Work Energy Power Module for IIT Main

WORK, ENERGY & POWER Syllabus : 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 gains some energy. Then the work is said to be done by the force. 

The work W done 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 = Fx i + Fy j + Fz 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 S2

W=





 F . dS

S1

Work depends on frame of reference. With change of frame of reference inertial force does not change while 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 ? Solution :  

The work done F . x

1





= 5i  2 j . 6 j = - 12 Joule

Illustration – 2 : A load of mass m = 3000 kg is lifted by a rope with an acceleration a = 2 m/s 2. 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 =

1 2 at tension in the rope T = 2

mg + ma

1   Work done = T.h = m(g +a)  at 2  2 

1



= 3000 (10 + 2)  x 2 x 1.52  2 

= 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 x

Ws = 

f

 kxdx x i

 Ws = 



1 k x f2  x i2 2



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

2

WORK DONE BY GRAVITY :

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 = F g . 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 deferent 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. Again displacement from the inertial frame of reference will be different from ground frame.

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.

3

Illustration – 3 : 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

Illustration – 4 : 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

v m 0 m

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.

ap  (a)

mg  g m

Pseudo force acting on the block = g (back wards)

4

Force of friction is mg ( acting backwards) From the plank frame time needed to stop the block is given by

a  2g 

O = V0  at  t=

V0 2g

Velocity of the plank during this time is Vp  u p  a p t = g

V0 V  0 2g 2

mg

 2  V0  2   V0     2  2   3V0  Displacement of the block = S =    8g 2 a       Work done by friction on the block = F.S. cos   mg.

(b)

ma p m

3 V02 3  1 =  mv20 8 g 8

From the Plank frame Work done by friction on smaller block = -mgl



0  V02  2g



 (c)

g 

mV02 2

work done by friction from the Plank frame = 

mV02 2

Final velocity of the block = Velocity of the plank =

V0 2

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

 vf2  vi2 + 2 ax W=





m v f2  vi2 1 1 = m v f2 m v i2 2 2 2

5

The quantity k =

1 m v2 is a scalar and is called the kinetic energy of the particle. 2

It is the

energy posses by the particle by virtue of its motion. Thus the equation takes the form W  K f  K i  K The work done by a force changes the kinetic energy of the particle. This is called the work Energy Theorem.

Illustration – 5 : 



The velocity of an 800 gm object changes from v 0 = 3 i - 4 j to v f = -6 j + 2 k m/s. What is the change in K.E of the body? Solution : Here m = 800gm = 0.8 kg 

vo =



3 2   4 2 = 5

vf 

  2  2 1    change in K.E = x 0.8  v f  v 0  2    

 6 2  2 2 =

=

40

1 x 0.8 x 40  25  6J 2

Illustration – 6 : 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

1 mv 2 = F.s =  N.s = mgs 2

 252 v2 s= = ~ 39 m 2 x 0.8 x 10 2g Illustration – 7 : 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)

6

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 =

1 mv 2 2

 (10 x 10 - F) x 20 =

1 x 10 x (10)2  F = 75 N 2

 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 =

iii)

1 2 kx 2

A charged body in any electrostatic field will have electrostatic potential energy. The change in 2



potential energy of a system associated with conservative internal force as U2-U1= - W=  F . dr 1

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

7

Where Wnet is the sum of work done by all the forces acting on the mass. If the system is subjected to 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.

Illustration – 8 : 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

v

 (K + U) = 0  Ki + Ui = Kf + Uf 

1 1 mv2 + mgH = mv' 2 + 0 2 2



1 1 2 (50)2 + 40 x 10 = v' 2 2

A



H

 (1250 + 400) x 2 = v' 2

B

 v' 2 = 3300

v'

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

8

Pav =

W t

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 =

Instantaneous power may also be written as P =

dW    F.v dt

dW dt

Since work and energy are

closely 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 =

dE . dt

Illustration – 9 : 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 =

1 1 m v 2 2  mv 1 2 2 2

 Power delivered by the engine



1 m v 2 2  v12 K 2 P=  t t



1 x 500 20 2  10 2 = 2 60





= 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 =

mv r

2

O



T

P

mg cos 

mg sin 

9

 T = mg cos +

mv 2 r

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 =

mv 2 min R

 v min =

 v 2min = gR

gR

Now conserving energy between the lowest and the highest point

1 1 mu 2min  mv 2min  mg 2 R  2 2  u 2min  gR  4gR  5gR

u min  5gR If u min  v=

5gR the particle will complete the circle. At u =

5gR , velocity at highest point is

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. Illustration – 10 : 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

h

 A

B  mg cos  mg sin 

u  g

10

A and B

1 1 mu2 = mv2 + mgh 2 2

 u2 = v2+ 2gh

 v2 = u2 - 2gh …. (i)

mv 2 T - mg cos = 

mv 2  T= mg cos + 

 mg = mg cos  +

mv 2 

 v2 = g l (1- cos)

From (i) and (ii)  cos =

2 3

…………….

u2 - 2gl (1 - cos) = gl (1 - cos)

2 3

  = cos-1  

putting the value of cos in equation

 

v2 = gl  1 

Equilibrium :

(ii)

2  g = 3 3

…………  v=

(ii)

g 3

As we have studied earlier a body is said to be in translational equilibrium if net

force acting on the body is Zero.

Fnet = 0  If the forces are Conservative F = -



dU dr

dU 0 dr

 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

Illustration – 11: The P.E of a Conservative system is given as U = 10 + (x-2)2. Find the equilibrium position and discuss type of equilibrium.

11

Solution:

For Equilibrium F = 0  F= -

dU  2( x  2 )  0 dx

x=2 and

d2U dx 2

0

 it is Stable equilibrium position at x= 2 and P.E at that position is 20 units.

*****

12

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

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 =

u2 1  =  mu 2  g 2 

 2     mg 

1 2 kx  2 

=

 2  kx 2  1 1 2 2     mu  kx  2   mg  mg  2

600 0.052 =  10 m . 0.015 x 10 [ 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 = 

k 5 x 3 3

13

as the work done is zero  force is perpendicular to the displacement

3  k  =-1 2 3

    k=2 EXAMPLE : 04

Power supplied to a particle of mass 2 kg varies with time as p =

3t 2 watt. Here 't' is in second. If 2

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

D) 2 2 m/s

C) 2 m/s

Solution : 2

kf - ki =

1  mv2 = 2

 P dt 0

2

 0

3 2 t dt 2

2

t3   v2 =    2  0

 v = 2 m/s

 m = 2 kg 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 =

u cos   cos 2

Wgravity =  K = =

u

u cos  / 2

1 1 mv2 - m (u cos)2 2 2



V

1  mu2 cos2  tan2 2 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)

14

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  Wair drag = mgh -

1 mu2 2

1 mu2 = (40 - 50) J = - 10 J 2

EXAMPLE : 07

The potential energy of particle of mass 'm' is given by U =

1 kx2 for x < 0 and U = 0 for x > 0. If total 2

mechanical energy of the particle is E. Then its speed at x =

A) zero

B)

2E M

C)

2E is k

E m

D)

E 2m

Solution : Potential energy of particle at x =



2E is zero  K.E = E k

1 mv2 = E or v = 2

2E m

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)

F 2F K2). When they are stretched by the same force :

18

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

19.

B) 2mas

C) 2a

as 2

D)

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

20.

B) x2

C) ln x

D) ex

A particle falls from rest under gravity. Its potential energy (PE) with respect to the ground and its kinetic energy (KE) are plotted against time (t). Choose the correct graph.

A) 21.

B)

C)

D)

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. 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

22.

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. 23.

Suppose y represents the work done and x the power, then dimensions of



 

 

A) M 1L2 T 4 B) M 2 L3T 2 C) M 2 L4 T 4 24.

25.





D) ML3 T 6



d2y dx 2

will be :

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

Identify the correct statement for a non-conservative force A) A force which is not conservative is called a non-conservative force

19

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 26.

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)

27.

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

28.

The force acting on a body moving along x-axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at

29.

A) x = x1

B) x = x2

C) both x1 and x2

D) neither x1 and x2

Displacement time graph of a particle moving in a straight line is as shown in figure. Select the correct alternative(s). 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 C) Work done by all the forces in region BC is negative D) Work done by all the forces in region OA is negative

KEY 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

27

28

29 20

B

C

A

B

B

D

B

A

C

B

B

A

B

B

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 a) 

C 2r

b)

Sol: Fcentipetal F =

C 2r

c) C x 2r

d) Zero

C mv 2 C  2 ; v =   Fdr  C r 2 dr  ; E1 = EK + v = C/2r – C/r = -C/2r r r r 

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

Sol: P = w/1 = (m/g) gh = 100 x 10 x 100 = 105w 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

Sol: P = Fv = constant or ma . at = constant or a2t = constant  S = 1/2at2 or S  at2 But a  1/ t

S  t2/ t or S  t3/2 4. A ball is dropped from a height of 10m. If 40% of its energy is lost un collision with the earth then after collision the ball will rebound to a height of a) 10m Sol: 

b) 8m

c) 4m

d) 6m

u1 h1 100 10  or  u2 h2 60 h 2

5. A particle moves under the influence of a force F = CX from X = 0 to X = X 1. The work done in this process will be a)

CX 12 2

Sol: W =

b) CX 12 x2

x2

x1

0

 Fdx 

1

 cx dx  2 Cx

c) CX 13

d) 0

2

21

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)

MgL 3

b) MgL

c)

MgL 9

d)

MgL 18

Sol: W = M/3 . g . 1/6 7. A body of mass 2kg moves under the influence of a force. Its position x changes with time according to the relation x = t3/3 where x is in meter and t in seconds. The work done by this force in first two seconds will be a) 1600 Joule

b) 160 Joule

c) 16 Joule

d) 1.6 Joule

Sol: W = ½ mv22 – ½ mv12

8. A man and a child are holding a uniform rod of length L in the horizontal direction in such a way that one fourth weight is supported by the child. If the child is at one end of the rod then the distance of man from another end will be a) 3L/4 Sol:

b) L/4

c) L/3

d) 2L/3

3w  L    x 4 2 

9. 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

Sol: P = Fv = 4500 x 2 = 9 kw 10. A body of mass m is accelerated to velocity v in time et1. The work done by the force as a function of time t will be a)

mv 2 t 2 2e 2

2

1  mv  2 b)   t 2 t 

Sol: Acceleration produced in a body a =

c)

mv 2 t 2t

d)

mvt2 2t

1 mv 2 2 1 v t ; W = ma2t2 = 2 t 12 2 t1

11. A motor of 100 HP is moving with a constant velocity of 72 km/hour. The forward force exerted by the engine of the car is a) 3.73 x 103 N

b) 3.72 x 102 N

c) 3.73 x 101 N

d) None of the above

Sol: F = P/v 22

12. The kinetic energy of a man is half the kinetic energy of a boy of half of his mass. If the man increases his speed by 1m/s, then his kinetic energy becomes equal to that of the boy. The ratio of the velocity of the boy and that the man is a) 2/1

b) 1/2

Sol: According to question

c) 3/4

d) 4/3

1 1 1 M 2 Mv 2  x  U  2 2 2 2 

13. A bomb of mass 9 kg explodes into 2 pieces of 3kg and 6kg. The velocity of 3 kg piece is 16 m/s. The kinetic energy of 6kg piece is a) 768 Joule

b) 786 Joule

c) 192 Joule

d) 687 Joule

1 2

Sol: m1v1 = m2v2; E K 2  m 2 v 22 14. The increase in the potential energy of a body of mass m, when it is carried from the surface of earth upto a height equal to the radius of earth Re, will be a) mgRe Sol:

b) mgRe/2

c) mgRe/4

d) 2mgRe

GMm mgR  2R 2

15. A person of mass 60kg carries a 15 kg body on the top of a building 10m high in 3 minutes. His efficiency is a) 40% Sol: M =

b) 30%

c) 20%

d) 10%

m x 100 Mm

16. A force F = (3x2 + 2x – 7)N acts on a 2 kg body as a result of which the body gets displaced form x = 0 to

x = 5m. The work done by the force will be

a) 35 Joule Sol: W =

b) 70 Joule

x2

s

x1

0



c) 115 Joule

d) 270 Joule



2  Fdx   3x  2x  7 dx

17. A 50 gm bullet moving with a velocity of 10 m/s gets embedded into a 950 gm stationary body. The loss in kinetic energy of the system will be a) 5% Sol:

b) 50%

c) 100%

d) 95%

m2 E x100  x100 E M1  m 2

23

18. A crane lifts 300 kg weight from earth’s surface upto a height of 2m in 3 seconds. The average power generated by it will be a) 1960 watt

b) 2205 watt

c) 4410 watt

d) 0 watt

Sol: P = w/t = mgh/t

19. A block of mass 16kg is moving on a frictionless horizontal surface with velocity 4m/s and comes to rest after pressing a spring. If the force constant of the spring is 100 N/m then the compression in the spring will be a) 3.2 m

b) 1.6 m

c) 0.6 m

d) 6.1 m

Sol: ½ mv2 = ½ kx2 20. 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 Sol: t =

b) 0 m x + 3 or x = (t – 3)2;

c) 3 m

d) 2 m

v = dx/dt

21. A 0 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be a) 108 x 100 x 10 Joule b)

108 x 10 Joule 100

c) 0 Joule

d)

100 x 10 Joule 108

Sol: W = Fd cos = Fd cos 900 = 0 22. A particle moves in a potential region given by u = 8x2 – 4x + 400 Joule. Its state of equilibrium will be a) x = 25 m

b) x = 0.25 m

c) x = 0.025 m

d) x = 2.5 m

Sol: F = - du/dx 23. Two men with weights in the ratio 5 : 3 run up a stair case in time in the ratio 11 : 9. The ratio of power of first to that of second is a) 15/11 Sol: P =

b) 11/15

c) 11/9

d) 9/11

mgh wh w   t t t

24. 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 24

Sol: mu + 2m x 0 = (m + 2m)v;

E K1  E K F E K1



E Ki 

E Ki 9



E Ki

8 9

25. The potential energy function of a diatomic molecule is given as u(r) =

a r

12



b r6

, where a and

b are positive constants and r is inter atomic distance. The equilibrium between two atoms is b a

1/ 6

a)   Sol:

a b

1/ 6

b)  

 b    2a 

1/ 6

c) 

 2a    b 

1/ 6

d) 

du 12a 6b   13  7  0 dr r r

26. A pump pulls 1000 kg water per minute from a 15 m deep well and provides 4 m/s velocity to it. The power of pump is (g = 10 m/s2) a) 2.6 kw Sol: P 

b) 2.6 w

c) 0.6 w

d) 0.6 kw

w mgh  1 / 2mv2  t t

27. A body weighing 80N is moved up a slope of angle 600 with the horizontal through a displacement of 1m. The energy loss due to friction is 20%. The energy gained by the body will be a) 32 3 J

b) 64 J

c) 40 3 J

d) 80 J

Sol: W = mg sin  d 28. For the path PQR in a conservator force field (figure) amounts work done in carrying a body from P to Q and from Q to R are 5 Joule and 2 Joule respectively. The work done in carrying the body from P to R will be a) 7 Joule b) 3 Joule c)

21 Joule

d) Zero Sol: WPR = WPQ + WQR 29. 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 25

Sol: E =

1 m1m 2 u1  u 2 2 2 M1  m 2

30. Two protons are situated at a distance of 100 fermi from each other. The potential energy of this system will be in ev a) 44

b) 1.44 x 103

c) 1.44 x 102

d) 1.44 x 104

Sol: U = kq2/r 31. In order to reduce the kinetic energy of a body to half its initial value, its speed will have to be changed by the following factor, of its initial speed a) 1/ 2 times

b)

Sol: E = ½ mv2;

v =

2 times

c) 1/2 times

d) 2 times

F

32. A body of mass M and moving with velocity u makes a head on elastic collision with another stationary body of m. If A = m/M, then the ratio (f) of the loss of energy of M to its initial energy will be a) f = A(A + 1)2 Sol: f =

b) f =

4Mm

M  m

2



A

A  1

2

c) f =

uA

A  1

2

d) f =

4A

A  12

4A

1  A 2

33. Two masses m1 = 2kg and m2 = 5kg are moving on a frictionless surface with velocities 10 m/s and 3 m/s respectively. m2 is ahead of m1. An ideal spring of spring constant k = 1120 N/m is attached on the backside of m2. The maximum compression of the spring will be, if on collision the two bodies stick together. a) 0.51 m

b) 0.062 m

c) 0.25 m

d) 0.72 m

Sol: E =

1 m1m 2 u1  u 2 2  1 kx 2 2 m1  m 2 2

34. A body at rest explodes all of a sudden in three equal parts. The moments of two parts are Pi and 2Pj and their kinetic energies are k1 and k2. If P3 and k3 are the momentum and kinetic energy respectively of the third part then the ratio k2/k3 will be a) 2/5

b) 3/5

c) 4/5

d) 1/5

Sol: Conceptual

26

35. A block falls down from a table 0.5m high. It falls on an ideal vertical spring of constant 4 x 102 N/m. Initially the spring is 25 cm long and its length becomes 10 cm after compression. The mass of the block is (g = 10m/s2) a) 0.5 kg

b) 2 kg

c) 1.2 kg

d) 0.9 kg

Sol: mgh = ½ kx2 36. The mass of a bucket full of water is 15 kg. It is being pulled up from a 15m deep well. Due to a hole in the bucket 6 kg water flows out of the bucket. The work done in drawing the bucket out of the well will be a) 900 joule Sol: W = mgh =

b) 1500 joule

c) 1800 joule

d) 2100 joule

15  9  12kg 2

37. A spring of force constant k is first stretched by a lens x and then again by a further length x. The work done in the first case is w1 and in the second case w2, then a) w2 = w1 Sol: w1 = ½ kx2,

b) w2 = 2w1

c) w2 = 3w1

d) w2 = 4w1

w3 = ½ k(2x2)

38. A 2k body is projected, at an angle of 300 with the horizontal, with a velocity of 10m/s. The kinetic energy of the body after 1 second will be a) 10 joule Sol: v =

b) 50 joule

c) 100 joule

d) 200 joule

u 2  g 2 t 2  2ugt sin 

39. A 10 kg block is pulled in the vertical plane along a frictionless surface in the form of an arc of a circle of radius 10m. The applied force is of 200N as shown in the figure. If the block started from rest to A, the velocity at B would be a) 1.732 m/s

b) 17.32 m/s

c) 173.2 m/s

d) none of these

Sol: ½ mx2 = 200 cos 300 x 5 3 40. A block of mass m is pushed towards a movable wedge of mass m and height h with a velocity u. All surfaces are smooth. When the block collides with the wedge, the velocity of centre of mass of block wedge system will be a) u

b)

u 1 

27

c) u(1 + )

d) 0

Sol: mu = (m + m)v cm 41. In the above problem, the minimum value of u for which the block will reach the top of the wedge, will be a)

 1 2gh 1    

b)  2gh

c)

2gh

d)

 1 2gh 1    

Sol: ½ mu2 = mgh + ½ (m + m)V2cm 42. 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 Sol: W = 2ahdgh – 2(ahdg h/2) = dgh2a 43. The human heart discharges 75cc of block through the arteries at each beat against an average pressure of 10cm of mercury. The pulse frequency of the heart is 72 per minute. The rate of working of heart is a) 2.35 w Sol: P = hdg

b) 3.29 w

c) 1.19 w

d) 9.11 w

dv dt

44. A block of mass 1kg is pulled up on an incline of angle 300 with the horizontal. The block moves with an acceleration of 1 m/s2. The power delivered by the pulling force at t = 4s will be a) 12 w

b) 36 w

c) 24 w

d) 48 w

Sol: F – mg sin  = ma or F = mg sin  + ma 45. A particle of mass m is moving in a circular path of constant radius r. The centripetal acceleration of the particle (ac) is varying with time t according to following relation ac = k2n2 where k is a constant. The power delivered to the particle by the forces acting on it will be a) mk2 r2 t2 Sol: ac = v2/r = k2n2;

b) m2k2 r2t2

c) m2k2 rt

d) mk2r2t

w = ½ mv 2/2 – ½ mv12 = ½ m k2r2t2 = 0; P = dw/dt 28

46. A block of mass 2kg is released from A on a track that is a on quadrant of a circle of radius 1m. It slides down the track and reaches B with a speed of 4m/s and finally stops at C at a distance of 3m from B. The work done against the force of friction is a) 2 joule

b) 5 joule

c) 10 joule

d) 20 joule



1

 1

1



Sol: W = mgh  mv2B    mv2B  mvC2  2 2   2  47. 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 M



a)   m gh 2 

M  m gh 2 

b)  

 

c) M 

m gh 2 

d) (M + m) gh

m  gh   w 2 

Sol: M  

48. A particle has shifted along some trajectory in the x – y plane from point r1  ˆi  2ˆj to another point r2  2ˆi  3ˆj .

During that time, the particle experiences the action of two forces

Fi  3ˆi  4ˆj and F2  2ˆi  7ˆj  2kˆ . The work done by the forces on the particle will be

a) 5 joule

b) -5 joule

c) 10 joule

d) -10 joule

Sol: F  F1  F2 49. A 2kg body is dropped from height of 1m on to a spring of spring constant 800 kg/m as shown in the figure.

A frictional force

equivalent to 0.4 kg wt acts on the body. The speed of the body just before striking the spring will be a) 1 m/s

b) 2 m/s

c) 3 m/s

d) 4 m/s

Sol: mgh = ½ mv2 + Ffr h 50. A shell is fired from a cannon with a velocity v and at an angle  from the horizontal to hit a target at a horizontal distance R. It splits in two equal parts at the highest point of its path. One part refracts its path and reaches back upto the cannon. The velocity of the second part just after the explosion will be a) 3/2 v cos 

b) 2 v cos 

c) 3 v cos 

d) 3 /2 v cos 

Sol: mv cos  = m/2 v cos + m/2 v 29

51. A block of mass 10 kg moving on a smooth surface with a speed of 30 m/s bursts into two equal parts. Both parts continue to move in the seme direction. If one of the parts moves at 40 m/s, the energy produce in the process is a) 200 J

b) 500 J

Sol: mv = m1 v1 + m2 v2;

c) 700 J

d) J

E = ½ m1v12 + ½ m2 v22 – ½ mv2

52. Two identical 5 kg blocks are moving with same speed of 2 m/s towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. The work done by the external forces is a) 0 Sol: As Fext = 0; 53.

b) 10 J

c) 20 J

d) none of these

Wext = F ext . S  0

In the above problem, the work done by the inertial forces is a) 0

b) 10 J

c) 20 J

d) none of these

Sol: Wint = ½ mv2 + ½ mv2 54.

The force-displacement curve for a body moving on a smooth surface under the influence of foce 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

55.

B) 4.5J

C) 2.25J

D) 9J

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

56.

B) 35m/s

C) 50m/s

D) 70m/s

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

57.

B) 245

2W

C) 235

2W

D) 235W

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

58.

B)

20 N

C) 12 N

D) 0

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

30

59.

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

60.

B) 5J

C) 7J

D) 9J

The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation t =

x +3, where x is in metres and t in seconds. The work done

by the force in first 6 seconds is A) 9J 61.

B) 6J

C) 0J

D) 3J

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

62.

A) mg 

B) -mg 

C) mgh

D) zero

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

63.

A) increases by mg l

B) increases by mg (l -h)

C) increases by mgh

D) increases by mg (2 l -h)

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 C) 1 D) 3

64.

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

65.

B) l-mg/k

C) l + mg/k

D) l - k/mg

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

31

66.

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

67.

B) 2m R g

C) 2/ m R g

D) 1/ m R g

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

68.

B) 48J

C) 84J

D) 36J

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)

69.

1 mgl 4

B)

1 mgl 8

C)

1 mgl 16

D)

1 mgl 32

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

70.

B) W2 = 2W1

C) W2 = 3W1

D) W2 = 4W1

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 1) zero

71.

B)

gl

C) 1.5gl

D)

2gl

A force F acting on a body depends on its displacement x as F xn. The power delivered by F will be independent of x if n is A) 1/3

72.

B) -1/3

C) 1/2

D) -1/2

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

73.

B) Wc = UA - UB

C) UA > UB

D) UB > UA

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

74.

B) 

C) Mv2

D) Mv2 /2

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 A)

75.

1 2 kx 2

B)

1 (kx) (s+x) 2

C)

1 kxs 2

D)

1 kxs  x  s  2

Force acting on a block moving along x-axis is given by :

32

 4   N F =    x2  2 

The block is displaced from x=-2m to x=+4m, the work done will be

75.

A) positive

B) negative

C) zero

D) may be positive or negative

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

77.

B) 0.4m

C) 0.6m

D) 0.8m

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) 78.

B)

C)

D)

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)

79.

A)

2 gR 5

B)

5 gR 2

C)

5 gR 7

D) none of these

A wind-powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical. For wind speed v, the electrical power output will be proportional to B) 2

A) v

C) 3

D) 4

KEY 54

55

56

57

58

59

60

61

62

63

64

65

66

67

68

A

B

D

B

B

A

C

C

B

D

B

B

C

A

D

69

70

71

72

73

74

75

76

77

78

79

C

A

B

B

A

A

B

B

B

B

C

33

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

2.

A) 0

2) Fh

C) 2Fh

D)

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)

3.

B)  = p2 / (2mg s)

C)  = p2 / (2g m2s)

D)  = p2 (2g m2s2)

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)

4.

2Pt m

B)

2Pt m

C)

Pt 2m

D)

Pt 2m

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)

mg k

B)

1 mg 2 k

C) 2

mg k

D) 4

mg k

34

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 in S. The height difference h between points A and B is

6.

A) h = S/

B) h = S

C) h = S2

D) h = S/2

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

7.

B) mg vt cos2

C) mg vt sin2

D) mg vt sin2

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

8.

1 2 kx 2

A) h' = h - 2d

B) h' = h - 2d -

C) h' = h - 2d + kx2

D) h' = h - 2d - kx2

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)

9.

2gl

B)

2 gl 3

C)

1 gl 3

D)

1 gl 2

If a constant power P is applied in a vehicle, then its acceleration increases with time according to the relation  P  t A) a =    2m 

 P  3/ 2 t B) a =    2m 

 P  1 / t C) a =    2m 

D) a =

P 2mt

35

10.

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

11.

A) mtg2 cos 

B) mtg2 cos  (sin  -  cos )

C) mtg2 sin 

D) mtg2 sin  (sin  -  cos )

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

12.

B)

2

C) 1/ 2

D) 0

A uniform rope of length '  ' and mass m hangs over a horizontal table with two third part on the 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

13.

B) 2/3 mgl

C) 1/3 mgl

D) 1/9 mgl

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

36

14.

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

15.

A) x

kh mg

B)

C) x

2kh mg

D)

xkh mg mg x kh

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

16.

B) 12m/s

C) 8m/s

D) 10m/s

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.

17.

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.

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`

C) 2m/s 18.

B) 2 2 m/s D) 4 2 m/s

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

19.

B) 15 N

C) 10 N

D) 25 N

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

37

the plane is (Neglect mass of string and pulley and friction in pulley).

A) 3/5m 20.

B) 4/5m

C) 6/5m

D) 3/2m

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 t as, ac = k2 r t2 where k is a constant. What is the power delivered to the particle by the forces acting on it? A) 2 pmk2r2t

21.

B) mk2r2t

C)

(mk 4 r 2 t s ) 3

D) zero

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 + ax2. Here k and a are positive constant. For x 0, the function form of the potential energy (x) of the particle is

A)

B)

c)

D)

KEY 1

2

3

4

5

6

7

8

9

10

11

C

C

A

C

B

C

A

B

D

B

A

12

13

14

15

16

17

18

19

20

21

A

C

C

D

A

B

A

A

B

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) F  kyˆi  kxˆj 

C) The force F is a conservative force 2.



B) F  kxˆi  kyˆj 

D) The force F is a non-conservative force

A sledge moving over a smooth horizontal surface of ice at a velocity v0 drives out on a horizontal road and comes to a halt as shown. The sledge has a length l, mass m and friction between runners and road is  A) No work is done by the friction to switch the sledge from ice to the road

38

B) A work of

1 mgl is done against friction while sledge switches completely on to road 2

 v 02 l    2 g 2   

C) The distance covered by the sledge on the road is 

 v 02 l    2 g 2   

D) Total distance moved by the sledge before stopping is  3.

A strip of wood of 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 A) The speed of the insect as seen from the ground is < B) The speed of the strip as seen from the ground is

1 M    t  M  m 

C) The speed of the strip as seen from the ground is

1 m    t  M  m 

D) The total kinetic energy of the system is

4.

1 t

1 1 (m + M)   2 t

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 

2

mv  B) the kinetic energy of the A-B system at maximum compression of the spring is   4   



C) the maximum compression of the spring is  v 



D) the maximum compression of the spring is  v 

5.

m  k  m 2k

   

Suppose a car is modeled as a cylinder moving with a speed v. If A is the area of cross section of the car and  is the density of air then

39

A) Power loss due to air resistance is C) drag force is 6.

1 Av3 2

B) power loss due to air resistance is Av3

1 Av2 2

D) drag force is Av2

A heavy mass M resting on the ground is connected to a lighter mass m through a light inextensible string passing over a light, frictionless pulley as shown in the figure. The string connected to mass M is loose. Let lighter mass m be allowed to fall freely through a height h such that the string becomes taut. If t is the time from this instant onwards when the heavier mass again makes contact with the ground and E is the change in kinetic then A) t =

7.

2m Mm

2h g

B) t =

2m Mm

2h g

C) E = -

1  Mm  2  v 2  M  m 

D) E = -

1 Mv 2 2

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

8.

B) F  t-1/2

C) F  v

D) F  v-1

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 ?

9.

A) F = 3R

B) F = 2R

C) a = 3P/m vmax

D) a = 2 P/m vmax

The net interacting force F between two particles is related to the distance x between them, as shown in figure. Then A) potential energy of the system decreases from x1 to x2 B) potential energy of the system increases from x2 to x3 C) potential energy of the system decreases from x3 to x4 D) kinetic energy increases from x1 to x2

10.

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 11.

The figure shows a potential energy function. In the region 0  r  R, U(r) is parabolic. In the region r  R, U(r) is an inversely proportional

40

function. Identify the correct conservative force function F(r) plot.

(A)

12.

(B)

(C)

(D)

The potential energy function between two atoms in a diatomic molecule can be plotted as shown in figure. Mark the correct statement(s): A) There is only one position of equilibrium B) There is two position of equilibrium C) There is only one position of stable equilibrium D) Both the positions are of stable equilibrium

13.

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

14.

B) It decreases

C) It increases

D) Nothing can be said

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 C) Both the forces perform positive work D) The net work done is equal to the increase in potential energy

15.

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 <  <

3 2

D) The value of N becomes zero only when  > /2

41

16.

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

mghv s

C) The frictional force exerting on the train is mf on the level track D) None of above is correct 17.

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 u  (14.4ˆi  4.2ˆj)ms 1 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 18.

A particle is moving in a conservative force field from point A to point 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

19.

B) Wc = UA - UB

C) UA > UB

D) UB > UA

At the position of stable equilibrium A)

20.

dU  0 only dx

B)

du dU d2U d2U = 0 and > 0 C) 0  0 and dx dx dx 2 dx 2

D) None of these

Choose the correct statement(s) related to the conservative force and potential energy. A) Potential energy decrease in the direction of conservative force B) Potential energy increase in the direction of conservative force C) Conservative force does work by lowering its potential energy D) Conservative force does work by raising its potential energy

KEY 1

2

3

4

5

6

7

8

9

10

AC

BCD

AC

BD

AC

BC

BD

AD

ABD

ABD

11

12

13

14

15

16

17

18

19

20

A

BC

B

ACD

ABD

ABC

ABC

BC

C

BC

COMPREHENSION TYPE QUESTIONS Passage I (Q.No: 1 to 7):

42

The potential energy of two atoms in a diatomic molecule is approximated by U(r) =

a b  6, 12 r r

where r is the spacing between atoms and a and b are positive constants. 1.

Find the force F(r) on one atom as a function of r: A) 0

2.

B)

B)

12a 6b  r 12 r 6

D)

12a 6b  r 13 r 7

C)

D)

Which is the most appropriate graph F(r) versus r:

A)

4.

C)

Which is the most appropriate graph U(r) versus r:

A)

3.

12a b  r 13 r 7

B)

C)

D)

Find the equilibrium distance between the two atoms: 1/ 6

A) 2a 5.

6.

B) 2a/b

C) 2a/5b

 2a  D)    b 

From the above conclusion can we predict about equilibrium state: A) the equilibrium is stable

B) the equilibrium is unstable

C) the equilibrium may be stable

D) the equilibrium may be unstable

What minimum energy must be added to the molecule to dissociate it, if the distance between the two atoms is equal to the equilibrium distance found in Q. 4 ? A) b2/a

7.

B) 2b2/a

C) b2/4a

D) 2a/b

For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is 1.13 x 1010

m and the dissociation energy is 1.54 x 10-18 J per molecule. Find the values of a and b:

A)

a = 6.67 x 10-138 J-m12

B)

b = 2.08 x 10-60 J m6 C)

a = 6.67 x 10-138 J-m12

a = 6.41 x 10-78 J-m6 b = 6.67 x 10-138 J-m12

D)

a=0

43

b = 6.41 x 10-78 J m6

b = 6.41 x 10-78 J m6

Passage II (Q.No: 8 to 11) : A cutting tool under microprocessor control has several forces acting on it.

One force is

 F  xy 2 ˆj , a force in the negative y-direction whose magnitude depend on the position of the tool. The constant is  = 2.50 N. Consider the displacement of the tool from the origin to the point x = 3.00 m, y = 3.00 m. 8.



Calculate the work done on the tool by F if this displacement is along the straight line y = x that connects these two points ? A) 2.50 J

9.

B) 500 J



C) 50.6 J

D) 2 J

Calculate the work done on the tool by F if the tool is first moved out along the x-axis to the point x = 3.00 m, y = 0 and then moved parallel to the y-axis to x = 3.00 m, y = 3.00 m. A) 67.5 J

10.

B) 85 J



C) 102 J

D) 7.5 J

Compare the work done by F along these two paths ? A) Work done on x-axis is zero B) Work done on y-axis is less than on y-axis C) Work done on x-axis is more than on y-axis but not zero D) Data insufficient

11.



What can you predict about F ? A) Force is non-conservative B) Force is conservative C) Force is neither conservative nor non-conservative D) Data insufficient to conclude

Passage III (Q.No: 12 to 16): A 1200 kg car is travelling at 7.5 m/s in a northerly direction on an icy road. It crashes into a 8000 kg truck moving in the same direction as the car with a velocity of 3.0 m/s before the collision. The speed of the car after the collision is 3.0 m/s in its original direction. 12.

Which of the following is true regarding the relationship between energy and momentum in the passage ? A) The collision is not perfectly elastic, both momentum and energy are not conserved B) The collision is inelastic, kinetic energy is conserved but momentum is not C) The collision is not perfectly elastic, momentum is conserved but total energy is not D) The collision is not perfectly elastic, momentum is conserved but kinetic energy is not

44

13.

What is the velocity of the truck after the collision ? A) 7.5 m/s

14.

B) 3.7 m/s

C) 3.0 m/s

D) 1.1 m/s

The car then proceeds to a garage. To get there, the driver turns off onto a smooth road with a coefficient of friction =  = 1/4. He then stops for a snack and then tries to drive off. what is the value of frictional force when the force the car exerts is 300 N ? A) 0 N

15.

B) 100 N

C) 300 N

D) 4000 N

After leaving the garage, the driver of the car follows the same road and eventually has to go up a hill. How does the frictional force on the car now compare to the value when the car was driving on level ground ? A) No change

B) It increased

C) It decreased

D) The direction of change depends on the angle of elevation 16.

If the car is moving up the hill at 5 m/s and the car is 40m up the hill as shown in the diagram, how much potential energy does the car possess at that point ? (g = 9.8 m/s2). A) 2.40 x 105 J

B) 2.40 x 104 J

C) 4.95 x 105 J

D) 4.95 x 104 J

KEY 1

2

3

4

5

6

7

8

9

10

11

D

A

D

D

A

A

C

C

A

A

A

12

13

14

15

16

***

45

MULTIPLE MATCHING TYPE QUESTIONS 1.

Match the following: 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)  F . dx , where F is conservative force





i) gravitational force 2.

Match the following: List - I

3.

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

Match the following: List - I

4.

List - II

a) stable equilibrium

e) PE in Max

b) unstable equilibrium

f) Fnet = 0

c)

dF 0 dx

g) PE is Min

d)

dF 0 dx

h) slope of F-x graph is +ve

Match the following: 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

KEY 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

***

46