best coaching institute for iit-jee booklet (work, energy and power)

7/29/2019 BEST COACHING INSTITUTE FOR IIT-JEE BOOKLET (WORK, ENERGY AND POWER)

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APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457,Webs i te : w w w.apex i it . co .in /

WORK, ENERGY AND POWER

WORK DONE BY A CONSTANT FORCE

Consider a particle, that undergoes a displacement r along a straight line while upon acted by a constant force

F that makes an angle with r as shown in figure.

The work W done by a constant force F acting on a particle is equal to the product of the component of F

along the direction the displacement of particle and the magnitude of the displacment )r( .Mathematically,

r.cosFW = (from Figure)

= rF. (Applying = cosabb.ar

)

Work is a scalar quantity. Its unit is Joule.

(i) Force does not work if point of application of force does not move ( r = 0)(ii) Work done by a force is zero if displacement is perpendicular to the force ( = 900)(iii) If the angle between the force and the displacement is acute ( < 900), we say that work done by the

force is positive.

(iv) If the angle between the force and the displacement is obtuse ( > 900), we say that work done bythe force is negative.

(v) Work done depends on the frame of reference. With the change of the frame of reference inertial

force does not change while displacement may change.

WORK DONE BY A VARIABLE FORCE

Consider a particle being displaced along the x-axis under the action of a varying force, as shown in the figure.

The particle is displaced in the direction of increasing x from x = xito x = x

f. In such a situation, we cannot use

W = (F cos )r to calculate the work done by the force because this relationship applies only when F is constantin magnitude and direction. However, if we imagine that the particle undergoes a very small displacement x,shown in the figure (1), then the x component of the force, F

x, is approximately constant over this interval, and

we can express the work done by the force for this small displacement as

W1

= Fxx

r

F cos

F


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Fx

Area = A = Fxx

Fx

xxfx

xi

(1)

Fx

xxfxi

(2)

Thus is just the area of the shaded rectangle in the figure (1). If we imagine that the Fx

versus x curve is divided

into a large number of such intervals, then the total work done for the displacement from xito x

fis approxi-

mately equal to the sum of a large number of such terms.

W = f

i

x

x

xxF

If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit

but the value of the sum approaches a definite value equal to the area under the curve bounded by Fx

and the x-

axis.

dxFxF

f

i

f

i

x

x

x

x

x

xx =0lim

This definite integral is numerically equal to the area under the Fx

versus x curve between xiand x

f. There-

fore, we can express the work done by Fx

for the displacement of the object from xito x

fas

W = dxF

f

i

x

x

x

when F

x= F cos is constant.

If more than one force acts on a particle, the total work done is just the work done by the resultant force. For

systems that do not act as particles, work must be found for each force separately. If we express the resultant

force in the x-direction as Fx, then the net work done as the particle moves from x

ito x

fis

Wnet

= ( )dxFf

i

x

x

x

Illustration:

A chain of mass m = 0.80 kg and length l = 1.5 m rests on a rough - surfaced table so that one of its

ends hangs over the edge. The chain starts sliding off the table all by itself provided the over hanging

part equals n = 1/3 of the chain length. What will be the total work performed by the friction forces

acting on the chain by the moment it slides completely off the table ?

Solution:

Slipping occurs when the weight of hanging part is just sufficient to overcome the frictional force

exerted by the table. Let be the coefficient of friction between chain and table.

Weight of hanging part = (weight of horizontal part)


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nmg = (1 n) mg

=n1

n

...(i)

Let x be the length of the hanging part at some time instant.

frictional force f(x) = (normal reaction)

=l

l mg)x(

The work done by the frictional force if the hanging part increases to (x + dx) is :

dW = f (x) dx

W = dW = l

l l

l

n

mg)x(dx

W = l

mgl

l

l

n

2

2

xx

W = mg

)n1(2

)n1( 2l

l

Substituting the value of from (i), we get :

W = 2

mg)n1(n l= 1.3 J.

WORK DONE BY A SPRING

Consider the situation shown in figure. One end of a spring is attached to a fixed vertical support and the other

end to a block which can move on a horizontal table. Let x = 0 denote the position of the block when the spring

is in its natural length. We shall calculate the work done on the block by the spring-force as the block movesfrom x = 0 to x = x

1.

x = 0

A

x = x1

A

The force on the block is k times the elongation of the spring. But the elongation changes as the block moves

and so does the force. We cannot take F out of the integration F rd . We have to write the work done duringa small interval in which the block moves from x to x + dx. The force in this interval is kx and the displacement

is dx. The force and displacement are opposite in direction.

So, F. rd = -F dx = - kx dxduring this interval. The total work done as the block is placed from x = 0 to x

1is

W =2

1

0

2

02

1

2

111

kxkxkxdx

xx

=

=

If the block moves from x = x1

to x = x2, the limits of integration are x

1and x

2and the work done is

W =

22

2

12

1

2

1kxkx


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Note that if the block is displaced from x1

and x2

and brought back to x = x1, the work done by the spring-

force is zero. The work done during the return journey is negative of the work during the onward journey.

The net work done by the spring-force in a round trip is zero.

Illustration:

A block of mass m has a velocity v0

when it just touches a spring. The block moves through a distance l before

it stops after. The spring constant is k. What is the work done on it by the spring force?

Solution :

The net force acting on the block by the spring is equal to Fspring

= kx

Work done by the spring = -Fspring

.ds

= -2

.2

0

kldxkx

l =

WORK DEPENDS ON THE FRAME OF REFERENCE

Work done by a force is given as

S.FW =

Since S displacement of point of application is frame dependent phenomena work is frame

dependent quantity, same force may do different work in different frames

Illustration :

Over a horizontal plank a small block of mass m is lying at rest. Now plank is moved withconstant acceleration a such that there is no relative motion between block and plank. Find the work

done by friction of plank on block in first t seconds.

(a) in ground frame

(b) in plank frame.

Solution:

(a) In ground frame

Friction force acting on the block (f) = ma

Displacement in first t second

2at2

1S =

Work done = F.S. = f

2

at2

1

=222 tma

2

1at

2

1.ma =

(b) In plank frame f = ma

But displacement s = 0

(because there is no relative motion between plank and block)

m

v0 k

ma

fa

f

F (=ma)P


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W = F.S.= f . s

= ma . 0

= 0

Note: While calculating net work done on a body in accelera ted frame work done by pseudo

force need to be taken into account .

WORK-ENERGY THEOREM

If the work done by the net force on a particle can be calculated for a given displacement, the change in the

particle's speed will be easy to evaluate.

As shown in the figure a particle of mass m moving to the right under the action of a constant net force F.

Because the force is constant, we know Newton's second law that the particle will move with a constant

acceleration a. If the particle is displaced a distance s, the net work done by the force F is

Wnet

= Fs = (ma)s

We found that the following relationships are valid when a particle moves at constant acceleration

m

vi vf

s

F

s = ;)(2

1tvv fi + a = t

vvif

where viis the speed at t = 0 and v

fis the speed at time t. Substituting these expressions

Wnet

= m tvvt

vvfi

if)(

2

1+

Wnet

=22

2

1

2

1if mvmv

Wnet

= Kf- K

i= K . . . . (i)

That is, the work done by the constant force Fnet

in displacing a particle equals the change in kinetic energy

of the particle.

Equation (i) is an important result known as the work-energy theorem. For convenience, it was derived

under the assumption that the net force acting on the particle was constant.

Now, we shall show that the work-energy theorem is valid even when the force is varying. If the resultant

force acting on a body in the x direction is Fx, then Newton's second law states that

Fx

= ma. Thus, we express the net work done as


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Wnet

= ( ) =f

i

f

i

x

x

x

x

x dxmadxF

Because the resultant force varies with x, the acceleration and speed also depend on x. We can now use the

following chain rule to evaluate Wnet

.

a =dx

dvv

dt

dx

dx

dv

dt

dv ==

Wnet

=22

2

1

2

1if mvmv

Illustration:

A bullet leaving the muzzle of a rifle barrel with a velocity v penetrates a plank and loses one fifth of

its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness

of the planks supposing average resistance to the penetration is same in both the cases.

Solution:Let R = resistance force offered by the planks,

t1

= thickness of first plank,

t2

= thickness of second plank.

For first p lank :

Loss in KE = work against resistance

2

1mv2

2

1m (

5

4v)2 = Rt

1

21 mv2 25

9 = Rt1

.................(i)

For second plank

2

1m (

5

4v)2 0 = Rt

2

2

1mv2

25

16= Rt

2.................(ii)

dividing (I) & (II) 21

t

t

= 16

9.

POTENTIAL ENERGY AND CONSERVATION OF MECHANICAL ENERGY

We define the change in potential energy of a system corresponding to a conservative internal force as

Uf- U

i= -W = -

f

i

rdF.

where W is the work done by the internal force on the system as the system passes from the initial configu-

ration i to the final configuration f.


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Note: We don't (or can't) define potential energy corresponding to a nonconservative internal force.

Suppose only conservative internal forces operate between the parts of the system and the potential energy U

is defined corresponding to these forces. There are either no external forces or the work done by the them is

zero. We have

Uf- U

i= -W = -(K

f- K

i)

or Uf+ Kf= Ui + Ki . . . . . (i)

The sum of the kinetic energy and the potential energy is called the total mechanical energy. We see from

equation (i) that the total mechanical energy of a system remains constant if the internal forces are conser-

vative and the external forces do not work. This is called the principle of conservation of mechanical energy.

The total mechanical energy K + U is not constant if nonconservative forces, such as friction, act between

the parts of the system. We can't apply the principle of conservation of energy in presence of nonconservation

forces. The work-energy theorem is still valid even in the presence of nonconservative forces.

Note: that only a change in potential energy is defined above. We are free to choose the zero potential

energy in any configuration just as we are free to choose the origin in space anywhere we like.

Brain Teaser: One persons says that the potiential energy of a particular book kept in an almirah is 20 J and

the other says it is 30 J. Is one of them necessarily wrong?

If nonconservative internal forces operate within the system, or external forces do work on the system, the

mechanical energy changes as the configuration changes. According to the work-energy theorem, the work

done by all the forces equals the change in the kinetic energy. Thus,

Wc

+ Wnc

+ Wext

= Kf- K

i

where the three terms on the left denote the work done by the conservative internal forces, nonconservative

internal forces and the external forces.

As Wc

= -(Uf- U

i),

We get Wnc + Wext = (Kf+ Uf) - (Ki + Ui)= E

f- E

i. . . . . (ii)

Where E = K + U is the total mechanical energy.

If the internal forces are conservative but external forces also act on the system and they do work, Wnc

= 0

and from (ii)

Wext

= Ef- E

i. . . . .(iii)

The work done by the external forces equals the change in the mechanical energy of the system.

Let us summarise the concepts developed so far in this chapter.

(1) Work done on a particle is equal to the change in its kinetic energy.

(2) Work done on a system by all the (external and internal) forces is equal to the change in its kinetic

energy.

(3) A force is called conservative if the work done by it during a round trip of a system is always zero. The

force of gravitation, Coulomb force, force by a spring etc. are conservative. If the work done by it

during a round trip is not zero, the force is nonconservative. Friction is an example of nonconservative

force.

(4) The change in the potential energy of a system corresponding to conservative internal forces is equal

to negative of the work done by these forces.

(5) If no external forces act (or the work done by them is zero) and the internal forces are conserva-

tive, the mechanical energy of the system remains constant. This is known as the principle of

conservation of mechanical energy.


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(6) If some of the internal forces are nonconservative, the mechanical energy of the system is not constant.

(7) If the internal forces are conservative, the work done by the external forces is equal to the change in

mechanical energy.

Brain Teaser: When you lift a box from the floor and put it on an almirah the potential energy of the box

increases, but there is no change in its kinetic energy. Is it a violation of conservation of

energy?

Illustration:

In the figure shown stiffness is k and mass of the block is m. The pulley is fixed. Initially

the block m is held such that, the elongation in the spring is zero and then released from

rest. Find the maximum elongation in the spring

Neglect the mass of the spring, pulley and that of the string.

Solution:

Let the maximum elongation in the spring be x, when the block is at position 2.

The displacement of the block m is also x. If E1

and E2

are the energies of the system

when the block is at position 1 and 2 respectively. ThenE1

= U1g

+ U1s

+ T1

where U1g

= gravitational P.E. with respect to surface S.

U1S

= P.E. stored in the spring.

T1

= initial K.E. of the block.

E1

= mgh1

+ 0 + 0 = mgh1

. . . . (i)

and E2

= U2g

+ U2s

+ T2

= mgh2

+ 02

1 2 +kx . . . . . (ii)

From conservation of energy E1

= E2

mgh1 = mgh2 +2

21 kx

mgx)hh(mgkx2

121

2 ==

x = 2mg/k

Illustration:

A block is placed on the top of a plane inclined at 37 with horizontal. The length of the plane is 5 m.

The block slides down the plane and reaches the bottom.

(a) Find the speed of the block at the bottom if the inclined plane is smooth.

(b) Find the speed of the block at the bottom if the coefficient of friction is 0.25

Solution:

Let h be the height of inclined plane

h = 5 sin 37 = 3 m(a) As the block slides down the inclined plane, it loses GPE and

gains KE.

Loss in GPE = gain in KE

mg (loss in height) = KEf KE

i

m

1

m

2h2

h1

S

m m

R

s

C

A

B

mgmg cos37

37

h37

v


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mgh =2

1mv2 0

v = gh2 = 38.92 = 7.67 m/s.

Note : 1. Loss in energy = initial energy - final energy

2. gain in energy = final energy - initial energy

(b) As the block comes down, it loses GPE. It gains KE and does work against friction.

loss in GPE = gain in KE + work done against friction

mgh = (1/2 mv2 0) + ( mg cos 37) s 3mg = 1/2 mv2 + (0.25) mg 4/5 5

v = g4 = 6.26 m/s

Illustration:A 1.0 kg block collides with horizontal light spring of force constant 2 N/m. The block compresses the

spring 4 m from the rest position. Assuming that the coefficient of kinetic friction between the block

and the horizontal surface is 0.25, what was the speed of the block at the instant of collision ?

Solution:

When the block compresses the spring, let x m be the amount of compression, i.e. x = 4m.

Let v = velocity of the block when it collides with the spring.

Loss in KE of the block = (gain in elastic potential energy of the spring) + (work done against friction)

21 mv2 0 =

21 kx2 + mg x

2

1mv2 =

2

1(2) (4)2 + 0.25 1 9.8 4

v2 = 51.6 v = 6.51 = 7.18 m/s

Illustration:

A pump is required to lift 1000 kg of water per minutes from a well 20 m deep and eject it at a rate of

20 m/s.

(a) How much work is done in lifting water ?

(b) How much work is done in giving it a KE ?

Solution:

(a) Work done in lifting water = gain in PE (potential energy)

work = 1000 g 20 = 1.96 105 J per minute

(b) Work done (per minute) in giving it KE = 1/2 mv2

= 1/2 (1000) (20)2 = 2 105 J per minute

R

s

C

A

B

mgmg cos37

37

h37

mgcos37

v


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Brain Teaser: A "Violation" of the Law of the Conservation of Energy

The following argument seems to prove a violation of the law of the energy conservation.

Suppose that a resting hand-cart of mass m is hit by and retains a projectile of the same mass.

The projectile had been flying horizontally before the collision at a velocity v in the same

direction as the hand-cart. As a result of this impact the hand-cart and projectile will together

set in motion at a initial velocity which can be found from the law of the conservation of

momentum

221

v

m

mvv == .

Hence, the kinetic energy of the hand-cart and projectile together is

42

22

2

2

1

mv

vm

W =

= ,

while before the collision the projectile had a kinetic energy of

2

2

mvW= .

i.e., twice as larger. Thus, after the collision half the energy has vanished altogether.

Can you say where it has gone?

TYPES OF POTENTIAL ENERGY

1. Elastic

We have seen that the work done by the spring force (of course conservative for an ideal spring) is

- 221 kx when the spring is stretched or compressed by an amount x from its unstretched or natural position.

Thus, U = -W = -

2

2

1kx

or U =2

2

1kx (k = spring constant)

Note that elastic potential energy is always positive.

2. Gravitational

The gravitational potential energy of two particles of masses m1 and m2 separated by a distance r is given by

U =r

mmG 21

Here, G = universal gravitation constant

= 6.67 10-11 22

kg

mN

If a body of mass m is raised to a height 'h' from the surface of earth, the change in potential energy of the

system (earth + body) comes out to be:


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U =

+

R

h

mgh

1 (R = radius of earth)

or U mgh if h


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power as t approaches zero.

i.e., P =dt

dW

t

W

t=

0lim

where we have represented the infinitesimal value of the work done by dW (even though it is not a change

and therefore not a differential).

P = vFdt

sdF

dt

dW.. ==

where we have used the fact thatdt

sdv = .

This SI unit of power is joule per second (J/s), also called watt(W)' (after James Watt);

1W = 1 J/s = 1 kg.m2/s3.

CONSERVATIVE AND NON-CONSERVATIVE FORCES

If the work done along a closed path is zero, the force is said to be conservative otherwise it is called non-

conservative. Under conservative forces, the work done depends only the initial and final positions and it ispath independent.Conservative forces are non-dissipative and we store work (or energy) whereas the non-conservative forces are

dissipative where in we do not store work. Examples of non-conservative force are friction, viscous force etc.

Three Types of Equilibrium

As we have studied in the chapter of 'Laws of motion' a body is said to be translatory equilibrium if net force

acting on the body is zero, i.e.,

0=netFIf the forces are conservative

F =drdU

and for equilibrium F = 0

So, ,0=dr

dUor 0=

dr

dU

i.e., at equilibrium position slope of U- r graph is zero or the potential energy is optimum (maximum or mini-mum or constant). Equilibrium are of three types, i.e. the situation where F = 0 and dU/dr = 0 can be obtained

under three conditions. These are stable equilibrium, unstable equilibrium and neutral equilibrium.

If 02

2

>dr

Udit is stable equilibrium

02

2

Rg2

Therefore, if ,gR5ugR2 R if u2 > 2gR. Thus, the particle, will

leave the circle when h > R or 900 < < 1800. This situation is shown in the figure.

gR5ugR2


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Illustration:

A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally

with speed ).(gl Find the speed of the particle and the inclination of the string to the vertical at the

instant of the motion when the tension in the string is equal to the weight of the particle.

Solution:Let tension in the string becomes equal to the weight of the particle when particle reaches the point B and

deflection of the string from vertical is . Resolving mg along the string and perpendicular to the string, weget net radial force on the particle at B i.e.

FR

= T - mg cos . . . . (i)If v be the speed of the particle at B, then

FR

=l

mv2

. . . . (ii)

From (i) and (ii), we get

T - mg cos = lmv 2

. . . . (iii)

Since at B, T = mg

mg(1 - cos ) =l

mv 2

v2 = gl(1 - cos ) . . . . .(iv)

Conserving the energy of the particle at point A and B, we have

22

02

1)cos1(

2

1mvmglmv +=

where v0

= gl and v = )cos1( gl

gl = 2gl(1 - cos) + gl(1 - cos) cos = 2/3 . . . . .(v)Putting the value of cos in equation (iv) we get

v =3

gl

A BODY MOVING INSIDE A HOLLOW TUBE

The same discussion holds good for this case, but instead of tension in the string we

have the normal reaction of the surface. If N is the normal reaction at the lowest

point, then

N - mg =r

mv 21; N = m

+ g

r

v 21

At the highest point of the circle,

N + mg =r

mv 22

N = m

g

r

v 22

A

BT

mg cos

mg sin

O

v1

O

N

mg

mg

N

O

v2


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The condition v1 rg5

All other equations (can be) similarly obtained by replacing tension T by reaction N.

BODY MOVING ON A SPHERICAL SURFACE

The small body of mass m is placed on the top of a smooth sphere of radius r.

If the body slides down the surface, at what point does it fly off the surface?Consider the point C where the mass is, at a certain instant. The forces are the

normal reaction R and the weight mg. The radial component of the weight is mg

cos acting towards the center. The centripetal force is

mg cos - R =r

mv 2

where v is the velocity of the body at O.

R = m

r

vg

2

cos . . . . . (i)

The body flies off the surface at the point where R becomes zero.

i.e., g cos =r

v 2

; cos =rg

v2

. . . . (ii)

To find v, we use conservation of energy

i.e., )(2

1 2 BNmgmv =

= mg(OB - ON) = mgr(1 - cos )v2 = 2rg(1 - cos )

2(1 - cos ) = rgv

2

. . . .(iii)

From equation (ii) and (iii) we get

cos = 2 - 2 cos ; 3 cos = 2

cos =3

2; = cos-1

3

2. . . . (iv)

This gives the angle at which the body goes off the surface.

The height from the ground of that point = AN = r(1 + cos)

= r r

3

5

3

21 =

+

Illustration:

A point mass m starts from rest and slides down the surface of a frictionless

solid sphere of radius R as shown in the figure. At what angle will this body

break off the surface of sphere? Find the velocity with which it will break off.

Solution:

Applying COE, we have at the point A and B,

N

A

mg

C

RB

O

m

RO

R


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m

R

O

R

N

m

A

B

v

we have mgR (1 cos ) = 2mv2

1. . (i)

Force equation gives, mg cos N = mv2/R . . (ii)N = 0 for break off.

v = cosgR . . . . (iii)

Putting it in (i)

We get cos = 2/3

Putting this in (iii) we get

v = gR3

2.


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APEX INSTITUTE - 62, Nitikhand-III, Indirapuram, Ghaziabad Ph.No.-+91-9910817866, 0120-4901457,

Webs i te : w w w.apex i it . co .in /

SOLVED OBJECTIVE

Example 1. A block of mass M is pulled a distance x along a horizontal table. The work done by

the weight is

(a) 0 (b) Mg x

(c)2

mgl(d) none of these.

Solution: (a) 0E.P = (because of no vertical movement) work done = 0.E.P =

Example 2. The potential energy of a particle varies with position x according to the relation

x27x2)x(U 4 = the point2

3x = is point of

(a) Unstable equilibrium (b) Stable equilibrium

(c) neutral equilibrium (d) none of these.

Solution : (b) 027x8dx

dUF 3 === (for equilibrium)

02

324x24

dx

Ud2

2

2

2

>

==

stable equilibrium.

Example 3. A conservative force F is acting on a body and body moves from Point A to B and

then B to A then work done by the force is

(a) 0W < (b) 0W >(c) 0W = (d) 0W .

Solution : (c) When conservative force applied, work done by the force do not depend on path.

Example 4. It is kinetic energy of a body increases by 21%, the momentum of a body increase by

(a) 10 (b) 11

(c) 9 (d) 12.

Solution : (a) 11 mK2P =

22 mK2P =

12 K21.1K =

22 K)21.1(m2P =

1.121.1P

P

1

2 ==

% change = 10%


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Example 5. If a compressed spring is dissolved in acid

(a) Total energy of molecules of acid increase

(b) The energy of the acid remains constant

(c) The energy of the acid decrease

(d) none of the above.

Solution : (a) Compressed spring has some potential energy. When this spring dissolved in acid

P.E. of spring transfer to molecules of acid.

Example 6. There are two identical mass less springs A and B of spring constants AK and BK

respectively and BA KK > . Then(a) If they are compressed to same distance, then work done on A > work done on B

(b) If they are compressed by same force work done on A < work done on B

(c) If they are compressed by same force work done on A> work done on B

(d) Both a and b are correct.

Solution : (d) Case I:

Compressed by same distance2

AA xK2

1W =

2BB xK

2

1W =

Q BA KK >

BA WW > .

Case II:

Compressed by same force

BBAA xKxKF ==

A

AK

Fx =

B

BK

Fx =

2

A

A2AAA

K

FK

2

1xK

2

1W

==

A

2

K

F

2

1=

B

2

BK

F

2

1W =

AB WW > .

Example 7. Force acting on a particle moving in a straight line varies with the velocity of the

particle as KVF = . Where K is constant. The work done by this force in time t is(a) KVt (b) K2V2t2

(c) K2Vt (d) KV2t


20/50

20



Solution : (d) KVF =P = KV.V = KV2

W = Pt

tKV2=

Example 8. A particle is projected with a velocity and making an angle with horizontal. Thenpower of the gravitational force(a) varies linearly with time

(b) is constant throughout

(c) is negative for complete path

(d) none of the above.

Solution : (a) jvivv yx +=r

gtcosvvy =

jmgF =

V.FP =

)jviv(.)jmg( yx +=

ymgv=

)gtcosv(mg =

)cosmgvtmg(P 2 = .

tP

Example 9. A block of mass M slides down a rough inclined plane of inclination with horizontal,with initial velocity v and from a height h. At the bottom point. Its velocity becomes

zero. Then the work done by the friction force in stopping the block.

(a)

+ mghmv

2

1 2(b) mghmv

2

1 2

(c) 22mv

2

1mgh (d) mghmv2 .

Solution: (a) fg WWK +=

f2

wmghmv2

1

+=

Work done by the friction force =

+ mghmv

2

1 2

Example 10. A ball is released from the top of a tower. The ratio of work done by force of gravity

in first, third and fifth second of the motion of ball is

(a) 1 : 3 : 5 (b) 5 : 3 : 1

(c) 1 : 5 : 9 (d) 9 : 5 : 1.


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21



Solution : (c) Work done by the force = E.Pdistance travelled in nth second

)1n2(g2

1uS

n+=

here u = 0

1.g2

1S

1=

S3

= 5.g2

1

9.g2

1S

5=

321 W;W;W 331 mgS;mgS;mgS== 1 : 5 : 9.


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22


SOLVED EXAMPLES (SUBJECTIVE)

Example 1. An object of mass 5 kg falls from rest through a vertical distance of 20 m and attains a

velocity of 10 m/s. How much work is done by the resistance of the air on the object?

(g = 10 m/s2).

Solution: Work done by all forces = change in K.E...EKWW gravityair =+

2

2

1mvmghWair =+

mghmvWair

= 22

1

20105101052

1=

JWair 750=

Example 2. A particle of mass m moves along a straight line on smooth horizontal plane, acted

upon by a force delivering a constant power P. If the initial velocity of the particle is

zero, then find its displacement as a function of time t.

Solution: Work done in time t = change in K.E.

2

2

1. mvtP =

2/12t

m

Pv = 2/1t

m

P2

dt

dx=

=t

0

2/1

x

0

dttm

P2dx

3tm

P2

3

2x =

Example 3. A block is projected horizontally on a rough hori-

zontal floor. The coefficient of friction between the

block and the floor is . The block strikes a lightspring of stiffness k with a velocity v

0. Find the maxi-

mum compression of the spring.

Solution: Since the block slides and the spring is com-

pressed through a distance x the net retarding

force acting on it

= F = (kx + N) = ( mg + kx) Work done by net force for the displace-ment x, W = F. dx

mk

v

mx


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23


W = x

dxF0

KE = +x

dxkxmg

0

)(

+=

2

kxmgxmv

2

10

220

x2 + 02 2

0= v

k

mx

k

mg

x =2

4422

0

2

222

k

mv

k

gm

k

mg+

x =

+ 11

2

0

g

v

m

k

k

mg

Example 4. A small block is projected with a speed v0on a horizontal

track which turns into a semi circle (vertical) of radius R.

Find the minimum value of V0

so that the body will hit

the point A after leaving the track at its highest point.

The arrangement is shown in the figure, given that the

straight part is rough and the curved part is smooth. The

coefficient of friction is .

Solution: Let the block escape the point at C with a velocity v horizontally. Since it hits the

initial spot A after falling through a height 2R we can write (2R) = (1/2)gt2

where t = time of its fall.

t = 2 gR/

the distance AB = 2v gR/

d = 2v gR/ . . . . . (i)

work energy theorem is applied to the motion of the body from A to B leads

KE = Wf

mgdmvmv = 212

0

2

1

2

1

v0

= gdv 221

+ . . . .(ii)

Energy conservation between B and C yields

)2(2

1

2

1 221 Rmgmvmv =

v1

= gRv 42 + . . . .(iii)

N

N

kx

mg

R

Ad

B

v1

v

C

Rv0

smoothRough surface A


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24


when the bead escapes C, its minimum speed v can be given as

mgR

mv=

2

( the normal contact force = 0)

v = gR . . . . (iv)

By using (iii) and (iv) we obtain

v1

= gR5 . . . . . (v)

using (i) and (iv) we obtain

d = ( ) Rg

RgR 22 =

. . . . .(vi)

Putting the values of v1

and d in (ii) we obtain

v0

= )2(25 RggR +

v0 = gR)45( +

Example 5. Two smooth balls of mass m1

and m2

connected by a light inextensible string are at the

opposite points of horizontal diameter of a smooth semi cylindrical surface of radius

R. If m1

is released, find its speed at any angular distance moved m2.

Solution: Let the ball m2

move through an angle , the mass m will fall through a distanceh

1= R.

The ball m2

rises through a height h2

as, h2

= R sinThe change in gravitational potential energy of m

1is

PE1 = -m1gh1 = = -m1 gR(since m1

loses its potential energy as it falls down).

The change in gravitational potential energy of m2

is

PE2

= m2gh

2= m

2gRsin

(Since m2

gains potential energy as it rises up)

The total change in gravitational potential energy= PE = PE

1+ PE

2

PE = -m1gR + m

2gR sin = gR(m

2sin - m

1) . . . . .(i)

= KE =2

)(

2

1

2

12

212

2

2

1

vmmvmvm

+=+ . . . . .(ii)

where v = speed of m1

and m2

at the positions as shown in the figure.

From the principle of conservation of energy we obtain,

KE + PE = 0 . . . . . (iii)Using (i), (ii) and (iii), we obtain,

0)sin()(2

121

2

21 =+ mmgRvmm

v =)(

)sin(2

21

21

mm

mmgR

+

.

R

m2

m2

m1 h2

v

m1

v h1


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25


Example 6: In Figure (a) and (b) AC, DG and GF are fixed inclined planes. BC = EF = x and AB =

DE = y. A small block of mass m is released from rest from the point A. Its slides

down AC and reaches C with a speed Cv . The same blocked is released from rest from

the point D; it slides down DGF and reaches the point F with speed Fv . The coefficient

of kinetic friction between the block and the surfaces AC and DGF is . Calculate Cvand

F

v .

R

s

C

A

B

(a)

mg

mg cos

x

y

(b)

x

y

x1 x2

s2

s1

G

D

E F

Solution:(a) ME at A = mgy + 0 and if Cv is the velocity at C,

So ME at C = 0 + (1/2) 2Cmv

Loss in ME = mgy (1/2) 2Cmv .

This loss in ME is equal to work done against friction i.e.,

scosmgmv)2/1(mgy 2C =

mxmgymv)2/1( 2C = g [as s/xcos = ]

or )xy(g2vc = ...(i)

(b) In this situation,

22112F sfsfmv)2/1(mgy +=

or21

2F scosmgscosmgmgymv)2/1( =

or )s/x(cosas][xxy[gv)2/1( 11212F == and )]s/x(cos 22=

or CF v)xy(g2v == [as xxx 21 =+ ] ...(ii)

Example 7. In the figure shown a massless spring of stiffness k and natural

length 0l is rigidly attached to a block of mass m and is invertical position. A wooden ball of mass m is released fromrest to fall under gravity. Having fallen a height h the ball strikesthe spring and gets stuck up in the spring at the top. What

should be the minimum value of h so that the lower block willjust lose contact with the ground later on? Find also the

corresponding maximum compression in the spring. Assume

thatk

mg4l0 >> . Neglect any loss of energy..

Solution: The minimum force needed to lift the lower block is equal to its weight. During upwardmotion the spring will get elongated. If elongation in the spring for just lifting the blockis 0x then

mgkx 0 =

m

k l0

h

m


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26


k

mgx 0 = ...(i)

From COE

20000 kx

2

1)xl(mg)hl(mg ++=+

200 kx2

1mgxmgh +=

k

gm

2

1

k

)mg(mgh

222

+=

k2

mg3h =

During down ward motion, suppose maximum compression in the spring is x From

COE.

200 kx

21)xl(mg)hl(mg +=+

2kx2

1mgxmgh += 2kx

2

1mgx

k2

mg3mg +=

222 xkkxmg2)mg(3 +=

0)mg(3mgkx2xk 222 =

2

222

k2

)mg(k12)mgk(4mgk2x

+=

2k2

mgk4mgk2 =

k

mg3x = .

Example 8. Two bodies A and B connected by a light rigid bar 10 m long

move in two frictionless guides as shown in the Figure. If B

starts from rest when it is vertically below A, find the

velocity of B when x = 6m. Assume kg200mm BA == and

mc = 100kg.Solution: At the instant, when the bar is as shown in the figure,

x2 + y2 = l2

0dt

dyy2

dt

dxx2 =+ ...(i)

dt

dyy

dt

dxx = ...(ii)

where =dt

dxvelocity of B and =

dt

dyvelocity of A.

c

10 m

A

BO

x


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27


Applying the law of conservation of energy,

loss of potential energy of A, if it is going down when the rod is vertical to the

position shown in the Figure

8.92002)810(gmA ==C moves down 6 m since B moves 6 m along x-axis

Loss of potential energy of 68.9100)mgh(C ==Total loss of potential energy = 200 68.910028.9 +

J9800108.9100 == .This must be equal to kinetic energy gained

Kinetic energy gained2

C2

BB2

AA )v(m2

1)v(m

2

1)v(m

2

1++=

222

dt

dx100

2

1

dt

dx200

2

1

dt

dy200

2

1

+

+

=

22

dt

dx150dt

dy100

+

=

22

dt

dx150dt

dx

y

x100

+

= from (ii)

22

dt

dx150

dt

dx

8

6100

+

=

2

dt

dx150

16

9100

+= 2

Bv

16

3300=

9800v16

3300 2B =

1B ms9.633

247

33

1698v ==

=

velocity of B at the required moment = 6.9 ms-1.

Example 9. A locomotive of mass m starts moving so that its velocity varies according to the law

s= , where is constant and s is the distance covered. Find the total work done

by all the forces acting on the locomotive during the first t seconds after the beginning

of motion.

Solution: Given v = sDifferentiating w.r.t. t, we get

==

s2dt

ds

s2

1

dt

dv 2/1

2s

s2

2=

=

acceleration,2

a2

=

Now force acting on the Locomotive,

c

A

BO

x

yl


28/50

28


2mmaF

2== .

Here u = 0

Now, using2at

2

1uts += , we have

4

tt

22

10s

222

2 =

+=

Work done,4

t

2mFsW

222

==

8

tm24

= .

Example 10. The kinetic energy of a particle moving along a circle of radius R depends on the

distance covered S as T 2S= , where is constant. Find the force acting on theparticle as a function of S.

Solution: K.E., T = 2S or22 Smv

2

1=

m

S2v

22 = ...(i)

Differentiating both sides w.r.t. t, we have

vm

S4

dt

dS

m

S4

dt

dvv2

=

=

= v

dt

dSQ

or tam

S2

dt

dv=

=

(Tangential component of acceleration)

Now, centripetal acceleration is

mR

S2

R

va

22

c

== [Using equation (i)]

Net acceleration of the particle is given by2c

2t aaa +=

222

mR

S2

m

S2

+

=

2

R

S1

m

S2

+

=

Now, force acting on the particle is given by,

2

2

R

S1

m

S2mmaF +

==

or 2

2

R

S1S2F += .


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29


Example 11. A 0.5 kg block slides from the point A (see figure) on a horizontal track with an initial

speed of 3m/s towards a weightless horizontal spring of length 1 m and force constant 2

Newton/m. The part AB of the track is frictionless and the part BC has the coefficients

of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are

2 m and 2.14 m respectively find the total distance through which the block moves

before it comes to rest completely (Take g = 10 m/s2).

A B D C

Solution: Suppose the block comes to rest at the point E i.e. let DE = x. The kinetic energy of the

block is spent in overcoming friction and compressing the spring through a distance

DE = x.

A B D C

m2 2.14 x

E

m m

Kinetic energy of the block2mv

2

1=

J25.235.02

1 2 == ...(i)

As the part AB of the track is frictionless, work done in moving from A to B is zero

Let normal reaction of the block = mg

Coefficient of friction =

Force due to friction along the track BC = mg

105.02.0 == 1 N

Distance through which the block moves against the frictional force

= 2.14 + x m

Work done by block against friction before it comes to rest)x14.2(mg +=

= ( 2.14 + x )J ...(ii)

Let the spring constant = k

Work done by the block in compressing the spring through distance x

2kx2

1=

22 xx22

1 == J ...(iii)

Adding (ii) and (iii) and equating it to (i), we get

25.2xx14.2 2 =++

or 011.0xx 2 =+

or 011x100x100 2 =+or 0)1x10)(11x10( =+


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30


10

11x = or

10

1x =

Since10

11x ,

m1.0101x ==

Restoring force of the spring = kx

N2.01.02 == ...(iv)Static frictional force of the block

105.022.0mgstatic == 1.1 N ...(v)

From (iv) and (v) it is clear that the static frictional force is grater than the restoring

force of the spring. Therefore the block will not move in the backward direction. Hence

the total distance through which the block moves before it comes to rest completely

2.00 + 2.14 + 0.10 = 4.24 metres.

Example 12. In a spring gun having spring constant

100 N/m, a small ball of mass 0.1 kgis put in its barrel by compressing thespring through 0.05 m as shown inthe figure.(a) Find the velocity of the ball whenspring is released.(b) Where should a box be placed on

ground so that ball falls in it, if the ball

leaves the gun horizontally at a height

of 2 m above the ground (g = 10 m/s2)

Solution: (a) When the spring is released its elastic potential energy is converted into kinetic

energy2

1mv2 , so

22 kx2

1mv

2

1=

v =2

5m/s.

(b) As vertical component of velocity of ball is zero. Time taken by the ball to reach the

ground ,2

gt2

1

h =

t =5

2

g

h2= seconds

So, the horizontal distance travelled by the ball in this time

d = v . t =5

2

2

5 = 1 m

v

h=2 m

d


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Example 13. A smooth, light horizontal rod AB can rotate about

a vertical axis passing through its end A. The rod is

fitted with a small sleeve of mass m attached to the

end A by a weightless spring of length l0and stiffness

k. What works must be performed to slowly get

this system going and reach the angular velocity .

Solution: The mass m rotates in a circle of radius l , which is the extended length of the spring.

Centripital force on m = k(l - l0) = m 2l

or, l =n1

0

l

where n =k

m 2

W =change in KE of m + energy stored in the spring

=2

1m2 2l +

20 )(k

2

1ll

=2

1m

2

00

2

220

n1

k

2

1

)n1(

+

l

ll

W =

+

22

2

2

nk

m

)n1(

k

2

1 l

Example 14. A particle is suspended by a string of length l. It isprojected with such a velocity v along the horizontalthat after the string become slack it flies through itsinitial position. Find v.

Solution: Let the velocity be v at B where the string becomeslack and the string makes angle with horizontal by the law of conservation ofenergy.

)++= sin1(mgvm2

1mv

2

1 22l (i)

or, v2 = v2 2gl (1+ sin ) (ii)By the dynamics of circular motion

mg sin =l

2vm

v2 = gl sin ...(iii)from equation (ii) and (iii) we get

g l sin = v2 2gl (1 + sin ) ...(iv)At B the particle becomes a projectile of velocity v at 90 - with the horizontal.Here, u

x= v sin & v

y= v cos

ax

= 0 & ay

= -g

l cos = v sin t (v)

t =

sinv

cosl& l (1+ sin ) = v cos

22

22

sinv

cosg

2

1

sinv

cos ll

2 sin3 + 3 sin2 - 1 = 0

m

0

A B

v

O

mgmg cos

mg sin

A v

B'v


32/50

32


sin =2

1is the acceptable solution

v2 = 2gl + 3gl 2

1=

2

g7 l v =

2

g7 l(from equation (iv))

Example 15. A system consists of two identical cubes each of mass m linked together

by a compressed massless spring of stiffness k. the cubes are connected

by a thread which is burnt at a certain moment. At what value of initial

compression of the spring will the lower cube bounce up after thethread is burnt.

Solution : Let us take horizontal line through the C.G. of lower cube as reference level. Let l

be the natural length of the spring.

T.E.initial

= )+ l(mgk2

1 2(i)

Let the spring extend by x after the thread is burnt

T.E.final

= )x(mgkx2

1 2 ++ l

kx = -k or k 2mg (-k not acceptable) kx = k 2mg.

Cube will bounce up when kx mg K 2mg mg

K 3mg = 3mg / k.

k

m

m

k

m

m


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33



OBJECTIVE

LEVEL - I

1. A force )jxiy(kF += , where k is a positive constant, acts on a particle moving in the xy

plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a,

0), and the parallel to the y-axis to the point (a, a). The total work done by the force on the

particle is

(a) 2ka2 (b) 2ka2

(c) 2ka (d) 2ka .

2. Supposing that the earth of mass m moves around the sun in a circular orbit of radius R,

the work done in half revolution is

(a) RR

mv2 (b) R2

R

mv2

(c) Zero (d) none of these

3. When water falls from the top of a water fall 100 m high

(a) it freezes (b) it warms up slightly

(c) it evaporates (d) there is no change in temperature.

4. A string of mass m end length l rests over a frictionless table with 1/4th of its length

hanging from a side. The work done in bringing the hanging part back on the table is(a) mgl/4 (b) mgl/32

(c) mgl/16 (d) none of these

5. A weight mg is suspended from a spring. If the elongation in the spring is x0, the elastic

energy stored in it is

(a) 0mgx2

1(b) 2mgx

0

(c) mgx0

(d) 0mgx4

1

6. Mechanical Energy is conserved under

(a) Conservative system of forces (b) disipative forces

(c) (a) and (b) (d) none of these

7. The same retarding force is applied to stop a train. If the speed is doubled then the dis-

tance will be

(a) eight times (b) doubled

(c) half (d) four times


34/50

34



8. Work done in time t on a body of mass m which is accelerated from rest to a speed v in

time t1

as a function of time t is given by

(a)

2

1 tt

v

m2

1

(b)

2

1 tt

v

m

(c)2

2

1

ttt

mv

2

1

(d)

2

2

1

2

tt

vm

2

1

9. A particle moves under the effect of a force F = kx2 from x = 0 to x = 4 the work done by

force is

(a)3

k8(b)

3

k32

(c)3

k64(d)

3

k128

10. Potential energy function describing the interaction between two atoms of a diatomic

molecule is

612 r

b

r

a)r(U =

Force acting between them will be zero when the distance between them would be

(a)

6/1

ba2

(b)

6/1

a2b

(c)

6/1

b

a

(d)

6/1

a

b

11. A ball is thrown up with a certain velocity at angle to the horizontal. The kinetic energyKE of the ball varies with horizontal displacement x as:

(a) (b)

(c) (d)

x

KE

O

x

KE

O

x

KE

O x

KE

O


35/50

35



12. A body m1

is projected upwards with velocity v1

another body m1

of same mass is projected

at an angle of 450. Both reach the same height. What is the ratio of their kinetic energies at

point of projection

(a) 1 (b) 1/2(c) 1/3 (d) 1/4

13. A 2kg block is dropped from a height of 0.4 m on a spring of force constant 2000N/m. The

max compression of the srpring is:-

(a) 0.1 m (b) 0.2 m

(c) 0.01 m (d) 0.02 m

14. The kinetic energy of body is increased by 300% its momentum will be increased by

(a) 100% (b) 200%

(c) 300% (d) 400%

15. A particle of mass M is moving in a horizontal circle of radius R under the centripetal

force equal to K/R2, where K is constant. The potential energy of the particle is

(a) K/2R (b) -K/2R

(c) K/R (d) -K/R


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36



LEVEL - II

1. An object of mass m is tied to a string of length l and a variable horizontal

force is applied on it, which initially, is zero and gradually increases

until the string makes an angle with the vertical. Workdone by theforce F is

(a) mgl(1 sin) (b) mgl(c) mgl(1 cos) (d) mgl(1 + cos )

2. A simple pendulum has a string of length l and bob of mass m. When the bob is at its lowest

position, it is given the minimum horizontal speed necessary for it to move in a circular path

about the point of suspension. The tension in the string at the lowest position of the bob is

(a) 3mg (b) 4mg

(c) 5 mg (d) 6 mg

3. A horse pulls a wagon with a force of 360 N at an angle of 60 with the horizontal at a speed

of 10 Km/hr. The power of the horse is

(a) 1000 W (b) 2000 W

(c) 500 W (d) 750 W.

4. A particle of mass m is fixed to one end of a light rigid rod of length l and rotated in a

vertical circular path about its other end. The minimum speed of the particle at its highest

point must be

(a) zero (b) lg

(c) lg5.1 (d) lg2

5. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched

length l. The system is rotated about the other end of the spring with an angular velocity ,in gravity free space. The increase in length of the spring will be

(a)k

l2m

(b) 2

2

m

m

-k

l

(c) 2

2

m

m

+

k

l(d) none of these

6. A particle of mass m is moving in a circular path of constant radius r such that its centripetal

acceleration acis varying with time as a

c= k2rt2, where k is a constant. The power delivered

to the particle by the forces acting on it is

(a) 2 mk2r2t (b) mk 2r2t

(c)3

1mk4r2t5 (d) 0


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TT

m

M

d

7. In the figure shown, the net work done by the tension when the bigger

block of mass M touches the ground is

(a) + Mgd

(b) (M + m) gd(c) Mgd

(d) zero

8. In the figure, a ball A is released from rest when the spring is at its

natural (unstretched) length. For the block B, of mass M to leave

contact with the ground at some stage, the minimum mass of A must

be

(a) 2 M

(b) M

(c) M/2

(d) a function of M and the force constant of the spring

9. A man pulls a bucket of water from a well of depth H. If the mass of the rope and that of the

bucket full of water are m and M respectively, then the work done by the man is

(a) (m + M)gh (b)

+ M

2

mgh

(c) gh2

Mm

+(d)

+2

Mm gh

10. A simple pendulum having a bob of mass m is suspended from the ceiling of a car used in a

stunt film shooting. The car moves up along an inclined cliff at a speed v and makes a jump

to leave the cliff and lands at some distance. Let R be the maximum height of the car from

the top of the cliff. The tension in the string when the car is in air is

(a) mg (b) mg R

mv2

(c) mg +

R

mv2

(d) zero.

11. A small block of mass m is kept on a rough inclined surface of inclination fixed in aelevator. 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 sin 2

B M

A


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12. Two equal masses are attached to the two ends of a spring of 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)2

1kx2 (b)

2

1kx2

(c)4

1kx2 (d)

4

1kx2

13. A bent bow used for shooting an arrow possesses :

(a) Kinetic Energy (b) Potential Energy

(c) Heat Energy (d) Chemical Energy.

14.A block of mass m moving with speed v compresses a spring through distance x before itsspeed is halved. What is the value of spring constant ?

(a) 2

2

x4

mv3(b) 2

2

x4

mv

(c) 2

2

x2

mv(d) 2

2

x

mv2

15. A ball falls under gravity from a height 10 m with an initial velocity v0. Ithits the ground,

losses 50% of its energy in collision and it rises to the same height. What is the value of v0?

(a) 14 m/s (b) 7 m/s

(c) 28 m/s (d) 9.8 m/s.


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LEVEL - 1(CBSE)(REVIEWYOURCONCEPTS)

1. A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of

7 N on a table with coefficient of kinetic fiction = 0.1. Compute the

(a) work done by the applied force in 10 s,

(b) work done by friction in 10s

(c) work done by the net force on the body in 10 s

(d) change in kinetic energy of the body in 10 s.

2. The potential energy function for particle executing linear

simple harmonic motion is given by 2/kx)x(V 2= . Where

k is the force constant of the oscillator. For k = 0.5 N m-1,the graph of )x(V versus x is shown in Figure. Show that

a particle of total energy 1 J moving under this potential

must turn back when it reaches x = 2 m.

3. A body is initially at rest. It undergoes one-dimensional motion with constant

acceleration. Show that power delivered to it at time t is proportional to t.

4. A body constrained to move along the z-axis of a coordinate system is subject to a

constant force F given by

Nk3j2iF ++=

where k,j,i are unit vectors along the x y and z-axis of the system respectively. What is

the work done by this force in moving the body a distance of 4 m along the z-axis?

5. A rain drop of radius 2 mm falls from a height of 500 m above the ground. It falls with

decreasing acceleration (due to viscous resistance of the air) until at half its original height,

it attains its maximum (terminal) speed, and moves with unifrom speed thereafter. What is

the work done by the gravitational force on the drop in the first and second half of its

journey? What is the work done by the resistive force in the entire journey if its speed on

reaching the ground is 10 ms-1

?

6. A particle of mass 0.5 kg travels in a straight line with velocity 2/3axv = where a = 5 12/1 sm .What is the work done by the net force during its displacement from x = 0 to x = 2 m?

7. A large family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal

surface at an average rate of 200 W per square meter. If 20% of this energy can be converted

to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this

area to that of the roof of a typical house.

V(x)

x


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8. A 1 kg block situated on a rough incline is connected to a

spring of spring constant 100 Nm-1 as shown in Figure. The

block is released from rest with the spring in the unstretched

position. The block moves 10 cm down the incline beforecoming to rest. Find the coefficient of friction between the

block and the incline. Assume that the spring has negligible

mass and the pulley is frictionless.

9. A pump on the ground floor of a building can pump up water to fill a tank of volume 30m3

in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30% , how

much electric power is consumed by the pump?

10. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform

speed of 7 ms-1. It hits the floor of the elevator (length of the elevator = 3m) and does not

rebound. What is the heat produced by the impact. Would your answer be different if the

elevator were stationary?

11. A particle of mass 0.01 kg travels along a space curve with velocity given by 1msk16i4 + .

After some time, its velocity becomes 1msj20i8 + due to the action of a conservative force.

Calculate the work done on particle during this interval of time

12. A particle moves in a straight line with retardation proportional to its displacement. Show

that its loss of kinetic energy for any displacement x is proportional to x2 .

13. A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to

its equilibrium position. This stretches the spring by a length d. If the same body attached to

the same spring is allowed to fall suddenly, what would the maximum stretching in this case

14. A spring obeys Hookes law with a force constant k. It required 4 J of work to stretch it

through 10 cm beyond its unstretched length. Calculate (a) value of force constant k and (b)

the extra work required to stretch it through additional 10 cm.

15. Show that the kinetic energy acquired by a body of mass m in traveling a certain distance

starting from rest, under a constant force is independent of m.

1 kg

37

k = 100 N/m


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LEVEL - II(BRUSHUPYOURCONCEPTS)

1. A ring of mass m slides on a smooth vertical rod ; attached to the ring is alight string passing over a smooth peg distant a from the rod, and at the other

end of the string is a mass M ( > m). The ring is held on a level with the peg

and released :

Show that it first comes to rest after falling a distance : 22 mM

mMa2

.

2. A small body A starts sliding from the height h down an

inclined groove passing into a half - circle of radius h/2 (see

figure). Assuming the friction to be negligible, find the velocity

of the body at the highest point of its trajectory (after breaking

off the groove).

3. An ideal massless spring can be compressed by 1 m by a

force of 100 N. This same spring is placed at the bottom of

a frictionless inclined plane which makes an angle = 300

with the horizontal. A 10 kg mass is released from rest at

the top of the incline and is brought to rest momentarily

after compressing the spring 2 meters.

(a) Through what distance does the mass slide before

coming to rest ?

(b) What is the speed of the mass just before it reaches the spring ?

4. A spring of mass m and stiffness k is fitted to a block of mass

M. The system is moving with a constant velocity v on a

smooth horizontal surface. If the system collides with a wall,

find the maximum compression of the spring before it re-

coils, assuming that the total energy is conserved.

5. A heavy particle hangs from a point O, by string of length a. It is projected horizontally with

a velocity v such that v2 = (2 + 3 ) ag, show that the string becomes slack when it has

described an angle cos-1(-1/ 3 ).

6. A stone with weight w is thrown vertically upward into the air with initial speed v0. If a

constant force f due to air drag acts on the stone throughout its flight.

(a) Show that the maximum height reached by the stone is]w/f1[g2

vh

2

0

+=

(b) Show that the speed of the stone upon impact with the ground is

2/1

0fw

fwvv

+

=

300

mM

v


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7. A heavy particle hangs by a inextensible string of length a from a fixed point and is then

projected horizontally with a velocity )gh2( . If (5/2)a > h >a, Prove that circular motion

ceases when the greatest height ever reaches by the particle above the point of projection is

(4a - h)(a + 2h)2/(27a2).

8. A particle is projected, along the inside of a smooth fixed sphere, from the lowest point, with

a velocity equal to the due to falling freely down the vertical diameter of the sphere. Show

that the particle will leave the sphere and afterwards pass vertically over the point of projec-

tion at a distance equal to (25/32) of the diameter.

9. Show that a particle projected with velocity )ag2( from the lowest point of a vertical

circle of radius a and moving inside it will just reach the end of the horizontal diameter;

while if projected with velocity )ag5( , it will just reach the highest point. Prove that the

reaction at any point in the first case is proportional to the depth below the horizontal diam-

eter and in the second case to the depth below the highest point.

10. An automobile of mass m accelerates, starting from rest, while the engine supplies

constant power P, show that

(a) The velocity is given as a function of time by 2/1)m/pt2(v =

(b) The position is given as a function of time by2/3

2/1

tm9

P8s

= .

11. A particle of mass m moves along a circle of radius R with a normal acceleration varying

with time as 2n bta = , where b is a constant. Find the time dependence of the power developed

by all the forces acting on the particle, and the mean value of this power developed by all the

forces acting on the particle, and the mean value of this power averaged over the first t

seconds after the beginning of motion.

12. Two bars of masses m1and m

2connected by a weightless spring of stiffness k (figure) rest on

a smooth horizontal plane.

1 2x

Bar 2 is shifted a small distance x to the left and then released. Find the velocity of the centre

of inertia of the system after bar 1 breaks off the wall.

13. In a certain two-dimensional field of force the potential energy of a particle has the form

22 yxU += , where and are positive constants whose magnitudes are diferent. Find out :(a) whether this field is central:

(b) what is the shape of the equipotential surfaces and also of the surfaces for which the magni-

tude of the vector of force F = const.


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14. One end of a spring of natural length h and spring constant k

is fixed at the ground and the other is fitted with a smooth

ring of mass m which is allowed to slide on a horizontal rod

fixed at a height h (see figure). Initially, the spring makes anangle of 37 with the vertical when the system is released

from rest. Find the speed of the ring when the spring becomes

vertial.

15. A car is travelling on a level road with speed v0

at the instant when the brakes lock, so the

tires slide rather than roll.

(a) Use the work energy theorem to calculate the minimum stopping distance of the car in

terms of v0, the acceleration of gravity g, and the coefficient of kinetic friction

kbetween

the tires and the road.

(b) The car stops in a distance of 98.3 m if v0

= 90 km/h. What is the stopping distance if

v0

= 60 km/h ? Assume that k

remains the same, so that the friction force remains the

same.

LEVEL - III(CHECKYOURSKILLS)

1. A body with zero initial velocity slips from the top of an inclined plane forming an angle with the horizontal. The coefficient of friction between the body and the plane increases

with the distance l from the top according to the law bl= . The body stops before it reachesthe end of the plane. Determine the time t from the beginning of motion of the body to the

moment when its comes to rest.

2. In the reference frame K two particles travel along the x-axis one of mass m1

with velocity

v1, and the other of mass m

2with velocity v

2. Find:

(a) The velocity v of the reference frame k in which the cumulative K.E. of these particles

is minimum.

(b) The cumulative K.E. of these particles in the k frame

3. A thin rim of mass m and radius r rolls down an inclinedplane of slope , winding thereby a thin ribbon of lineardensity (shown in the figure). At the initial moment, the rim

is at a height h above the horizontal surface.

Determine the distance s from the foot of the inclined plane

at which the rim stops, assuming that the incline plane

smoothly changes into the horizontal plane.

4. A small particle of mass m initially at A (see Figure) slides down a frictionless surface

AEB. When the particle is at the point C, show that the angular velocity and the force

h

r


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exerted by the surface are

r

sing2 = and = sinmg3F

5. A chain AB of length l is loaded in a smooth horizontal tube so that

its fraction of length h hangs freely and touches the surface of the

table with its end B. At a certain moment, the end A of the chain is

set free, with what velocity will this end of the chain slip out of the

tube?

6. Figure shows a smooth track, a part of which is a circle of a radius

R. A block of mass m is pushed against a spring constant k fixed

at the left end and is then released. Find the initial compression of

the spring so that the block presses the track with a force mg when

it reaches the point P, where the radius of the track is horizontal.

7. Three identical spring A, B and C each of natural length l and

spring constant K are connected to a point mass m as shown in the

figure. A and B are horizontal and C is vertically fixed with rigid

supports. What is the work done by the external agent in slowly

lowering the mass m till it attains equilibrium when the springs A

and B make an angle 2 sin1 3/5 between them. Neglect the masses

of the springs.

8. A small sphere tied to the string of length 0.8m is describing a vertical circle so that the

maximum and minimum tensions in the strings are in the ratio 3:1. The fixed end of the

string is at a height of 5.8m above ground.

(a) Find the velocity of the sphere at the lowest position.

(b) If the string suddenly breaks at the lowest position, when and where will the sphere hit

the ground? (take = 10 m/s2)

9. A body of mass m was slowly hauled up the hill. (fig) by a

force F which at each point was directed along a tangent to thetrajectory. Find the work performed by this force, if the height

of the hill is h, the length of its base l, and coefficient of

kinetic friction k.

10. A small bar resting on a smooth horizontal plane is attached by

threads to a point P (fig.) and by means of a weightless pulley, to

a weight B possessing the same mass as the bar itself. Besides,

the bar is also attached to a point O by means of a light non-

deformed spring of length l0

= 50 cm and the stiffness

K = 5 mg/l0

, where m is the mass of the bar. The thread PA having

E

m

l

h

B

l0

P

O

A

m

BA r

E

C

A

h

B


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m

O

l0

m m

l

N

F

mg

R

B /4

m1

been burned, the bar starts moving. Find its velocity at the moment when it is breaking off

the plane.

11. A horizontal plane supports a plank with a bar of mass m = 1.0kg placed on it and attached by a light elastic non-deformed

cord of length l0

= 40 cm to a point O (fig.). The coefficient of

friction between the bar and the plank equals k = 0.20. The

plank is slowly shifted to the right until the bar starts sliding

over it. It occurs at the moment when the cord deviates from

the vertical by an angle = 300. Find the work that has beenperformed by that moment by the friction force acting on the

bar in the reference frame fixed to the plane.

12. A block of mass m connected with a light spring of natural

length l0

and stiffness k as shown in the figure. The spring

is relaxed the block is pulled by an external force slowly.

Find the work done by the external agent till the block

break off the surface.

= 1

0kl

mgGiven .

13. The figure shows a ball A of mass m connected to a light spring

of stiffness k. Another identical ball B is connected with the ball

A by a light inextensible string as shown in the figure. Other end

of the spring is fixed. Initially the spring is in relaxed position.A vertical force F acts on B such that the balls move slowly.

What is the work done by the force in pulling the ball B till that

ball A reaches at the top of the cylindrical surface the ball A

remains in contact with the surface and coefficient of friction

between the surface and the ball A is .

14. A weightless horizontal rigid rod along which two ball of the

same mass m can move without friction rotates at a constant

angular velocity about a vertical axle. The ball are connectedby a weightless spring of rigidity k, whose length in the

undeformed state is l0. The ball which is closer to the vertical

axle is connected to it by a similar spring. Determine the lengths of the springs. Under what

conditions will the balls move in the circle ?

15. Two blocks P and Q of mass 2 m and m respectively are

connected by a massless string and are at rest as shown in

figure all pulleys are ideal and the surface is frictionless. Find

the velocity of the block P at point A and B when the system

is released from rest. [at A, thread from P to pulley is vertical]

A


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LEVEL - IV (IIT-JEE PROBLEMS)

(JUDGEYOURSELFAT JEE-LEVEL)

1. A particle is suspended vertically from a point O by an inextensible massless string

of length L. A vertical line AB is is at a distance of LS from O as shown. The object

is given a horizontal velocity u. At some point, its motion ceases to be circular and

eventually the object passed through the line AB. At the instant of crossing AB, its

velocity is horizontal. . Find u. [I.I .T. 1999]

2. A cart is moving along x direction with a velocity of 4 m/s. A person on the cart throws a stone with

a velocity of 6 m/s relative to himself. In the frame of reference of the cart, the stone is thrown in y-

z plane making an angle of 300 with vertical z-axis. At the highest point of its trajectory, the stone

hits an object of equal mass hung vertically from branch of a tree by means of a string of length L.

A completely inelastic collision occurs in which the stone gets embedded in the object. Determine

(a) The speed of the combined mass immediately after the collision with respect to an observer

on the ground.

(b) The length L of the string such that the tension in the string becomes zero when the string

becomes horizontal during the subsequent motion of the combined mass.

[ I.I .T. 1997 New]

3. A spherical ball of mass m is kept at the highest point in the space

between two fixed, concentric spheres A and B (see Figure). Thesmaller sphere A has a radius R and the space between the two

spheres has a width d. The ball has a diameter very slightly less

than d. All surfaces are frictionless. The ball is given a gentle

push (towards the right in the figure). The angle made by the

radius vector of the ball with the upward vertical is denoted by (Shown in the figure).

(a) express the total normal reaction force exerted by the spheres on the ball as a function of

angle .(b) let N

Aand N

Bdenote the magnitudes of the normal reaction forces on the ball exerted by the

spheres A and B, respectively. Sketch the variations of NA and NB as functions of cos in therange 0 < by drawing two separate graphs in your answer book, taking cos on thehorizontal axes.

[I.I .T. 2002]

L

L8

O A

u

B

RO

d

Sphere B

Sphere A


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SOLUTION (OBJECTIVE)

LEVEL - I

1. (c)2. (c)

3. (b)

4. (b)

5. (a)

6. (a)

7. (d)

8. (d)

9. (c)

10. (a)

11. (c)12. (b)

13. (a)

14. (a)

15. (c)

LEVEL - II

1. (c)

2. (d)

3. (c)

4. (a)

5. (b)

6. (b)

7. (d)

8. (c)

9. (b)

10. (d)

11. (c)

12. (d)

13. (b)14. (a)

15. (a)


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SOLUTION (SUBJECTIVE)

LEVEL - I (CBSE)

1. (a) 882 J (b) -247 J

(c) 635 J (d) 635 J Work done by the net force on a body equals change in its K.E.

4. 12 J

5. 0.082 J in each half, 0.163 J

6. 50 J

7. (a) 200 m2 (b) comparable to the roof of a large house of dimension 14 m 14 m

8. 0.125 9. 43.6 KW

10. 8.82 J for both cases.

11. 0.96 J 13. 2 d

14. (a) 800 N (b) 12 J

LEVEL - II

2. v =3

gh

3

2

3. (a) 4 m (b) s/m52

4. vk3

mM3 +

11. P = mRat

=2

mRat


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12. )mm/(kmxv 212c +=

13. (a) No;

(b) ellipses whose ratio of semiaxes is = /b/a ; also ellipses, but with a/b = / .

14.m

k

4

h

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