WORK, POWER AND ENERGY 1

EXERCISE – 1: BASIC OBJECTIVE QUESTIONS

 3a  4 a  7 a
Work
W  (3iˆ  4j)
ˆ  (aˆi)  (3iˆ  4j)
ˆ  ( aˆj)
 
2
1. A force of 10 iˆ  3 ˆj  6 kˆ N acts on a body of 5 kg
 3a  4 a  7 a
and displaces it from A  6 iˆ  5 ˆj  3 kˆ  m to B W1  W2
4. The net work done by the tension in the figure when
10 iˆ  2 ˆj  7 kˆ  m . The work done is the bigger block of mass M touches the ground is:
(a) zero (b) 121 J
(c) 100 J (d) 221 J
Ans. (b)

Sol. F  (10iˆ  3jˆ  6k)Nˆ
 
r1  6iˆ  5jˆ  3k,
ˆ r  10iˆ  2jˆ  7kˆ
2
  
Δr  r2  r1  4i  7 ˆj  10kˆ
ˆ
 
W  F.Δr  (10iˆ  3 ˆj  6kˆ)  (4iˆ  7ˆj  10kˆ)
 40  21  60  121J
2. A body is under the action of two equal and opposite
forces, each of 3 N. The body is displaced by 2m. The
work done is: (a) + Mgd (b) – (M + m)gd
(a) + 6 J (b) – 6 J (c) – mgd (d) zero
(c) 0 (d) none of the above Ans. (d)
Ans. (c) Sol. Work by tension T = Same in both strings in
Sol. magnitude
On M : T  d  cos180   Td
On m : T  d  cos 0   Td
Total work by Tension
W  Td  Td  0
5. A ball of mass 5 kg experiences a force F = 2 x2 + x.
Fnet  0
Work done in displacing the ball by 2 m from origin
  
W  F  Δr  0.Δr  0 is:
3. A particle is moved from (0, 0) to (a, a) under a force 22 44
 (a) J (b) J
F  (3i  4 j ) from two paths. Path 1 is OP and path 2 3 3
is OQP. Let W1 and W2 be the work done by this 32 16
(c) J (d) J
force in these two paths. Then: 3 3
y Ans. (a)
P (a, a) Sol. F  2x 2  x
W  Fdx    2 x 2  x  dx
2

2
45° x  x3  1 2 2 1
 2     x 2   (8)  (4)
O Q  3 0 2
0 3 2
(a) W1 = W2 (b) W1 = 2W2 16 22
(c) W2 = 2W1 (d) W2 = 4W1  2  J
3 3
Ans. (a)

Sol. F  3iˆ  4jˆ

W1  F  Δrˆ  (3iˆ  4j)
ˆ  ( aˆi  aˆj)
WORK, POWER AND ENERGY 2

6. The relationship between force and position is shown Ans. (b)

in figure (in one dimensional case). The work done Sol. F  Cx given,
by the force in displacing a body from x = 1 cm to x  0, x  x1 (Variable force)
x = 5 cm is: x1
x1  x2  1
W   Fdx  C     Cx12
0
 2 0 2
10. A particle of mass 0.5 kg is displaced from position
 
r1 (2, 3, 1) to r2 (4, 3, 2) by applying a force of
magnitude 30 N which is acting along iˆ  ˆj  kˆ  .
(a) 20 erg (b) 60 erg The work done by the force is
(c) 70 erg (d) 700 erg (a) 10 3 J (b) 30 3 J
Ans. (a)
(c) 30 J (d) none of these
Sol. Work done from x=1cm to x=5cm
Ans. (b)
Area under F – x graph will be work done
Sol. m = 0.5 kg
W  (1 10)  1 20  1 20  1 10  20 erg 
r1  2iˆ  3 ˆj  kˆ
7. Under the action of a force, a 2 kg body moves such

that its position x as a function of time t is given by r2  4iˆ  3 ˆj  2kˆ
t3 F = 30 N
x= , where x is in metre and t in second. The work
3 The work done is the dot product of force and
done by the force in the first two seconds is: displacement.
 
(a) 1600 J
(c) 16 J
(b) 160 J
(d) 1.6 J
  
r2  r1  4iˆ  3 ˆj  2kˆ  2iˆ  3 ˆj  kˆ 
Ans. (c)  2iˆ  0 ˆj  kˆ
3 
t dx | F | 30 N
Sol. m  2kg , x  , v  t2 Since v 
3 dt
 iˆ  ˆj  kˆ 
 a  2t ,
dv
since a  . F  30
12  12  12

3

30 ˆ ˆ ˆ
i  jk 
dt
  
W  Fdx  ma  dx  ma 
dx
dt
 dt W  F .  r2  r1  
30 ˆ ˆ ˆ
3
 
i  j  k . 2iˆ  0iˆ  kˆ 
2 2 2
W   mavdt   2  2t  t 2 dt  4  t 3 dt 60 30 90 3
0 0 0      30 3 J
3 3 3 3
4
  t 4   16J
2
W 11. A box is dragged across a floor by a rope which
4 0
makes an angle of 45° with the horizontal. The
8. A particle moves along the x-axis from x = 0 to
tension in the rope is 100 N while the box is dragged
x = 5 m under the influence of a force given by
10 m. The work done is:
F = 7 – 2x + 3x2. Work done in the process is:
(a) 607.1 J (b) 707.1 J
(a) 70 (b) 270
(c) 1414.2 J (d) 900 J
(c) 35 (d) 135
Ans. (b)
Ans. (d)
1
W  Fdx    7  2 x  3 x 2  dx Sol. W  Fs cos   100 10   707.1J
5
Sol.
0 2
 7( x)50   x 2    x 3   35  25  125  135J
5 5
12. A horizonal force F pulls a 10 kg carton across the
0 0
floor at a constant speed. If the coefficient of sliding
9. A particle moves under a force F = Cx from x = 0 to friction between carton and floor is 0.50, the work
x = x1. The work done is: done by F in moving the carton through 5 m is: [take
Cx12 g =10 m s -2 ]
(a) Cx12 (b)
2
(a) 196 J (b) 210.5 J
(c) 0 (d) Cx13 (c) 245 J (d) 254 J
WORK, POWER AND ENERGY 3

Ans. (d) 1 1 5 3
Sol. W  F.s and F=μmg W  2    10  10  5 3J
10 2 10
 W = μmgs = 0.5  10  9.8  5  245J 15. A mass M is lowered with the help of a string by a

13. The work done by a force F  6 x3 iˆ   N is distance x at a constant acceleration
g
. The
2
displacing a particle from x = 4 m to x = –2 m is magnitude of work done by the string will be:
(a) – 240 J
1 2
(b) 360 J (a) Mgx (b) Mgx
2
(c) 420 J
1 2
(d) will depend upon the path (c) Mgx (d) Mgx
Ans. (b) 2
 Ans. (c)
Sol. F  6 x3iˆ
  Sol. Tension in string by law of motion
 dW   F . dr
2
W    6 x iˆ  . dxiˆ  dyjˆ  dzkˆ 
3

4
2
  6x dx
3

6
 16  256
4
6
   240  360 J
4 mg  T  ma
14. A body of mass 500 g is taken up an inclined plane of mg mg
length 10 m and height 5 m, and then released to slide T  mg  ma  mg  
2 2
down to the bottom. The coefficient of friction work  T  d  cos θ
between the body and the plane is 0.1. What is the
mg mgx
amount of work done by friction in the round trip? W   x  cos180  
2 2
(a) 5 J (b) 15 J
16. The work done by pseudo forces is
5
(c) 5 3 J (d) J (a) positive (b) negative
3 (c) zero (d) all of these
Ans. (c) Ans. (d)
Sol. Sol. Pseudo force can do positive, negative and zero work,
Depends in direction of displacement of object from
SA accelerating frame.
17. A block of mass m is pulled along a horizontal
surface by applying a force at an angle q with the
horizontal. If the block travels with a uniform
velocity and has a displacement d and the coefficient
of friction is m, then the work done by the applied
Given, force is
m  500gm
μ  0.1
1 500 5 3  10
Force of friction  μmg cos θ   
10 1000 10
1 1 5 3
Work by friction     10  10  mgd  mgd cos 
10 2 10 (a) (b)
cos    sin  cos    sin 
Work for Round trip is double. (Same amount)
WORK, POWER AND ENERGY 4

 mgd sin   mgd cos  25 1

(c) (d) cos   
cos    sin  cos    sin  50 2
Ans. (b)   60
Sol. Fcosθ  f Thus, force makes angle of 60° with direction of
N  Fsin θ  mg motion.
F cos θ  μ(mg - Fsinθ) Kinetic Energy
μmg 20. The P.E. and KE of a helicopter flying horizontally at
F
cosθ+μsinθ a height 400 m are in the ratio 5 : 2. The velocity of
 Work  F  d  cos θ helicopter is
μmg (a) 28 m/s (b) 14 m/s
W  d  cos θ (c) 56 m/s (d) 30 m/s
cosθ+μsinθ
Ans. (c)
18. A uniform chain of length L and mass M is lying on a
U 5
smooth table and one third of its length is hanging Sol. H  400m and 
vertically down over the edge of the table. If g is K 2
acceleration due to gravity, work required to pull the mgH 5 2gH 5
   2 
hanging part on to the table is 1 2 2 v 2
mv
MgL 2
(a) MgL (b)
3 2 2  10  400  2
v 2  2gH  v
MgL MgL 5 5
(c) (d)
9 18 40 2m / s  56m / s

Ans. (d)
Sol. Let at any time hanging part is x.
21.  
A 120 g mass has a velocity v  2 iˆ  5 ˆj ms 1 at a

Force required to pull it up, certain instant. K.E. of the body at that instant is
Mgx (a) 3.0 J (b) 1.74 J
dF  (c) 4.48 J (d) 5.84 J
L
Total work, Ans. (b)

L Sol. m  120gm and v  2iˆ  5iˆ m/s
L
Mgx Mg  x 2  3 MgL2 
W 3
 dx     v  (2) 2  (5) 2  29
0 L L  2 0 18L
MgL 1 1 120
W KE  mv 2    29  1.74J
18 2 2 1000
19. A car covers a distance of 10 km along an inclined 22. A body is moving under the action of a force.
plane under the action of a horizontal force of 5 N. Suddenly, force is increased to such an extent that its
The work done on car is 25 kJ. The inclination of the kinetic energy is increased by 100%. The momentum
plane to horizontal is: increases by:
(a) 0° (b) 30° (a) 100% (b) 60%
(c) 60° (d) 90° (c) 40% (d) 20%
Ans. (c) Ans. (c)
Sol. p2
Sol. K , p  2mK
2m
K   2K (since there is100% )
p'  2mK '  2m2K
p' 2m  2 K p'-p 2p-p
  2  
p 2mK p p
W  (F cos θ)10  10 3
Δp 2p  p
% is increase 100   100  41.4%
25 10  5cos θ 10 10
3 3 p p
WORK, POWER AND ENERGY 5

23. A man has a box of weight 10 kg. The energy of the  v =2V
box, when the man runs with a constant velocity of Also,
2 m/sec along with the box behind the bus, is: K man '  K boy
(a) 10 joule (b) 30 joule
1 1 1 M
(c) 20 joule (d) 2 joule  M(V  1) 2  mv 2  × 4V 2
2 2 2 2
Ans. (c)
(V+1) 2
Sol. m  10kg v  2m / s   2  V  1  2V
(V) 2
1 1
KE  mv 2   10  (2) 2  20J 1 2 1
2 2 V   2  1m / s
24. What is the shape of the graph between the speed and 2 1 2 1
kinetic energy of a body? and v  2V  2  ( 2  1)m / s
(a) straight line (b) hyperbola
27. An object moving horizontally with kinetic energy of
(c) parabola (d) exponential
800 J experiences a constant opposing force of 100 N
Ans. (c)
while moving from a to b (where ab = 2m). The
1 energy of particle at b is:
Sol. K  mv 2 K  v2
2 (a) 700 J (b) 400 J
25. If the linear momentum is increased by 50%, then (c) 600 J (d) 300 J
kinetic energy will be increased by: Ans. (c)
(a) 50% (b) 100% Sol. Now the work done by the opposing force for
(c) 125% (d) 25% travelling the distance 2m = 100  2  200J
Ans. (c) This will resist the forward motion of the object and
50 p2 therefore, it will lose K.E.
Sol. p'  p  p  1.5p, K
100 2m So, the TE at point b will be = 800 - 200 = 600 J
K  p '2 (1.5p) 2 (as the object is moving on horizontal plane and there
 2   2.25
K p p2 is no change in P.E.)
K   2.25K
28. A particle moves on a rough horizontal ground with
3
ΔK 2.25K  K some initial velocity v 0 . If th of its K.E. is lost in
 100  100  125% 4
K K
friction in time t0, the coefficient of friction between
26. A running man has half the KE that a boy of half his
mass has. The man speeds up by 1m/s and then has the particle and the ground is
the same KE as that of boy. The original speeds of v v
(a) 0 (b) 0
man and boy in m/s are: 2 gt0 4 gt0
(a) ( 2  1), ( 2 1) (b) ( 2  1), 2( 2  1) 3 v0 v0
(c) (d)
(c) 2, 2 (d) ( 2  1), 2( 2  1) 4 gt0 gt0
Ans. (b) Ans. (a)
Sol. Let V be the velocity of man and v be the velocity of 1 2
Sol. KEi  mv0
the boy 2
1 M 1 1 3 1
K man   K boy , m  , M is mass of man KE f  mv 2  mv02   mv02
2 2 2 2 4 2
1 1 2 1 1 2
K man   M  V 2 mv   mv0
2 2 4 2
1 1 v
K boy   m  v 2  mv 2 v 0
2 2 2
1 v  u  at
K man   K boy
2 v0 v
 v0  at  a   0
1 1 1 1 M 2 2t0
 MV 2    m×v 2  × ×v 2
2 2 2 4 2 Let friction coefficient is μ
WORK, POWER AND ENERGY 6

a = μg v 20m / s
given as  a    2m / s 2
a v t 10 s
  0
g 2 gt0 Thus, force acting on the body = F = ma = 100 N
1 2
Work Energy Theorem Distance travelled in given time = s = at  100m
2
29. A particle of mass 0.l kg is subjected to a force which  
Thus, work done in given time  F . s  104 J
varies with distance as shown in figure. If it starts its
32. The displacement of a body of mass 2 kg varies with
journey from rest at x = 0, its velocity at x = 12 m is
time t as s = t2 + 2t, where s is in meters and t is in
seconds. The work done by all the forces acting on
the body during the time interval t = 2s to t = 4s is
(a) 36 J (b) 64 J
(c) 100 J (d) 120 J
Ans. (b)
Sol. s = t 2 + 2t
(a) 0 m/s (b) 40 m/s ds
v  2t  2
(c) 20 2 m / s (d) 20 m/s dt
Ans. (b) By work energy theorem
Sol. M  0.1kg , let V=velocity at x = 12(m) Wext force  K
1
by work - Energy Theorem K  m  v 2f  vi2 
1 1 2
Wnet =ΔK  MV 2  M(0) 2

1
2 2
1 1

1
2
 2

 2  2  4   2    2  2   2 
2

(12  4)  10  80    V 2 = 100 - 36 = 64 J
2 2 10
(Work from F-x area) 33. An object of mass m is allowed to fall from rest along
a rough inclined plane. The speed of the object on
V 2  1600
reaching the bottom of the plane is proportional to
V  40m / s
(a) m 0 (b) m
30. What average force is necessary to stop a bullet of
mass 20 gm and speed 250 m/sec as it penetrates (c) m 2
(d) m 1
wood to a distance of 12 cm: Ans. (a)
Sol.
(a) 3.4 × 103 newton
(b) 5.2 × 103 newton
(c) 4.0 × 103 newton
(d) 3.6 × 103 newton
Ans. (b)
Sol. 5.2  103 newton since, m  0.02kg , u  250m/s
v  0m/s, s  0.12m
mu 2 h h
u 2  2as  F  ma   sin  l 
2s l sin 
31. How much work must be done by a force on 50 kg From work energy theorem we get,
body in order to accelerate it from rest to 20 m/s in 1 2  h 
mv  mgh   mg cos   
10 s? 2  sin  
(a) 103 J (b) 104 J h
v  2 gh  2 g cos  
(c) 2 × 103 J (d) 4 × 104 J sin 
Ans. (b)
v  2 gh  2 gh cot 
Sol. Acceleration that causes the given rise in speed is
v  m0
WORK, POWER AND ENERGY 7

34. A spring of spring constant 1000 N/m is compressed v 2  72 km / hr  20 m / s

through 5 cm and is used to push a metal ball of mass 1 1
0.1 kg. The velocity with which the metal ball moves W  K   mv 2   m×u 2
2 2
is
1
(a) 5 m/s (b) 7.5 m/s W  1000   202  10 2   500  300
2
(c) 10 m/s (d) 2.5 m/s
Ans. (a)  1.5 105 J
Sol. applying work energy theorem 37. The work done in time t on a body of mass m which
Wext force  K is accelerated from rest to a speed v in time t1 as a

1 2 1 2 function of time t is given by:

kx  mv 1 v v
2 2 (a) m t 2 (b) m t 2
2 t1 t1
k 1000
v x 0.05  5m / s 2
m 0.1 1  mv  2 1 v2 2
(c)   t (d) m t
35. A block of mass 4 kg falls from a height of 3 m on a 2  t1  2 t12
spring of force constant l500 N/m. Calculate Ans. (d)
maximum compression of spring (g = 9.8 N/kg) Sol. By work Energy Theorem
(a) 1.35 m (b) 0.42 m
v
(c) 0.735 m (d) 0.676 m W  K, v  0  a  t1  a 
t1
Ans. (b)
Sol. M  4 kg, h  3 m, K  1500 N / m v  vt
velocity in t sec v  0  at  0   1   t 
By conservation of mechanical energy law,  t1  t1
2
1 vt 1 v2
W  ΔK   m   1   m 2 t 2
2  t1  2 t1
38. What average force is necessary to stop a bullet of
mass 20 gm and speed 250 m/sec as it penetrates
wood to a distance of 12 cm:
3 3
(a) 3.4 × 10 newton (b) 5.2 × 10 newton
3 3
(c) 4.0 × 10 newton (d) 3.6 × 10 newton
Ans. (b)
1 Sol. By work Energy Theorem,
mg (h  l )   K  ( l ) 2
2 W  K, W  F  d
1 1 1 20
1
mgh  mg l   K  (l ) 2
2
F d 
2
m v 2  mu 2  
2 2 1000
 02  2502 
4  9.8  3  4  9.8  l 12 1 20
F    (250) 2
1 100 2 1000
 1500  l 2
2  5.21103 N (Retarding opposite)
Solving quadratic equation
l  0.42 m
36. A truck weighing 1000 kg changes its speed from
36 km/h to 72 km/h in 2 minutes. Thus, the work
done by the engine on the truck is:
5 5
(a) 2.5 × 10 J (b) 3.5 × 10 J
5 5
(c) 1.5 × 10 J (d) 5.5 × 10 J
Ans. (c)
Sol. M  1000 kg
v1  36 km / hr  10m/s
WORK, POWER AND ENERGY 8

39. A particle at rest on a frictionless table is acted upon 1 2 1 2

Wint  Wext  mv2  mv1
by a horizontal force which is constant in magnitude 2 2
and direction. A graph is plotted of the work done on 1 1
Wint   10    2  3    2  5 
2 2
the particle W, against the speed of the particle v. If 2 2
there are no frictional forces acting on the particle,
Wint  6 J
the graph will look like:
(a) U  Wint
=6J
41. A block of mass 0.5 kg has an initial velocity of 10
W m/s. down an inclined plane of angle 30°, the
coefficient of friction between the block and the
inclined surface is 0.2. The velocity of the block after
v it travels a distance of 10 m is:
(b) (a) 17 m/s (b) 13 m/s
(c) 24 m/s (d) 8 m/s
W Ans. (b)
Sol.

v
(c)

Fnet = mg sin30 - μmg cos30

v
anet  5  3
(d)
From A to B
Given: u = 10 m/s v = ?
W s = 10 m
 
a  5  3 m / s2
v
 u 2  2as  13m / s
Ans. (d)
Sol. By work energy Theorem. 42. A block is moved from rest through a distance of 4 m
Work by all forces = change in its K.E. along a straight-line path. The mass of the block is 5
kg and the force acting on it is 20 N. If kinetic energy
1 1
W  F  d   mv 2   (0) 2 acquired by the block be 40 J, at what angle to the
2 2
path is the force acting?
1 1 m
F  d  mv 2 F    v 2 (a) 30o (b) 60o
2 2 d (c) 45 o
(d) 0o
F  v2 graph will be parabola. Ans. (b)
40. A body of mass 2 kg is moved from a point A to a Sol. Fscosθ = K.E {from Work-Energy theorem}
point B by an external agent in a conservative force 40
field. If the velocity of the body at the points A and B cos  
20  4
are 5 m/s and 3 m/s respectively and the work done   60
by the external agent is –10 J, then the change in
potential energy between points A and B is
(a) 6 J (b) 36 J
(c) 16 J (d) none of these
Ans. (a)
Sol. Work done by all the forces = change in K.E
WORK, POWER AND ENERGY 9

43. A particle is moving in a conservative force field 46. A meter stick of mass 400 g is pivoted at one end and
from point A to point B. UA and UB are the potential displaced through an angle 60°. The increases in its
energies of the particle at point A and B and WC is the potential energy is:
work done by conservative forces in the process of (a) 1 J (b) 10 J
taking the particle from A to B: (c) 100 J (d) 1000 J
(a) WC = UB – UA (b) WC = UA – UB Ans. (a)
(c) UA > UB (d) UB > UA Sol. M  400gm, θ  60
Ans. (b) ΔW  ΔU  mgh
Sol. WA  B =ΔWC  ΔU
ΔWC    U B  UA   U A  U B
44. Work done by the conservative forces on a system is
equal to
(a) the change in kinetic energy of the system
(b) negative of the change in potential energy of the
system
(c) the change in total mechanical energy of the
system
(d) none of these Using COM concept,
Ans. (b) ΔW  ΔU  mgl (1  cos θ)
Sol. Wconservative + WNC + Wothers = KE 400 50
ΔW   10   (1  cos 60)  1J
But work done by conservative force is equal to the 1000 100
change in Potential energy of the system. This is 47. A man weighing 60 kg lifts a body of mass 15 kg to
because a conservative force is one in which work is the top of a building 10 m high in 3 minutes. His
done in moving a particle b/w two positions. efficiency is
(a) 20% (b) 10%
Potential Energy
(c) 30% (d) 40%
45. If we shift a body in equilibrium from A to C in a Ans. (a)
gravitational field via path AC or ABC Sol. M  60kg, Δm  15kg, H  10m, t  3min
Wout
χ
w in
ΔmgH Δm 15
χ  
(M  Δm) gh (M  Δm) 60  15
1
 χ   100%  20%
(a) the work done by the force F for both paths will 5
be same 48. A spring of spring constant 8 N/cm has an extension
(b) WAC > WABC of 5 cm. The minimum work done in joule in
increasing the extension from 5 cm to 15 cm is
(c) WAC < WABC
(a) 16 J (b) 8 J
(d) None of the above (c) 4 J (d) 32 J
Ans. (a) Ans. (b)
 h  Sol. K  8 N / m  800 N / m
Sol. WAC  F   cos  90     Fh
 sin   1
W K  x 22  x12 
WAB  Fa cos 90  0 2
WBC  mgh 1  15  2  5 2 
  800       8J
WABC  0  mgh  Fh. 2  100   100  

49. The potential energy of a certain spring when

WORK, POWER AND ENERGY 10

stretched through a distance ‘S’ is 10 joules. The 51. Two springs have their force constants k1 and k2.
amount of work (in joule) that must be done on this Both are stretched till their elastic energies are equal.
spring to stretch it through an additional distance ‘S’ Then, ratio of stretching forces F1/F2 is equal to:
will be: (a) k1: k2 (b) k2 : k1
(a) 30 (b) 40 (c) k1 : k2 (d) k 22 : k12
(c) 10 (d) 20
Ans. (c)
Ans. (a)
Sol. Let for same potential energy, elongations of springs
1
Sol. W1  U1   K  (S) 2  10 be x1 and x2 and their respective force constants be
2 k1 and k2
1 1 1 1 1
W2  U 2   K(2 S) 2  K  (S) 2  3  KS2 The we have k1 x12  k 2 x22
2 2 2 2 2
U 2  3U  3  10  30 J
x12 k 2
Work  30 J Hence  i 
x22 k1
50. The force required to stretch a spring varies with the
distance as shown in the figure. If the experiment is F1 k1 x1 k1 k k1
   2 
performed with the above spring of half the length, F2 k2 x2 k2 k1 k2
the line OA will: 52. On changing the length of a spring by 0.1 m there is a
change of 5 J in its potential energy. The force
constant of the spring is:
(a) 80 Nm-1 (b) 10.0 Nm-1
-1
(c) 90 Nm (d) 1000 Nm-1
Ans. (d)
(a) shift towards F–axis
Sol. x = 0.1 m
(b) shift towards X–axis
(c) remain as it is 1 2
kx  PE
(d) become double in length 2
Ans. (a) 1
k  0.1  5
2

Sol. 2
K  1000 N / m
53. A rod of mass m and length l is lying on a horizontal
table. Work done in making it stand on one end will
be:
mg
(a) mgl (b)
Equation of straight line. 2
y = mx + c mg
(c) (d) 2 mgl
F = kx 4
mk Ans. (b)
where m is slope. Sol. By using centre of mass concept work is required to
change pot. Energy.
1
Stiffness constant (k)  . l
l Centre of mass rises by 
2
kl = Constant
Mgl
k1l1  k 2l2 W  U 
2
k1l1 k1l1
k2    2k1
l2 l1
2
 y  2kx
 Slope increases.
 Shift towards F axis.
WORK, POWER AND ENERGY 11

Conservation of Mechanical Energy 1 1

mgd   Kd 2  m(0) 2
54. A toy gun uses a spring of very large value of force 2 2
constant k. When charged before triggering in the 2 mg
d   2  (d )
upward direction, the spring is compressed by a small K
56. A sphere of mass 2 kg is moving on a frictionless
distance x. If mass of shot is m, on being triggered it
horizontal table with velocity v. It strikes with a
will go up to a height of:
spring (force constant = 1 N/m) and compresses it by
kx 2 x2
(a) (b) 4 m. The velocity (v) of the sphere is:
mg kmg
(a) 4 m/s (b) 2 2 m/s
kx 2 (kx ) 2
(c) (d) (c) 2 m/s (d) 2 m/s
2mg mg
Ans. (b)
Ans. (c) Sol. Mass of sphere = 2 kg
Sol. By mechanical energy conservation law, k = 1 N/m
x = 4m
KE of sphere = PE of spring
1 2 1 2
mv  kx
2 2
1 1
 2  v 2   1  4 
2

2 2
v 2  8  v  2 2m / s
57. An elastic string of unstretched length L and force
constant k is stretched by a small length x. It is
further stretched by another small length y. The work
done in the second stretching is
1 1
(a) ky 2 (b) k  x 2  y 2 
2 2
1 1
(c) ky  2 x  y  (d) k  x  y 
2

2 2
1 2
kx  mgh Ans. (c)
2 Sol. Work done = Change in Potential energy
kx 2
h 1 ky
W  k  x  y   x 2    2 x  y 
2
2mg
2   2
55. A body is attached to the lower end of a vertical spiral 58. A block of mass 0.5 kg has an initial velocity of 10
spring and it is gradually lowered to its equilibrium m/s down an inclined plane 30°, the coefficient of
position. This stretches the spring by a length d. If the friction between the block and the inclined surface is
same body attached to the same spring is allowed to 0.2. The velocity of the block after it travels a
fall suddenly, what would be the maximum stretching distance of 10 m figure is
in this case?
(a) d (b) 2d
1
(c) 3d (d) d
2
Ans. (b)
Sol. For slow release force is balanced by gravity,
Kd  mg
(a) 17 m/s (b) 13 m/s
mg
 d (c) 24 m/s (d) 8 m/s
K Ans. (b)
When suddenly released the mass goes beyond
Sol. m  0.5 kg v 0  10 m / s
equilibrium till its velocity becomes zero.
Then by mechanical energy conservation law. Acceleration a  ( g sin θ  μg cos θ)
WORK, POWER AND ENERGY 12

 2 3 1 3 1
a  10  sin 30     10    K.E. at that time K,   3  (29.4) 2
10 2  2
  2 10 
 1296.5J
a  5  1.732  3.27 m / s2
62. A body of mass 2 kg moves down the quadrant of a
v 2  u 2  2as circle of radius 4 m. The velocity on reaching the
v 2  (10) 2  2  3.27  10  165.4 lowest point is 8 m/s. What is work done against
v  13 m / s(approx) friction?

59. A coconut of mass 1.0 kg falls to earth from a height

of 10 m. The kinetic energy of the coconut, when it is
4 m above ground is:
(a) 0.588 joule (b) 58.8 joule
(c) 5.88 joule (d) 588 joule
Ans. (b) (a) 14.4 J (b) 28.8 J
Sol. Height of fall  10  4  6m (c) 64 J (d) Zero
v 2  u 2  2 as  v 2  0  2  g  H Ans. (a)
Sol. Initial velocity = 0
v  2 gH  2  9.8  6
By WNC  ΔK  ΔU
1 2 1
KE  mv   1 (2  9.8  6)
2 2
 9.8  6  58.8 J
60. Calculate the K.E. and P.E. of the ball halfway up,
when a ball of mass 0.1 kg is thrown vertically
–1
upwards with an initial speed of 20 m/s .
(a) 10 J, 20 J (b) 10 J, 10 J
(c) 15 J, 8 J (d) 8 J, 16 J
Ans. (b)
1 1 
Sol. m  0.1 kg v 0  20 m / s WNC   mv2 - mu 2    U f  U i 
2 2 
By mechanical energy conservation
1 
1 1 WNC    2  82  0   (2  9.8  4)
 m v 2  m(0) 2  mgH (max height)  2 
2 2
Initial KE. = potential Energy  64  78.4  14.4 J
1 1 63. The KE of a 500-gram stone is 100 J. Against a force
KE    (20) 2  20 J
2 10 of 50 N, how long will it travel?
1 (a) 0.2 s (b) 0.1 s
KE of half height   initial  10J (c) 0.3 s (d) 0.4 s
2
Ans. (a)
P.E.  20  10  10 J
61. If a body of mass 3 kg is dropped from top of a tower Sol. F  50 N m  0.5 kg v 0  ? K  100 J
of height 250 m, then its kinetic energy after 3 sec. 1
100   0.5  (v) 2  v  20 m / s
will be 2
(a) 1126 J (b) 1048 J 50
a  100 m / s 2  v  u  at
(c) 735 J (d) 1296.5 J 0.5
Ans. (d) 0  20  (100)  t
Sol. m  3kg
20
Under free fall velocity after 3 seconds. t   0.2sec
100
v = 0 + gt = 9.8  3=29.4m/s 64. If water falls from a dam into a turbine wheel 19.6 m
WORK, POWER AND ENERGY 13

below, then velocity of water at turbine, is Find the work performed by this force:
(Take g = 9.8 m/s2)
(a) 9.8 m/s (b) 19.6 m/s
(c) 39.2 m/s (d) 98.0 m/s
F h
Ans. (b)
Sol. By mechanical Energy Conservation law, m
Loss of potential energy = gain in KE
1 (a) mgl (b) – mgl
mgh  mv 2  v  2gh
2 (c) mgh (d) zero
v  2  9.8  19.6  19.6 m / s Ans. (c)
65. Three particles A, B and C are projected from the top Sol. Friction is absent. Therefore, work will be required to
of a tower with the same speed. A is thrown straight change its gravitational potential energy change
upwards B straight down and C horizontally. They hit (slowly moved).
the ground with speeds vA, vB and vC, then which of W = mgh
the following is correct:
(a) vA = vB > vC (b) vA = vB = vC
(c) vA > vB = vC (d) vB > vC > vA
Ans. (b)
Sol. By mechanical energy conservation law, all bodies 68. A particle is released from the top of two inclined
lose same P.E. So, all will have same K.E. rough surfaces of height ‘h’ each. The angle of
Hence the velocities, inclination of the two planes are 30° and 60°
v A  v B  vc respectively. All other factors (e.g. coefficient of
66. A pendulum of length 2 m left at A. When it reaches friction, mass of block etc.) are same in both the
B, it loses 10% of its total energy due to air cases. Let K1 and K2 be kinetic energies of the
resistance. The velocity at B is: particle at the bottom of the plane in two cases. Then
(a) K1 = K2 (b) K1 > K2
A (c) K1 < K2 (d) data insufficient
Ans. (c)
Sol. K.E = mgh - Wf (Work done against friction)
B
(a) 6 m/s (b) 1 m/s K1 = mgh - (m mg cos θ) s
(c) 2 m/s (d) 8 m/s h
K1  mgh-μmg cosθ 
Ans. (a) sinθ
Sol. Length = 2m K1  mgh  μmghcotθ
10 % of potential equation is lost and of potential
K1 = mgh -μ mghcot30
Energy is converted into K.E.
 mgh  μmgh 3
μmgh
K 2 = mgh   K 2  K1
3
69. A particle is released from a height H. At certain
height its kinetic energy is two times its potential
energy. Height and speed of particle at that instant are
90 9 H 2 gH H gH
K.E. at B   ( mg  h)   m  9.8  2
100 10 (a) , (b) , 2
3 3 3 3
1 9
 m v 2   m  9.8  2 2H 2 gH H
2 10 (c) , (d) , 2 gH
3 3 3
 v  6m / s Ans. (b)
67. A body of mass m was slowly pulled up the hill by a Sol. Total mechanical = mg H
force F which at each point was directed along the PE 1
tangent to the trajectory. All surfaces are smooth. 
KE 2
WORK, POWER AND ENERGY 14

KE + PE = mg H
2PE + PE = mgH
mgH
PE 
3
(b)
H
 height from the ground at this instant is
3
2
 KE   mgH  
3
1 2 2 (c)
mv  mgH
2 3
gH
v2
3
70. A body is falling with velocity 1 m/s at a height 3 m
(d)
from the ground. The speed at height 2 m from the
ground will be:
(a) 4.54 m/s (b) 1 m/s
(c) 6 m/s (d) 5.32 m/s
Ans. (a)
Sol. g  1 m / s Ans. (c)
Sol. As slope of graph is positive and constant up to a
h = 3m
certain distance and then it becomes zero. So from
v 2  u 2  2 g  S f  Si 
dU
F  , up to distance a, force (F) is constant and
v 2  1  2   g   2  3 
2
dx
becomes zero afterwards
v 2  1  2 g 1
73. The diagrams represent the potential energy (U) of a
v 2  19.6  1  20.6 function of the inter-atomic distance r. Which
v  20.6  4.54 m / s diagram corresponds to stable molecules found in
nature.
71. If v be the instantaneous velocity of the body dropped
(a)
from the top of a tower, when it is located at height h,
then which of the following remains constant?
v2
(a) gh + v2 (b) gh +
2
v2
(c) gh – (d) gh – v2
2
Ans. (b) (b)
2
v
Sol. gh + Since, Total Energy per unit mass is always
2
constant.
Potential Energy Graphs
72. The potential energy of a particle is represented in the (c)
figure. The force acting on the system will be
represented by

(a)
(d)
WORK, POWER AND ENERGY 15

m v2 
P  gh  
t  2 
2 v2 
 200   9.8  10  
1 2 
v2
100  98   v  2m / s
2
Ans. (a) 76. A machine gun fires 360 bullets per minute, with a
Sol. When distance between atoms is large then velocity of 600 m/s. If the power of the gun is 5.4
interatomic force is very weak. When they come kW, mass of each bullet is (assume 100% efficiency)
closer, force of attraction increases and at a particular (a) 5 kg (b) 0.5 kg
distance force becomes zero. When they are further (c) 5 g (d) 0.5 g
brought closer, force becomes repulsive in nature. Ans. (c)
Sol. N = 360 t = 60 sec
V  600 m / s P  5400watt
1
mV 2  N 1 m  (600)2  360
P 2  5400  
t 2 60
1
74. The potential energy of a particle varies with distance m kg  5g
200
x as shown in the graph.

77. A train of mass 100 ton is moving up an incline of 1

in 100 at a constant speed of 36 km ph. If the friction
per ton is 100 N, then power of the engine is
(a) 198 kW (b) 96 kW
(c) 298 kW (d) 398 kW
Ans. (a)
Sol. v  36 km / hr
The force acting on the particle is zero at
5
(a) C (b) B  36   10 m / s
18
(c) B and C (d) A and D
Ans. (c) m  100  103 kg
dU F  (mg sin θ  f )
Sol. F   it is clear that slope of U-x curve is zero at
dx P  F  v  (mg sin θ  f )  v,f  100  100
point B and C. so force is zero at point B and C. Since, it is given that friction per ton is 100N.
Power = 198 kW
78. The power of a water pump is 2 kW. If g = 10 m/s2,
75. A pump of 200 W power is lifting 2 kg water from an the amount of water it can raise in one minute to a
average depth of 10 m per second. Velocity of water
height of 10 m is
delivered by the pump is (g = 9.8 m/s2) (a) 2000 litre (b) 1000 litre
(a) 3 m/s (b) 2 m/s
(c) 100 litre (d) 1200 litre
(c) 4 m/s (d) 1 m/s
Ans. (d)
Ans. (b)
Sol. P  2000watt, h  10 m
Sol. P  200 W m  2 kg, h  10 m
W mgh
1 P  , t  60sec, m  ?
mgh  m v 2 t t
W 2
P  m  10  10
t t 2000 
60
m  1200 kg  1200 litre
79. A man is riding on a cycle with velocity 7.2 km/hr up
WORK, POWER AND ENERGY 16

a hill having a slope 1 in 20. Total mass of the man

and cycle is 100 kg. The power of man is:
(a) 98 W (b) 49 W
(c) 196 W (d) 147 W
Ans. (a)
Sol. v  7.2 km / hr
1
sin θ 
20

m  100 kg
1 5
P  100  9.8   7.2   98 W
20 18

80. Power applied to a particle varies with time as

P = (3t2 – 2t + 1) W, where t is in second. Find the
change in its kinetic energy between time t = 2s and
t = 4s.
(a) 32 J (b) 46 J
(c) 61 J (d) 102 J
Ans. (b)
Sol. P  3t 2  2t  1
K  W   Pdt

ΔK   Pdt

 3t  2t  1dt
t4
 2
t 2

4
 t3  2 4
K  3    t 2   [t ]42
3
 2 2 2

 56  12  2  46 J