solving Eqs. (1) , (2) and (3), we get cos 0.86 or 30
u sin 50 50 1
2 2
y=H= 31.25m
2g 2 10 1
1 w 2
9. a) v0 wh fh
2 g
1 w 2 2
b) v0 v f 2h
2 g
10 Self explanatory.
NARAYANA IIT ACADEMY – INDIA
Work Power Energy 55
PREVIOUS YEARS IIT QUESTIONS
1. A particle is suspended vertically from a point O by an inextensible massless string of length L. A vertical
line AB is is at a distance of L/8 from O as shown. The object is given a horizontal velocity u. At some
point, its motion ceases to be circular and eventually the object passed through the line AB. At the
instant of crossing AB, its velocity is horizontal. Find u. [1999]
2. A spherical ball of mass m is kept at the highest point in the space
between two fixed, concentric spheres A and B (see Figure). The smaller
Sphere B
sphere A has a radius R and the space between the two spheres has a
width d. The ball has a diameter very slightly less than d. All surfaces
are frictionless. The ball is given a gentle push (towards the right in the d
O
figure). The angle made by the radius vector of the ball with the upward R
vertical is denoted by (Shown in the figure).
Sphere A
(a) express the total normal reaction force exerted by the spheres on the ball
as a function of angle .
(b) let NA and NB denote the magnitudes of the normal reaction forces on the ball exerted by the spheres
A and B, respectively. Sketch the variations of NA and NB as functions of cos in the range 0
< by drawing two separate graphs in your answer book, taking cos on the horizontal axes.
[2002]
SOLUTIONS
C v
90-
D Q
1. Now, we have following equations
mg
L + Lsin
L
L 8
P u
Lcos
mv 2
(1) TQ 0 Therefore, mg sin ...... (1)
L
(2) v 2 u 2 2gh u 2 2gL(1 sin ) ...... (2)
1
(3) QD (range ) ........ (3)
2
3 3
u gL 2
2
NARAYANA IIT ACADEMY – INDIA
56 Work Power Energy
d
(a) h R 1 cos
2
2.
v
mg
d
v 2 2gh 2 R 1 cos g ........ (1)
2
N A 0 and NB = mg (2 –3 cos )
The correspondig graphs are as follows
NA NB
5mg
mg mg
+1
cos cos
–1 2/3 –1 2/3 +1
NARAYANA IIT ACADEMY – INDIA
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