Work, Energy and Conservation

1
WORK, POWER & ENERGY
Kinetic and potential energy; Work and Power, Conservation of mechanical
energy, work energy principle.
WORK : .
When a force is applied at a point and the point gets some displacement along the direction of force
then work is said to be done by the force. The work done, W by a constant force

F when its point of
application undergoes a displacement

S is measured as
W = F

. S

= |

F | |

S | cos
Where  is the angle between

F and

S . Work is a scalar quantity and its SI unit is N-m or joule (J).
Only the component (Fcos) of the force F which is along the displacement contributes to the work
done. If

F = F x

i + F y

j + F z

k and

S = x i + y j + z k then W =

F . S

= Fx x+Fyy+Fz z
Positive and Negative work : The work is said to be positive if the angle  is acute ( < 900) and
negative if the angle  is obtuse ( > 900). If the angle between

F and

S is 900 then work done by the force is
zero.
If the force is variable then the work done by the variable force is given by dW =

F .

dS or
W = 
2 
1
S
S
F.dS
Work depends on frame of reference. With change of frame of reference displacement may change,
so the work done by a force will be different in different frames.
Illustration – 1 :
A particle of mass 2 kg moves under the action of a constant force

F = 5i  2 j  N. If its
displacement is 6 j m. What is the work done by the force

F ?
S Y L L A B U S

2
Solution :
The work done

F .

x
= 5i  2 j  . 6 j = - 12 Joule
A load of mass m = 3000 kg is lifted by a rope with an acceleration a = 2 m/s2. Find the work done
during the first one and a half seconds from the beginning of motion.
Solution :
The height to which the body is lifted during the first 't' second is h =
2
1
at2 tension in the rope T = mg
+ ma
 Work done = T.h = m(g + a) 


 2
2
1
at
= 3000 (10 + 2)   


 2 2 1 5
2
1
x x .
= 81 KJ
WORK DONE BY A SPRING FORCE :
Whenever a spring is stretched or compressed, the spring force always tend to restore it to the
equilibrium position. If x be the displacement of the free end of the spring from its equilibrium position then,
the magnitude of the spring force is FS = - kx
The negative sign indicates that the force is restoring.
The work done by the spring force for a displacement from xi to xf is given by
Ws =  
f
x
xi
kxdx  Ws =  2 2 
2
1
f i  k x  x
WORK DONE BY FRICTION :
Work done by friction may be zero, positive or negative depending upon the situations:
 When a block is pulled by a force F and the block does not move, the work done by friction is zero.
 When a block is pulled on a stationary surface, the work done by the kinetic friction is negative.
 When one block is placed on another block and is pulled by a force then friction force does negative
work on top block and positive work on the lower block
WORK DONE BY GRAVITY :

3
Here the force of gravity is Fg = - mg j and the displacement is given by

S = x i + y j + z k
 Work done by gravity is Wg = g

F .

S = - mg y
y = yf  yi = - h
 Wg = + mgh
If the block moves in the upward direction, then the work done by gravity is negative and is given by
Wg = - mgh.
DEPENDENCE OF WORK ON FRAME OF REFERENCE :
Work depends upon the frame of reference from where it is calculated. As the displacement as well as
force, depends upon the different frames of reference. Therefore, the work also changes.
For example :- If you calculate work from a non inertial frame work due to pseudo force has to be included.
CONSERVATIVE AND NON CONSERVATIVE FORCES :
In Conservative force field the work done by the force is independent on path followed and depends
only on initial and final co-ordinates. Such forces are known as Conservative forces.
Examples are gravitational, electrostatic forces.
If the work done depends on path followed. Such forces are called non- Conservative forces.
Example is frictional force.
A train is moving with a constant speed "v". A box is pushed by a worker applying a force "F" on the
box in the train slowly by distance "d" on the train for time "t". Find the work done by "F" from the train frame
as well as from the ground frame.
Solution :
As the box is seen from the train frame the displacement is only 'd' if the force direction is same as the
direction of motion of the box.
Then the work done = F.d = Fdcos00 = Fd
= Fdcos1800 = -Fd (if the displacement on the train is opposite to 'F')
As the box is seen from ground frame,
the displacement of the box = vt + d (if the displacement is along the direction of motion of the train )
= d - vt (if the displacement is opposite to direction of motion of the train)
then work done = F. (vt + d) = Fvt + Fd OR = F.(d-vt) = Fd – Fvt

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A block is (mass m) placed on the rough surface of a plank (mass m) of coefficient
of friction "" which in turn is placed on a smooth surface. The block is given a velocity v0
with respect to the plank which comes to rest with respect to the plank. Find the
a) The total work done by friction in the plank frame.
b) The work done by friction on the smaller block in the plank frame.
c) Find the final velocity of the plank
Solution :
The acceleration of the plank = Friction force applied by the block on the plank / mass of the plank.
g
m
mg
ap  


(a) Pseudo force acting on the block = g (back wards)
Force of friction is mg ( acting backwards)
From the plank frame time needed to stop the block is given by
0 = V  at 0 a  2g
 t =
g
V
2
0
Velocity of the plank during this time is Vp up apt  
=
2 2
0 0 V
g
V
g 


Displacement of the block = S =
g
V
a
V
V


 





 











8
3
2
2 2
0
2
2 0
0
Work done by friction on the block =  1
8
3 2
0 

  
g
V
F.S.cos mg. =
2
0 8
3
 mv
(b) From the Plank frame
Work done by friction on smaller block = -mgl
2 g
0 V2
0
 

  
2
mV
g
2
0   
 work done by friction from the Plank frame =
2
mV2
0 
(c) Final velocity of the block
= Velocity of the plank =
2
0 V
m
m 0 v
mg
p ma
m

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WORK ENERGY THEOREM :
Now we have to study which physical quantity changes when work is done on a particle. If a
constant force F acts through a displacement x, it does work W = Fx

2 2
f i v  v + 2 ax
 W =
 
2
2 2
f i m v  v
=
2
1
m
2
f v -
2
1
m
2
i v
The quantity k =
2
1
m v2 is a scalar and is called the kinetic energy of the particle. It is the energy
posses by the particle by virtue of its motion.
Thus the equation takes the form W Kf Ki K    
The work done by a force changes the kinetic energy of the particle. This is called the work -Energy
Theorem.
The velocity of an 800 gm object changes from

v0 = 3 i - 4 j to

vf = -6 j + 2 k m/s. What is the
change in K.E of the body?
Solution :
Here m = 800gm = 0.8 kg

vo =  32   4 2 = 5    2 2   6  2

vf = 40
 change in K.E =
2
1
x 0.8
 



 




  2
0
2
vf v = x 0.8 x 40 25 6J
2
1
 
The coefficient of sliding friction between a 900 kg car and pavement is 0.8. If the car is moving at 25
m/s along level pavement, when it begins to skid to a stop, how far will it go before stopping?
Solution :
Here m = 900kg  = 0.8, v = 25 m/s S =?
K.E = work done against friction
2
2
1
mv = F.s =  N.s = mgs
 s =
g
v
2
2
=
 
2 0 8 10
25 2
x . x
~ 39 m

6
An object of mass 10kg falls from rest through a vertical distance of 20m and acquires a velocity of 10
m/s. How much work is done by the push of air on the object? (g = 10 m/s2)
Solution :
Let upward push of air be F
 The resultant downward force = mg - F
As work done = gain in K.E
(mg - F) x S =
2
2
1
mv
 (10 x 10 - F) x 20 =
2
1
x 10 x (10)2  F = 75 N
 Work done by push of air = 75 x 20 = 15 Joule
This work done is negative.
POTENTIAL ENERGY :
Potential energy of any body is the energy possessed by the body by virtue of its position or the state
of deformation. With every potential energy there is an associated conservative force. The potential energy
is measured as the magnitude of work done against the associated conservative force
du = -F.dr
 
for example :
(i) If an object is placed at any point in gravitational field work is to be done against gravitational field
force. The magnitude of this work done against the gravitational force gives the measure of
gravitational potential energy of the body at that position which is U = mgh. Here h is the height of
the object from the reference level.
(ii) The magnitude of work done against the spring force to compress it gives the measure of elastic
potential energy, which is U =
2
1
k x2
(iii) A charged body in any electrostatic field will have electrostatic potential energy. The change in
potential energy of a system associated with conservative internal force as U2-U1= - W=  
2
1
F . dr
CONSERVATION OF MECHANICAL ENERGY :
Change in potential energy U = - WC where WC is the work done by conservative forces. From work
energy theorem
Wnet = k
Where Wnet is the sum of work done by all the forces acting on the mass. If the system is subjected to

7
only conservative forces then Wnet = WC = k
  U = - k  U + k = 0
The above equation tells us that the total change in potential energy plus the total change in kinetic
energy is zero, if only conservative forces are acting on the system.
 (k+U) = 0 or E = 0 where E = k + U
 When only conservative forces act, the change in total mechanical energy of a system is zero.
i.e if only conservative forces perform work on and within a system, the total mechanical
energy of the system is conserved.
 kf + Uf - (ki + Ui) = 0
 kf + Uf = ki + Ui
 E = 0, integrating both sides E = constant.
A projectile is fired from the top of a 40m. high cliff with an initial speed of 50 m/s at an unknown
angle. Find its speed when it hits the ground.
Solution :
Taking ground as the reference level we can conserve the mechanical energy between the points A
and B
  (K + U) = 0  Ki + Ui = Kf + Uf

2
1
mv2 + mgH =
2
1
mv' 2 + 0

2
1
(50)2 + 40 x 10 =
2
1
v' 2
 (1250 + 400) x 2 = v' 2
 v' 2 = 3300
v' ~ 58 m/s
POWER : .
Power is defined as the rate at which work is done. If an amount of work W is done in a time
interval t, then average power is defined to be
Pav =
t
W


The S.I. unit of power is J/S or watt (W). Thus 1 W = 1 J/S
The instantaneous power is the limiting value of Pav as t  0 that is P =
dt
dW
Instantaneous power may also be written as P =
 
 F . v
dt
dW
Since work and energy are closely
v'
H
A 
v
B

8
related, a more general definition of power is the rate of energy transfer from one body to another, or the rate
at which energy is transformed from one form to another, i.e. P =
dt
dE
.
A car of mass 500 kg moving with a speed 36km/hr in a straight road unidirectionally doubles its
speed in 1 minute. Find the average power delivered by the engine.
Solution :
Its initial speed V1 = 10 m/s then V2 = 20 m/s
 k =
2
1
m
2
1
2
2 2
1
v  mv
 Power delivered by the engine
P =
 
t
m v v
t
K





2
1
2
2 2
1
=
 
60
500 20 10
2
1 2 2 x 
= 1250 W.
MOTION IN A VERTICAL CIRCLE :
A particle of mass 'm' is attached to a light and inextensible string.
The other end of the sting is fixed at O and the particle moves in vertical
circle of radius 'r' equal to the length of the string as shown in the fig. At the
point P, net radial force on the particle is T-mg cos.
 T - mg cos =
r
mv2
 T = mg cos +
r
mv2
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  =  for critical situation T = 0 and  = 
 mg =
R
mv min
2

2
min v = gR
 vmin = gR
Now conserving energy between the lowest and the highest point
2
1
mu mv mg R min min 2
2
2 1 2  
O
T
 P
mg cos
mg sin 

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 u gR gR gR min 2   4  5
umin  5gR
If umin  5gR the particle will complete the circle. At u = 5gR ,velocity at highest point is v =
gR and tension in the string is zero.
If u < 5gR , the tension in the string become zero before reaching the highest point and at that point the
particle will leave the circular path. After leaving the circle the particle will follow a parabolic path.
Above conditions are applicable even if a particle moves inside a smooth spherical shell of radius R.
The only difference is that the tension is replaced by the normal reaction N.
A heavy particle hanging from a fixed point by a light inextensible string of length l is projected
horizontally with speed g . Find the speed of the particle and the inclination of the string to the vertical at
the instant of the motion when the tension in the string is equal to the weight of the particle.
Solution :
Let T = mg at an angle  as shown in figure
h = l (1 - cos)
Conserving mechanical energy between
A and B
2
1
mu2 =
2
1
mv2 + mgh
 u2 = v2+ 2gh  v2 = u2 - 2gh …. (i)
T - mg cos =

mv2
 T= mg cos +

mv2
 mg = mg cos  +

mv2
 v2 = g l (1- cos) ……. (ii)
From (i) and (ii) u2 - 2gl (1 - cos) = gl (1 - cos)
 cos =
3
2
  = cos-1 



3
2
putting the value of cos in equation …… (ii)
v2 = gl 

 

3
2
1 =
3
g
 v =
3
g
Equilibrium : As we have studied earlier a body is said to be in translational equilibrium if net force acting
on the body is Zero.

A u  g
mg cos
mg sin 
B
h




10
Fnet = 0
 If the forces are Conservative F = -
dr
dU
 0
dr
dU

 At Equilibrium slope of U and r graph is Zero (or) Potential energy either maximum or
minimum or constant at that position.
At the stable equilibrium position P.E is minimum
At the unstable equilibrium position P.E is maximum
The P.E of a Conservative system is given as U = 10 + (x-2)2. Find the equilibrium position and
discuss type of equilibrium.
Solution :
For Equilibrium F = 0
 F = - 2(x 2) 0
dx
dU
   
 x = 2
and 0
dx
d U
2
2

 it is Stable equilibrium position at x= 2 and P.E at that position is 20 units.
* * * *
WORKED OUT OBJECTIVE PROBLEMS : .
Example : 01
A particle moves with a velocity 5 i - 3 j + 6 k m/s under the influence of a constant force

F =
10 i  10 j  20 k  N. The instantaneous power applied to the particle is
A) 200 J/S B) 40 J/S C) 140 J/S D) 170 J/S
Solution :
P =

F .

V = (5 i - 3 j +6 k ) . (10 i + 10 j +20 k )
= 50 - 30 + 120 = 140 J/S
Example : 02

11
A 15 gm ball is shot from a spring gun whose spring has a force constant of 600 N/m. The spring is
compressed by 5 cm. The greatest possible horizontal range of the ball for this compression is
(g = 10 m/s2)
A) 6.0 m B) 12.0 m C) 10.0 m D) 8.0 m
Solution :
R max =
g
u 2
=  


 






mg
mu
2
2
1 2 =
mg
kx
mg
kx
2
2 2
2
1
  


 










 2 2 kx
2
1
mu
2
1

=
 
m
. x
.
10
0 015 10
600 0 05 2
 .
[Note : The actual value of 'u' will be less than the calculated value as some part of 1/2kx2 is used up in doing work against gravity
when the spring regains its length]
Example : 03
Force acting on a particle is (2 i + 3 j ) N. work done by this force is zero, when a particle is moved on
the line 3y + kx = 5 Here value of k is
A) 3 B) 2 C) 1 D) 4
Solution :
Force is parallel to the line y = 3/2 x + c
and the given line can be written as y =
3
5
3
 x 
k
as the work done is zero
 force is perpendicular to the displacement
 



 



2 3
3 k
= - 1
 k = 2
Example : 04
Power supplied to a particle of mass 2 kg varies with time as p =
2
3 2 t
watt. Here 't' is in second. If
velocity of particle at t = 0 is v = 0. The velocity of particle at time t = 2 second will be
A) 1 m/s B) 4 m/s C) 2 m/s D) 2 2 m/s
Solution :
kf - ki = 
2
0
P dt 
2
1
mv2 = 
2
0
2
3
t2 dt  v2 =
2
0
3
2 






 t
 m = 2 kg  v = 2 m/s

12
Example : 05
A particle of mass 'm' is projected with velocity 'u' at an angle  with horizontal. During the period
when the particle descends from highest point to the position where its velocity vector makes an angle /2
with horizontal, work done by the gravity force is
A) 1/2 mu2 tan2 /2 B) 1/2 mu2 tan2 
C) 1/2 mu2 cos2  tan2 /2 D) 1/2 mu2 cos2 /2 sin2 
Solution :
As horizontal component of velocity does not change v cos /2 = ucos 
v =
2


cos
u cos
Wgravity =  K =
2
1
mv2 -
2
1
m (u cos)2
=
2
1
mu2 cos2  tan2
2

Example : 06
A body of mass 1 kg thrown upwards with a velocity of 10 m/s comes to rest (momentarily) after
moving up 4 m. The work done by air drag in this process is (g = 10 m/s2)
A) 10 J B) - 10 J C) 40 J D) 50 J
Solution :
From work energy theorem Wgr + Wair drag =  k
 - mgh + Wair drag = 0 -
2
1
mu2
 Wair drag = mgh -
2
1
mu2 = (40 - 50) J = - 10 J
Example : 07
The potential energy of particle of mass 'm' is given by U =
2
1
kx2 for x < 0 and U = 0 for x > 0. If total
mechanical energy of the particle is E. Then its speed at x =
k
2E is
A) zero B)
M
2E
C)
m
E
D)
m
E
2
Solution :
Potential energy of particle at x =
k
2E
is zero  K.E = E

/ 2
u cos 
V
u

13

2
1
mv2 = E or v =
m
2E
Example : 08
A block is suspended by an ideal spring of force constant k. If the block is pulled down by applying a
constant Force 'F' and if maximum displacement of block from its initial position of rest is  then
A)
K
F
<  <
K
2F
B)  =
K
2F
C) Work done by force F is equal to F D) Increases in energy stored in spring is
2
1
k2
Solution :
If the mass of the hanging block be 'm' then elongation of spring is
k
mg
.
Due to the applied force the additional stretching is 
 F  + mg =
2
1
K
K
m g
K
mg
2
2 2 2
 



 
=
2
1
K
K
m g
K
mg
K
m g
2
2 2 2
2
2
2 2
  


 

 
  
=
2
1
K 2 + mg   =
K
2F
.
Example : 09
A stone is projected at time t = 0 with a speed V0 and an angle  with the horizontal in a uniform
gravitational field. The rate of work done (P) by the gravitational force plotted against time (t) will be as
A) B) C) D)
Solution :
Rate of work done is the power associated with the force. It means rate of work done by the
gravitational force is the power associated with the gravitational force. Gravitational force acting on the block
is equal to its weight mg which acts vertically downwards.
Velocity of the particle (at time t) has two components,
(i) a horizontal component v cos and
(ii) a vertically upward component (v sin - gt)
P
O t
P
O t
P
O t O t
P

14
Hence, the power associated with her weight mg will be equal to p = m

g .

v = -mg (v sin - gt)
This shows that the curve between power & time will be straight line having positive slope but negative
intercept on Y-axis.
Hence (D) is correct.

15
SINGLE ANSWER TYRE .
LEVEL - I .
1. Two springs A and B(KA = 2KB) are stretched by applying forces of equal magnitudes at the four
ends. If the energy stored in A is E, that in B is
a) E/2 b) 2E c) E d) E/4
2. Two equal masses are attached to the two ends of a spring constant K. The masses are pulled out
symmetrically to stretch the spring by a length x over its natural length. The work done by the spring
on each mass is
a) ½ Kx2 b) -1/2 Kx2 c) ¼ Kx2 d) -1/4 Kx2
3. The negative of the work done by the conservative internal forces on a system equals the change in
a) total energy b) kinetic energy c) potential energy d) none of these
4. The work done by the external forces on a system equals the change in
5. The work done by all the forces (external and internal) on a system equals the change in
6. ________ of a two particle system depends only on the separation between the two particles. The
most appropriate choice for the blank space in the above sentence is
a) kinetic energy b) total mechanical energy
c) potential energy d) total energy
7. A small block of mass ‘m’ is kept on a rough inclined surface of inclination  fixed in an elevator. The
elevator goes up with a uniform velocity ‘v’ and the block does not slide on the wedge. The work
done by the force of friction on the block in time ‘t’ will be
a) zero b) mgvt cos2 c) mgvt sin2 d) mgvt sin2
8. A block of mass ‘m’ slides down a smooth vertical circular track. During the motion, the block is in
a) vertical equilibrium b) horizontal equilibrium
c) radial equilibrium d) none of these
9. A particle is rotated in a vertical circle by connecting it to a string of length ‘l’ and keeping the other
end of the string fixed. The minimum speed of the particle when the string is horizontal for which the
particle will complete the circle is
a) gl b) 2gl c) 3gl d) 5gl
10. Consider two observers moving with respect to each other at a speed v along a straight line. They
observe a block of mass m moving a distance ‘l’ on a rough surface. The following quantities will be
same as observed by the two observers
a) kinetic energy of the block at time t b) work done by friction
c) total work done on the block d) acceleration of the block
11. A particle of mass ‘m’ is attached to a light string of length ‘l’, the other end of which is fixed. Initially
the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to
complete a circle
a) the string becomes slack when the particle reaches its highest point

16
b) the velocity of the particle becomes zero at the highest point
c) the kinetic energy of the ball in initial position was ½ mv2 = mgl
d) the particle again passes through the initial position
12. The kinetic energy of a particle continuously increases with time
a) the resultant force on the particle must be parallel to the velocity at all instants
b) the resultant force on the particle must be at an angle less than 900 all the time
c) its height above the ground level must continuously decrease
d) the magnitude of its linear momentum is increasing continuously
13. One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block
placed on a smooth horizontal surface. In a displacement, the work done by the spring is ½ kx2. The
possible cases are
a) the spring was initially compressed by a distance x and was finally in its natural length
b) it was initially stretched by a distance x and finally was in its natural length
c) it was initially in its natural length and finally in a compressed position
d) it was initially in its natural length and finally in a stretched position
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 20J 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 20J in the above is 1s
d) the work done by the force of gravity is –20J in the above 1s
15. A particle of mass 0.25kg moves under the influence of a force F = (2x-1). If the velocity of the particle
at x = 0 is 4m/s. its velocity at x = 2m will be
A) 4 2 m/s B) 2 2 m/s C) 8m/s D) 6m/s
16. Work done to accelerate a car from 10 to 20m/s compared with that required to accelerate it from 0 to
10m/s is
A) twice B) three times C) four times D) same
17. Two springs have their force constant as K1 and K2 (K1 > K2). When they are stretched by the same
force :
A) no work is done in case of both the springs B) equal work is done in case of both the springs
C) more work is done in case of second spring D) more work is done in case of first spring
18. The kinetic energy K of a particle moving in a straight line depends upon the distance s as K = as2
where a is a constant. The force acting on the particle is
A) 2as B) 2mas C) 2a D) 2 as
19. A particle moves in a straight line with a retardation proportional to its displacement. Its loss of
kinetic energy for any displacement x is proportional to
A) x B) x2 C) ln x D) ex
20. Choose the wrong option
A) If conservative forces are doing negative work then potential energy will increase and kinetic
energy will decrease.
B) If kinetic energy is constant it means work done by conservative forces is zero.

17
C) for change in potential energy only conservative forces are responsible, but for change in kinetic
energy other than conservative forces are responsible
D) all of the above are wrong
21. Instantaneous power of a constant force acting on a particle moving in a straight line under the action
of this force :
A) is constant B) increases linearly with time
C) decreases linearly with time D) either increases or decreases linearly with time.
22. Suppose y represents the work done and x the power, then dimensions of
2
2
dx
d y
will be :
A) 1 2 4 M L T     B) 2 3 2 M L T     C) 2 4 4 M L T      D) 3 6 ML T  
23. Choose the correct statement Work done by a variable force
A) Is defined as F .S B) Is independent of path
C) Is always dependent on the initial and final positions
D) None of these
24. Identify the correct statement for a non-conservative force
A) A force which is not conservative is called a non-conservative force
B) The work done by this force depends on the path followed
C) The word done by this force along a closed path is zero
D) The work done by this force is always negative
25. The figure shows a plot of potential energy function, u(x) = kx2 where x is the
displacement and k is a constant. Identify the correct conservative force
function F(x)
26. A plot of velocity versus time is shown in figure. A single force acts on the body.
Find correct statement
A) In moving from C to D, work done by the force on the body is positive
B) In moving from B to C, work done by the force on the body is positive
C) In moving from A to B, the body does work on the system and is negative
D) In moving from O to A, work done by the body and is negative
LEVEL - II .
1. A particle of mass m is moving in a circular path of radius r under the influence of centripetal force F
= C/r2. The total energy of the particle is

18
a)
2r
C
 b)
2r
C
c) C x 2r d) Zero
2. Water from a stream is falling on the blades of a turbine at the rate of 100kg/sec. If the height of the
stream is 100m then the power delivered to the turbine is
a) 100 kw b) 100 w c) 10 kw d) 1 kw
3. A body is being moved along a straight line by a machine delivering a constant power. The distance
covered by the body in time t is proportional to
a) 1 b) t3/2 c) t3/4 d) t2
4. A ball is dropped from a height of 10m. If 40% of its energy is lost on collision with the earth then
after collision the ball will rebound to a height of
a) 10m b) 8m c) 4m d) 6m
5. A particle moves under the influence of a force F = CX from X = 0 to X = X1. The work done in this
process will be
a)
2
CX2
1 b) 2
CX1 c) 3
CX1 d) 0
6. A uniform chain of mass M and length L lies on a horizontal table such that one third of its length
hangs from the edge of the table. The work done is pulling the hanging part on the table will be
a)
3
MgL
b) MgL c)
9
MgL
d)
18
MgL
7. An electric motor produces a tension of 4500N in a load lifting cable and rolls it at the rate of 2m/s.
The power of the motor is
a) 9 kw b) 15 kw c) 225 kw d) 9 x 103 HP
8. The relation between time and displacement of a particle moving under the influence of a force F is t
= x +3 where x is in meter and t in second. The displacement of the particle when its velocity is zero
will be
a) 1 m b) 0 m c) 3 m d) 2 m
9. A moving particle of mass m collides head on with another stationary particle of mass 2m. What
fraction of its initial kinetic energy will m lose after the collision?
a) 9/8 b) 8/9 c) 19/18 d) 18/19
10. Two particles each of mass m and traveling with velocities u1 and u2 collide perfectly inelastically.
The loss of energy will be
a) ½ m(u1 – u2)2 b) ¼ m(u1 – u2)2 c) m(u1 – u2)2 d) 2m(u1 – u2)2
11. A liquid in a U tube is changed from position (a) to position (b) with
the help of a pump. The density of liquid is d and area of cross
section of the tube is a. The work done in pumping the liquid will
be
a) dgha b) dgh2a
c) 2gdh2a d) 4dgh2a
12. A man pulls a bucket full of water from a h metre deep well. If the mass of the rope is m and mass of
bucket full of water is M, then the work done by the man is

19
a) m gh
2
M




 b) gh
2
M m



 
c) gh
2
m
M 



 d) (M + m) gh
13. The force-displacement curve for a body moving on a smooth surface
under the influence of force F acting along the direction of displacement s
has been shown in fig. If the initial kinetic energy of the body is 2.5J. its
kinetic energy at s = 6m is
A) 7J B) 4.5J C) 2.25J D) 9J
14. A bullet, moving with a speed of 150m/s, strikes a wooden plank. After passing through the plank its
speed becomes 125m/s. Another bullet of the same mass and size strikes the plank with a speed of
90m/s. It speed after passing through the plank would be
A) 25m/s B) 35m/s C) 50m/s D) 70m/s
15. A man of mass 60kg climbs a staircase inclined at 450 and having 10steps. Each step is 20cm high. He
takes 2 seconds for the first five steps and 3 seconds for the remaining five steps. The average power
of the man is
A) 245W B) 245 2 W C) 235 2 W D) 235W
16. The potential energy of a particle moving in x-y plane is given by U = x2 + 2y. The force acting on the
particle at (2, 1) is
A) 6N B) 20 N C) 12 N D) 0
17. Water is flowing in a river at 20m/s. The river is 50m wide and has an average depth of 5m. The
power available from the current in the river is
A) 0.5MW B) 1.0MW C) 1.5MW D) 2.0MW
18. A 5kg brick of dimensions 20cm x 10cm x 8cm is lying on the largest base. It is now made to stand
with length vertical. If g = 10m/s2, then the amount of work done is
A) 3J B) 5J C) 7J D) 9J
19. A body of mass m was slowly pulled up the hill by a force F which at each
point was directed along the tangent of the trajectory. All surfaces are
smooth. Find the work performed by this force
A) mg  B) -mg
C) mgh D) zero
20. A rope ladder with a length l carrying a man of mass m at its end, is attached to the basket of a
balloon of mass M. The entire system is in equilibrium in air. As the man climbs up the ladder into the
balloon, the balloon descends by height h. Then the potential energy of man
A) increases by mg l B) increases by mg (l -h)
C) increases by mgh D) increases by mg (2 l -h)
21. Two springs s1 and s2 have negligible masses and the spring constant of s1 is one-third that
of s2. When a block is hung from the springs as shown, the springs came to the equilibrium
again. The ratio of work done is stretching s1 to s2 is
A) 1/9
B) 1/3

20
C) 1
D) 3
22. A light spring of length l and spring constant 'k' it is placed vertically. A small ball of mass m falls
from a height h as measured from the bottom of the spring. The ball attaining to maximum velocity
when the height of the ball from the bottom of the spring is
A) mg/k B) l-mg/k C) l + mg/k D) l - k/mg
23. A block of mass 1kg is permanently attached with a spring of spring constant k = 100N/m. The
spring is compressed 0.20m and placed on a horizontal smooth surface. When the block is released, it
moves to a point 0.4m beyond the point when the spring is at its natural length. The work done by the
spring in changing from compressed state to the stretched state is
A) 10J B) -6J C) -8J D) 18J
24. A chain of length l and mass m lies on the surface of a smooth sphere of radius R with one end tied on
the top of the sphere. If  = R/2, then the potential energy of the chain with reference level at the
centre of sphere is give by
A) m R g B) 2m R g C) 2/ m R g D) 1/ m R g
25. If the force acting on a particle is given by F = 2i + xyj + xz2k, how much work is done when the
particle moves parallel to Z-axis from the point (2, 3, 1) to (2, 3, 4) ?
A) 42J B) 48J C) 84J D) 36J
26. A uniform chain of length '  ' and mass m is placed on a smooth table with one-fourth of its length
hanging over the edge. The work that has to be done to pull the whole chain back onto the table is
A)
4
1
mgl B)
8
1
mgl C)
16
1
mgl D)
32
1
mgl
27. A spring, which is initially in its unstretched condition, is first stretched by a length x and then again
by a further length x. The work done in the first case is W1 and in the second case is W2
A) W2 = W1 B) W2 = 2W1 C) W2 = 3W1 D) W2 = 4W1
28. A particle of mass m is fixed to one end of a light rigid rod of length '  ' 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) gl C) 1.5gl D) 2gl
29. A particle is moving in a conservative force field from point A to B. UA and UB are the potential
energies of the particle at points A and B and Wc is the work done in the process of taking the particle
from A to B.
A) Wc = UB - UA B) Wc = UA - UB C) UA > UB D) UB > UA
30. A force is given by Mv2/r when the mass moves with speed v in a circle of radius r. The work done by
this force in moving the body over upper half circle along the circumference is
A) zero B)  C) Mv2 D) Mv2 /2
31. A moving railway compartment has a spring of constant 'k' fixed to its front wall. A boy in the
compartment stretches this spring by distance x and in the mean time the compartment moves by a
distance s. The work done by boy w.r.t earth is

21
A) 2 kx
2
1
B)
2
1
(kx) (s+x) C) kxs
2
1
D) kxs x s
2
1
 
32. Force acting on a block moving along x-axis is given by : F = N
x 2
4
2  


 




The block is displaced from x=-2m to x=+4m, the work done will be
A) positive B) negative
C) zero D) may be positive or negative
33. The system is released from rest with both the springs in unstretched positions. Mass of each block is
1 kg and force constant of each springs is 10 N/m. Extension of horizontal spring in equilibrium is:
A) 0.2m B) 0.4m C) 0.6m D) 0.8m
34. In a projectile motion, if we plot a graph between power of the force acting on the projectile and time
then it would be like :
A) B)
C)
D)
35. A golfer rolls a small ball with speed u along the floor from point A. If x =
3R, determine the required speed u so that the ball returns to A after rolling
on the circular surface in the vertical plane from B to C and becoming a
projectile at C. (Neglect friction)
A) gR
5
2
B) gR
2
5
C) gR
7
5
D)
none of these
LEVEL - III .
1. A block m is pulled by applying a force F as shown in fig. If the block has moved up
through a distance 'h', the work done by the force F is
A) 0 2) Fh C) 2Fh D)
2
1
Fh
2. 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
A)  = p/(2mg x) B)  = p2 / (2mg s) C)  = p2 / (2g m2s) D)  = p2 (2g m2s2)
3. A body of mass m moves from rest, along a straight line, by an engine delivering constant power P.
the velocity of the body after time t will be
A)
m
2Pt
B)
m
2Pt
C)
2m
Pt
D)
2m
Pt

22
4. The spring shown in fig has a force constant k and the mass of block is m. Initially, the spring is
unstretched when the block is released. The maximum elongation of the spring on the releasing the
mass will be
A)
k
mg
B)
2
1
k
mg
C) 2
k
mg
D) 4
k
mg
5. A skier starts from rest at point A and slides down the hill, without
turning or braking. The friction coefficient is . When he stops at point B,
his horizontal displacement is S. The height difference h between points
A and B is
A) h = S/ B) h = S
C) h = S2 D) h = S/2
6. A block of mass m starts at rest at height h on a frictionless inclined
plane. The block slides down the plane travels a total distance d
across a rough horizontal surface with coefficient of kinetic friction
k and compresses a spring with force constant k, a distance x before
momentarily coming to rest. The spring then extends and the block travels back across the rough
surface, sliding up the plane. The maximum height h' that the block reaches on its return is
A) h' = h - 2d B) h' = h - 2d -
2
1
kx2 C) h' = h - 2d + kx2 D) h' = h - 2d - kx2
7. A chain of length 3  and mass m lies at the top of smooth prism such that its
length  is one side and 2  is on the other side of the vertex. The angle of
prism is 1200 and the prism is not free to move. If the chain is released. What
will be its velocity when the right end of the chain is just crossing the
top-most point?
A) 2g l B) g l
3
2
C) g l
3
1
D) g l
2
1
8. If a constant power P is applied in a vehicle, then its acceleration increases with time according to the
relation
A) a = t
2m
P
 


 


B) a = 3/ 2 t
2m
P
 


  

C) a = 1/ t
2m
P
 


 


D) a =
2mt
P
9. A body of mass m slides downward along a plane inclined at an angle . The coefficient of friction is
. The rate at which kinetic energy plus gravitational potential energy dissipates expressed as a
function of time is
A) mtg2 cos  B) mtg2 cos  (sin  -  cos )
C) mtg2 sin  D) mtg2 sin  (sin  -  cos )
10. The potential energy for a force field F is given by U(x, y) = sin (x + y). The force acting on the
particle of mass m at (0, /4) is
A) 1 B) 2 C) 1/ 2 D) 0
11. A uniform rope of length '  ' and mass m hangs over a horizontal table with two third part on the

23
table. The coefficient of friction between the table and the chain is . The work done by the friction
during the period the chain slips completely off the table is
A) 2/9 mgl B) 2/3 mgl C) 1/3 mgl D) 1/9 mgl
12. A particle is moving in a force field given by potential U = - (x + y + z) from the point (1, 1, 1) to (2, 3,
4). The work done in the process is
A) 3 B) 1.5 C) 6 D) 12
13. A compressed spring of spring constant k releases a ball of mass m. If the
height of spring is h and the spring is compressed through a distance x, the
horizontal distance covered by ball to reach ground is
A) x
mg
kh
B)
mg
xkh
C) x
mg
2kh
D)
x kh
mg
14. A block of mass m = 2kg is moving with velocity vo towards a
massless unstretched spring of force constant K = 10 N/m.
Coefficient of friction between the block and the ground is  = 1/5.
Find maximum value of vo so that after pressing the spring the block
does not return back but stops there permanently.
A) 6 m/s B) 12m/s C) 8m/s D) 10m/s
15. Potential energy of a particle moving along x-axis under the action of only conservative forces is
given as : U = 10 + 4 sin(4x). Here U is in Joule and x in meters. Total mechanical energy of the
particle is 16J. Choose the correct option.
A) At x = 1.25m, particle is at equilibrium position. C) both A and B are correct
B) Maximum kinetic energy of the particle is 20J D) both A and B are wrong.
16. A system shown in figure is released from rest. Pulley and spring is massless and
friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the
contact with ground is (Take force constant of spring k = 40 N/m and g = 10 m/s2)
A) 2 m/s` B) 2 2 m/s
C) 2m/s D) 4 2 m/s
17. Two blocks of masses m1 = 1 kg and m2 = 2 kg are connected by a
non-deformed light spring. They are lying on a rough horizontal
surface. The coefficient of friction between the blocks and the surface is
0.4 what minimum constant force F has to be applied in horizontal direction to the block of mass m1
in order to shift the other block? (g = 10 m/s2)
A) 8 N B) 15 N C) 10 N D) 25 N
18. A block of mass m is attached with a massless spring of force constant k. The
block is placed over a rough inclined surface for which the coefficient of friction
is  = ¾. The minimum value of M required to move the block up the plane is
(Neglect mass of string and pulley and friction in pulley).

24
A) 3/5m B) 4/5m C) 6/5m D) 3/2m
MULTIPLE ANSWER TYPE QUESTIONS .
1. The potential energy U for a force field F 
is such that U = - kxy, where k is a constant
A) j ˆ kx i ˆ
F ky 

B) j ˆ ky i ˆ
F kx 

C) The force F 
is a conservative force D) The force F 
is a non-conservative force
2. Two blocks A and B each of mass m are connected by a light
spring of natural length L and spring constant k. The blocks are
initially resting on a smooth horizontal floor with the spring at
its natural length, as shown. A third identical block C, also of
mass m, moves on the floor with a speed v along the line joining A to B and collides with A. Then
A) the kinetic energy of the A-B system at maximum compression of the spring is zero
B) the kinetic energy of the A-B system at maximum compression of the spring is








4
mv2
C) the maximum compression of the spring is
 


 


k
m
v
D) the maximum compression of the spring is
 


 


2 k
m
v
3. The kinetic energy of a body moving along a straight line varies directly with time t. If the velocity of
the body is v at time t, then the force F acting on the body is such that
A) F  t1/2 B) F  t-1/2 C) F  v D) F  v-1
4. A car of mass m is moving on a level road at a constant speed vmax while facing a resistive force R. If
the car slows down to vmax/3, then assuming the engine to be working at the same power, what force
F is developing and what is the acceleration a of the car ?
A) F = 3R B) F = 2R C) a = 3P/m vmax D) a = 2 P/m vmax
5. The potential energy of a particle of mass 1 kg moving in xy plane is given by U = 10 - 4x - 3y
The particle is at rest at (2, 1) at t = 0. Then
A) velocity of particle at t = 1 is 5 m/s B) the particle is at (10m, 7m) at t = 2s
C) work done during t = 1s to t = 2s is 75J D) the magnitude of force acting on particle is 5N
6. A block of mass m is gently placed on a vertical spring of stiffness k. Choose the correct statement
related to the mechanical energy E of the system.
A) It remains constant B) It decreases C) It increases D) Nothing can be said
7. A spring of stiffness k is pulled by two forces FA and FB as shown in the figure
so that the spring remains in equilibrium. Identify the correct statement (s) :
A) The work done by each force contributes into the increase in potential
energy of the spring
B) The force undergoing larger displacement does positive work and the force undergoing smaller
displacement does negative work

25
C) Both the forces perform positive work
D) The net work done is equal to the increase in potential energy
8. A particle of mass m is released from a height H on a smooth curved surface
which ends into a vertical loop of radius R, as shown in figure.
If  is the instantaneous angle which the line joining the particle and the centre
of the loop makes with the vertical, the identify the correct statement(s) related to
the normal reaction N between the block and the surface.
A) The maximum value N occurs at  = 0 B) The minimum value of N occurs at N = 
C) The value of N becomes negative for /2 <  <
2
3
D) The value of N becomes zero only when  > /2
9. An engine is pulling a train of mass m on a level track at a uniform speed v. The resistive force offered
per unit mass is f
A) Power produced by the engine is mfv
B) The extra power developed by the engine to maintain a speed v up a gradient of h in s is
s
mghv
C) The frictional force exerting on the train is mf on the level track
D) None of above is correct
10. A particle of mass 5 kg moving in the x-y plane has its potential energy given by U = (-7x + 24y) J,
where x and y are in metre. The particle is initially at origin and has a velocity 1 ms ) j ˆ 2 . 4 i ˆ
u (14.4   

A) The particle has a speed of 25 ms-1 at t = 4 s B) The particle has an acceleration of 5 ms-2
C) The acceleration of the particle is perpendicular to its initial velocity
D) None of the above is correct
* * * * *

26
MULTIPLE MATCHING TYPE QUESTIONS : .
1.
List - I List - II
a) Area under F - S e) Change in KE
b) Work energy theorem f) negative of work done to gravitational force
c) change in PE g) work done by F
d) conservative force h)   dx . F 
, where F is conservative force
i) gravitational force
2.
List - I List - II
a) KE e) depends on frame of reference
b) work done f) defined for conservative force only
c) PE g) independent on frame of reference
d) spring PE h) same for either compression or elongation for
same distance
3.
List - I List - II
a) stable equilibrium e) PE in Max
b) unstable equilibrium f) Fnet = 0
c) 0
dx
dF

g) PE is Min
d) 0
dx
dF

h) slope of F-x graph is +ve
4.
List - I List - II
a) work done by frictional force e) indepent of path
b) work done by electrostatic force f) non-conservative
c) work done by gravitational force for
closed loop
g) depends on path
d) for slowly moving body, wc + wn.c equal
to
h) define PE
i) zero
* * * * *

27
ASSERTION AND REASON : .
Read the assertion and reason carefully to mark the correct option out of the options given below: (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If assertion is false but reason is true.
1. Assertion : A person working on a horizontal road with a load on his head does no work.
Reason : No work is said to be done, if directions of force and displacement of load are perpendicular to each other.
2. Assertion : Work done by friction on a body sliding down an inclined plane is positive.
Reason : Work done is greater than zero, if angle between force and displacement is acute or both are in same direction.
3. Assertion : When a gas is allowed to expand, work done by gas is positive.
Reason : Force due to gaseous pressure and displacement (of piston) are in the same direction.
4. Assertion : The instantaneous power of an agent is measured as the dot product of instantaneous velocity and the force acting on it at that instant.
Reason : The unit of instantaneous power is watt.
5. Assertion : The change in kinetic energy of a particle is equal to the work done on it by the net force.
Reason : Change in kinetic energy of particle is equal to the work done only in case of a system of one particle.
6. Assertion : A spring has potential energy, both when it is compressed or stretched.
Reason : In compressing or stretching, work is done on the spring against the restoring force.
7. Assertion : Comets move around the sun in elliptical orbits. The gravitational force on the comet due to sun is not normal to the comet’s velocity but the work done by the gravitational force over every complete orbit of the comet is zero.
Reason : Gravitational force is a non conservative force.
8. Assertion : The rate of change of total momentum of a many particle system is proportional to the sum of the internal forces of the system.
Reason : Internal forces can change the kinetic energy but not the momentum of the system.
9. Assertion : Water at the foot of the water fall is always at different temperature from that at the top.
Reason : The potential energy of water at the top is converted into heat energy during falling.
10. Assertion : The power of a pump which raises 100 kg of water in 10sec to a height of 100 m is 10 KW.
Reason : The practical unit of power is horse power.
11. Assertion : According to law of conservation of mechanical energy change in potential energy is equal and opposite to the change in kinetic energy.
Reason : Mechanical energy is not a conserved quantity.
12. Assertion : When the force retards the motion of a body, the work done is zero.
Reason : Work done depends on angle between force and displacement.
13. Assertion : Power developed in circular motion is always zero.

28
Reason : Work done in case of circular motion is zero.
14. Assertion : A kinetic energy of a body is quadrupled, when its velocity is doubled.
Reason : Kinetic energy is proportional to square of velocity.
15. Assertion : Work done by or against gravitational force in moving a body from one point to another is
independent of the actual path followed between the two points.
Reason : Gravitational forces are conservative forces.
16. Assertion : Graph between potential energy of a spring versus the extension or compression of the
spring is a straight line.
Reason : Potential energy of a stretched or compressed spring, proportional to square of extension or
compression.
17. Assertion : Heavy water is used as moderator in nuclear reactor.
Reason : Water cool down the fast neutron.
18. Assertion : If two protons are brought near one another, the potential energy of the system will
increase.
Reason : The charge on the proton is C 19 1.6 10   .
19. Assertion : In case of bullet fired from gun, the ratio of kinetic energy of gun and bullet is equal to
ratio of mass of bullet and gun.
Reason : In firing, momentum is conserved.
20. Assertion : Power of machine gun is determined by both, the number of bullet fired per second and
kinetic energy of bullets.
Reason : Power of any machine is defined as work done (by it) per unit time.
21. Assertion : Mountain roads rarely go straight up the slope.
Reason : Slope of mountains are large therefore more chances of vehicle to slip from roads.
* * * * *

29
IIT TRACK QUESTIONS : .
1. A ball hits the floor and rebounds after inelastic collision. In this case [IIT 1986]
(a) The momentum of the ball just after the collision is the same as that just before the collision
(b) The mechanical energy of the ball remains the same in the collision
(c) The total momentum of the ball and the earth is conserved
(d) The total energy of the ball and the earth is conserved
2. 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, the work
required to pull the hanging part on to the table is [IIT 1985]
(a) MgL (b) MgL/3 (c) MgL/9 (d) MgL/18
3. A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to
2  K / r , where K is a constant. The total energy of the particle is [IIT 1977]
(a)
r
K
2
(b)
r
K
2
 (c)
r
K
 (d)
r
K
4. 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 is in seconds. The work done
by the force in the first 6 seconds is [IIT 1979]
(a) 9 J (b) 6 J (c) 0 J (d) 3 J
5. A force F  K(yi  xj) (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 then
parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is [IIT 1998]
(a) 2  2Ka (b) 2 2Ka (c) 2  Ka (d) 2 Ka
6. If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an
object of mass m raised from the surface of earth to a height equal to the radius of the earth R, is
[IIT 1983]
(a) mgR
2
1
(b) 2 mgR (c) mgR (d) mgR
4
1
7. A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force,
then [IIT 1973]
(a) Lorry will come to rest in a shorter distance (b) Car will come to rest in a shorter distance
(c) Both come to rest in a same distance (d) None of the above
8. A particle free to move along the x-axis has potential energy given by ( ) [1 exp( ) ] 2 U x  k  x for
  x  , where k is a positive constant of appropriate dimensions. Then [IIT-JEE 1999]
(a) At point away from the origin, the particle is in unstable equilibrium
(b) For any finite non-zero value of x, there is a force directed away from the origin
(c) If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin
(d) For small displacements from x = 0, the motion is simple harmonic
9. 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 [IIT 1984]

30
(a) 1 / 2 t (b) 3 / 4 t (c) 3 / 2 t (d) 2 t
10. A shell is fired from a cannon with velocity v m/sec at an angle  with the horizontal direction. At the
highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path
to the cannon and the speed in m/sec of the other piece immediately after the explosion is [IIT 1984]
(a) 3v cos (b) 2v cos (c) cos
2
3
v (d) cos
2
3
v
11. Two particles of masses 1 m and 2 m in projectile motion have velocities 1 v

and 2 v

respectively at
time t = 0. They collide at time 0 t . Their velocities become ' 1 v

and ' 2 v

at time 0 2t while still moving
in air. The value of | ( ' ') ( ) 1 1 2 2 1 1 2 2 m v  m v  m v  m v | is [IIT-JEE Screening 2001]
(a) Zero (b) 1 2 0 (m m )gt (c) 1 2 0 2(m m )gt (d) 1 2 0 ( )
2
1
m m gt
12. A particle is placed at the origin and a force F  kx is acting on it (where k is positive constant). If
U(0)  0 , the graph of U(x) versus x will be (where U is the potential energy function) [IIT-JEE
(Screening) 2004]
(a) (b) (c) (d)
* * * * *
x
U(x)
x
U(x)
x
U(x)
x
U(x)

31
SINGLE ANSWER TYRE .
LEVEL - I .
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
D
C
A
B
C
C
D
C
D
AD
BD
AB
B
A
16
17
18
19
20
21
22
23
24
25
26
B
C
A
B
D
B
A
C
B
B
A
LEVEL - II .
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
A
A
B
D
A
D
A
B
B
B
B
C
A
B
D
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
B
B
A
C
B
D
B
B
C
A
D
C
A
B
A
31
32
33
34
35
A
B
B
B
B
LEVEL - III .
1
2
3
4
5
6
7
8
9
10
11
C
C
A
C
B
A
B
D
B
A
A
12
13
14
15
16
17
18
C
C
D
A
B
A
A
MULTIPLE ANSWER TYPE QUESTIONS .
1
2
3
4
5
6
7
8
9
10
AC
BD
BD
AD
ABD
B
ACD
ABD
ABC
ABC
MULTIPLE MATCHING TYPE QUESTIONS .
1
2
3
4
a-eg, b-e, c-fh, d-i
a-e, b-e, c-ef, d-gh
a-fg, b-efh, c-g, d-eh
a-fg, b-eh, c-i, d-i
ASSERTION AND REASON .
1
2
3
4
5
6
7
8
9
10
11
A
D
A
B
C
A
C
D
A
B
C
12
13
14
15
16
17
18
19
20
21
D
D
A
A
D
C
B
A
A
A
IIT TRACK QUESTIONS : .
1
2
3
4
5
6
7
8
9
10
11
12
C
D
B
C
C
A
C
D
C
A
C
A

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