Notes on Work Energy Power for IIT(Mains+ Advance)

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 | cosθ Where θ 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 (Fcosθ) 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 θ is acute (θ < 900_{) and negative if the angle} θ _{is obtuse (}θ _{> 90}0_{). If the angle between}

→

F

and

→

S is 900 _then

work done by the force is zero.

If the force is variable then the work done by the variable force is given by dW = →

F

.dS→ or W =

∫

→ → 2 1 S S

dS

.

F

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

=

(

5

i

−

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

=

(

5

i

−

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/s2_{. Find the work}

done during the first one and a half seconds from the beginning of motion.

Solution :

The height to which the body is lifted during the first 't' second is h = 2 1

at2 _{tension in the}

rope T = mg + ma

∴

Work done = T.h = m(g +a)













2

2

1

at

= 3000 (10 + 2)

_



2

( )

1

5

2

_



2

1

.

x

x

= 81 KJ

WORK DONE BY A SPRING FORCE :

Whenever a spring is stretched or compressed, the spring force always tend to restore it to the equilibrium position. If x be the displacement of the free end of the spring from its equilibrium position then, the magnitude of the spring force is FS = - kx

The negative sign indicates that the force is restoring.

Ws =

−

∫

f x i x

kxdx

⇒ Ws =

(

2 2

)

2 1 i f x x k − −

WORK DONE BY FRICTION :

Work done by friction may be zero, positive or negative depending upon the situations: ΠWhen 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.

Ž When one block is placed on another block and is pulled by a force then friction force does negative work on top block and positive work on the lower block

WORK DONE BY GRAVITY :

Here the force of gravity is Fg = - mg

j

and the displacement is given by

→

S = ∆x

i

+ ∆y

j

+ ∆z

k

∴ Work done by gravity is Wg = g →

F

. → S = - mg ∆y ∆y = yf - yI = - h ∴ Wg = + mgh

If the block moves in the upward direction, then the work done by gravity is negative and is given by Wg = - mgh.

DEPENDENCE OF WORK ON FRAME OF REFERENCE :

Work depends upon the frame of reference from where it is calculated. As the displacement as well as force, depends upon the 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.

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 "µ" which in turn is placed on a smooth surface. The block is given a velocity

v0 with respect to the plank which comes to rest with respect to the plank. Find the

a) The total work done by friction in the plank frame.

b) The work done by friction on the smaller block in the plank frame. c) Find the final velocity of the plank

m

m

v

0

Solution :

The acceleration of the plank = Friction force applied by the block on the plank / mass of the plank.

g m

mg a_p = µ =µ

(a) Pseudo force acting on the block = µg (back wards) Force of friction is µmg ( acting backwards)

From the plank frame time needed to stop the block is given by O =

V

₀

+

at

(

a

=

−

2

µ

g

)

⇒ t =

g

V

µ

2

0

Velocity of the plank during this time is Vp =up +apt

=

2

2

0 0

V

g

V

g

=

µ

µ

Displacement of the block = S =

g V a V V µ =             ×       − 8 3 2 2 02 2 0 2 0

Work done by friction on the block =

( )

1

8

3

02

₋

µ

µ

=

π

g

V

.

mg

cos

.

S

.

F

= 20 8 3 mv −

(b) From the Plank frame

Work done by friction on smaller block = -µmgl

g

2

V

0

₀2

µ

−

−

=

l

⇒

2

mV

g

2 0

=

µ

l

⇒ work done by friction from the Plank frame =

2

mV

₀2

−

(c) Final velocity of the block = Velocity of the plank =

2 0 V mg µ p ma m

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

Q

v

_f2

=

v

_i2 + 2 ax ∴ W =

(

)

2 2 2 i f mv −v = 2 1 m

v

_f2- 2 1 m

v

_i2 The quantity k = 2 1

m v2 _{is a scalar and is called the kinetic energy of the particle. It is}

the energy posses by the particle by virtue of its motion. Thus the equation takes the form

W

=

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 → 0

v

= 3

i

- 4

j

to → f

v

= -6

j

+ 2

k

m/s. What is the change in K.E of the body?

Solution: Here m = 800gm = 0.8 kg → o

v

=

3

2

+

( )

−

4

2 = 5

v

→_f

=

( ) ( )

−

6

2

+

2

2 =

40

∴

change in K.E = 2 1 x 0.8         − → → 2 0 2 v v_f = x0.8x

(

40 25

)

6J 2 1 _{− =} 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 :

K.E = work done against friction 2 2 1 mv = F.s = µ N.s = µmgs ⇒ s = g v µ 2 2 =

( )

10 8 0 2 25 2 x . x ~ 39 m 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₎

Solution :

Let upward push of air be F

∴

The resultant downward force = mg - F As work done = gain in K.E

(mg - F) x S = 2 2 1 mv

∴

(10 x 10 - F) x 20 = 2 1 x 10 x (10)2 ⇒ _{F = 75 N} ∴ Work done by push of air = 75 x 20 = 15 Joule This work done is negative.

POTENTIAL ENERGY :

Potential energy of any body is the energy possessed by the body by virtue of its position or the state of deformation. With every potential energy there is an associated conservative force. The potential energy is measured as the magnitude of work done against the associated conservative force

du = -Fr.drr

For Example :

(i) If an object is placed at any point in gravitational field work is to be done against gravitational field force. The magnitude of this work done against the gravitational force gives the measure of gravitational potential energy of the body at that position which is U = mgh. Here h is the height of the object from the reference level.

ii) The magnitude of work done against the spring force to compress it gives the measure of elastic potential energy, which is U =

2 1

k x2

iii) A charged body in any electrostatic field will have electrostatic potential energy. The change in potential energy of a system associated with conservative internal force as U2

-U1= - W=

−

∫

2

1

F

. dr

CONSERVATION OF MECHANICAL ENERGY :

Change in potential energy ∆U = - WC where WC is the work done by conservative

forces. From work energy theorem Wnet = ∆k

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

subjected to 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

Q

∆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

∴

∆(K + U) = 0 ⇒ Ki + Ui = Kf + Uf

H A _θ

⇒ 2 1 mv2 _{+ mgH =} 2 1 mv' 2 _{+ 0} ⇒ 2 1 (50)2 _{+ 40 x 10 =} 2 1 v' 2 ⇒ (1250 + 400) x 2 = v' 2 ⇒ v' 2 _{= 3300} v' ~ 58 m/s

POWER

Power is defined as the rate at which work is done. If an amount of work ∆W is done in a time interval ∆t, then average power is defined to be

Pav = t W

∆ ∆

The S.I. unit of power is J/S or watt (W). Thus 1 W = 1 J/S

The instantaneous power is the limiting value of Pav as ∆t

→

0 that is P = dt dW

Instantaneous power may also be written as P =

=

→

F

.

→

v

dt

dW

Since work and energy are closely related, a more general definition of power is the rate of energy transfer from one body to another, or the rate at which energy is transformed from one form to another, i.e. P =

dt dE

. 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 = 2 1 m ₂2 ₁2 2 1 mv v −

P =

(

)

t

v

v

m

t

K

∆

−

=

∆

∆

22 12

2

1

=

(

)

60

10

20

500

2

1

_{2 2}

−

x

= 1250 W.

MOTION IN A VERTICAL CIRCLE :

A particle of mass 'm' is attached to a light and inextensible string. The other end of the sting is fixed at O and the particle moves in vertical circle of radius 'r' equal to the length of the string as shown in the fig. At the point P, net radial force on the particle is T-mg cosθ.

∴T - mg cosθ =

r

mv

2 ⇒ T = mg cosθ +

r

mv

2

The particle will complete the circle if the string does not slack even at the highest point (θ = π). Thus, tension in the string should be greater than or equal to zero (T > 0) at θ = π for critical situation T = 0 and θ = π

∴ mg =

R

mv

2min _⇒ 2 min

v

=

gR

⇒

v

_min = gR

Now conserving energy between the lowest and the highest point 2 1

_{( )}

R mg mv mu_{min min} 2 2 1 2 2 _{= +} ⇒

u

_min2

=

gR

+

4

gR

=

5

gR

gR u_min = 5

If u_min ≥ 5gR the particle will complete the circle. At u = 5gR , velocity at highest point is v = gR and tension in the string is zero.

O T P θ

θ

cos

mg

θ

sin

mg

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 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 T = mg at an angle θ as shown in figure h = l (1 - cosθ)

Conserving mechanical energy between A and B 2 1 mu2 ₌ 2 1 mv2 _{+ mgh} ⇒ u2 _{= v}2_{+ 2gh} ⇒ _v2 _{= u}2 _{- 2gh …. (i)} T - mg cosθ =

l

2

mv

⇒ T= mg cosθ +

l

2

mv

⇒ mg = mg cos θ +

l

2

mv

⇒ v2 _{= gl (1- cos}θ_{) ………. (ii)}

From (i) and (ii) u2 _{- 2gl (1 - cos}θ_{) = gl (1 - cos}θ₎ ⇒ cosθ = 3 2 ⇒ θ = cos-1













3

2

putting the value of cosθ in equation ………… (ii) v2 _{= gl}











 −

3

2

1

= 3 l g ⇒ v =

3

l

g

θ A u= gl

θ

cos

mg

θ sin mg B h

•

•

•

Equilibrium :

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 = -dr dU

⇒ 0 dr dU ₌

∴ At Equilibrium slope of U and r graph is Zero (or) Potential energy either maximum or minimum or constant at that position.

At the stable equilibrium position P.E is minimum At the unstable equilibrium position P.E is maximum

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.

Solution:

For Equilibrium F = 0 ∴ F = - 2(x 2) 0 dx dU _{=− − =} ⇒ x = 2 and

0

dx

U

d

2 2

<

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

WORKED OUT OBJECTIVE PROBLEMS

EXAMPLE : 01

A particle moves with a velocity 5

i

- 3

j

+ 6

k

m/s under the influence of a constant force

→

F

=

(

10

i

+

10

j

+

20

k

)

N. The instantaneous power applied to the particle is

A) 200 J/S B) 40 J/S C) 140 J/S D) 170 J/S Solution : P = →

F

.

V

→ = (5

i

-

3

j

+6

k

) . (10

i

+ 10

j

+20

k

) = 50 - 30 + 120 = 140 J/S EXAMPLE : 02

A 15 gm ball is shot from a spring gun whose spring has a force constant of 600 N/m. The spring is compressed by 5 cm. The greatest possible horizontal range of the ball for this compression is (g = 10 m/s2₎ A) 6.0 m B) 12.0 m C) 10.0 m D) 8.0 m Solution : R max =

g

u

2 =



_



_













mg

mu

2

2

1

2 = mg kx mg kx 2 2 2 2 1 ₌          













2

₌

2

kx

2

1

mu

2

1

Q

=

(

)

m x . . 10 10 015 0 05 0 600 2 ₌ .

[ Note : The actual value of 'u' will be less than the calculated value as some part of 1/2kx2 _is

used up in doing work against gravity when the spring regains its length]

EXAMPLE : 03

Force acting on a particle is (2

i

+ 3

j

) N. work done by this force is zero, when a particle is moved on the line 3y + kx = 5 Here value of k is

A) 3 B) 2 C) 1 D) 4

Solution :

and the given line can be written as y =

3 5 3 +

−k x

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

∴











−













3

2

3

k

= - 1 ⇒ k = 2 EXAMPLE : 04

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

2

3

t

2

watt. Here 't' is in second. If velocity of particle at t = 0 is v = 0. The velocity of particle at time t = 2 second will be A) 1 m/s B) 4 m/s C) 2 m/s D) 2

2

m/s Solution : kf - ki =

∫

2 0

dt

P

⇒ 2 1 mv2 ₌

∫

2 0

2

3

t2 _dt ⇒ _v2 ₌ 2 0 3

2

_



_











_t

Q

m = 2 kg ⇒ v = 2 m/s EXAMPLE : 05

A particle of mass 'm' is projected with velocity 'u' at an angle θ with horizontal. During the period when the particle descends from highest point to the position where its velocity vector makes an angle θ/2 with horizontal, work done by the gravity force is

A) 1/2 mu2 _tan2 θ_{/2 B) 1/2 mu}2 _tan2θ

C) 1/2 mu2 _cos2θ _tan2θ_{/2 D) 1/2 mu}2 _cos2θ_{/2 sin}2 θ

Solution :

As horizontal component of velocity does not change v cos θ/2 = ucos θ v =

2

θ

θ

cos

cos

u

Wgravity = ∆ K = 2 1 mv2 _- 2 1 m (u cosθ)2 = 2 1 mu2 _cos2θ _tan2 2 θ θ 2 / θ

θ

cos

u

V

u

EXAMPLE : 06

A body of mass 1 kg thrown upwards with a velocity of 10 m/s comes to rest (momentarily) after moving up 4 m. The work done by air drag in this process is (g = 10 m/s2₎

A) 10 J B) - 10 J C) 40 J D) 50 J

Solution :

From work energy theorem Wgr + Wair drag = ∆ k ⇒ - mgh + Wair drag = 0 - 2 1 mu2 ⇒ Wair drag = mgh - 2 1 mu2 _{= (40 - 50) J = - 10 J} EXAMPLE : 07

The potential energy of particle of mass 'm' is given by U = 2 1

kx2 _{for x < 0 and U = 0 for x >}

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

k

E

2

is A) zero B)

M

E

2

C)

m

E

D)

m

E

2

Solution :

Potential energy of particle at x =

k

E

2

is zero ∴ K.E = E ⇒ 2 1 mv2 _{= E or v =}

m

E

2

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 δ then A) K F < δ < K F 2 B) δ = K F 2

D) Increases in energy stored in spring is 2 1

kδ2

Solution :

If the mass of the hanging block be 'm' then elongation of spring is k mg

. Due to the applied force the additional stretching is δ

∴ F δ + mgδ = 2 1 K K g m K mg 2 2 2 2 −       _+δ = 2 1 K

K

g

m

K

mg

K

g

m

2

2

2 2 2 2 2 2

−









_δ

+

δ

+

= 2 1 K δ2 _{+ mg}δ ⇒ δ ₌ K F 2 . EXAMPLE : 09

A stone is projected at time t = 0 with a speed V0 and an angle θ with the horizontal in a

uniform gravitational field. The rate of work done (P) by the gravitational force plotted against time (t) will be as

A) B) C) D)

Solution :

Rate of work done is the power associated with the force. It means rate of work done by the gravitational force is the power associated with the gravitational force. Gravitational force acting on the block is equal to its weight mg which acts vertically downwards.

Velocity of the particle (at time t) has two components, (i) a horizontal component v cosθ and

(ii) a vertically upward component (v sinθ - gt) P O t P O t P O t O t P

Hence, the power associated with her weight mg will be equal to p = m →

g

.→v = -mg (v sinθ - gt)

This shows that the curve between power & time will be straight line having positive slope but negative intercept on Y-axis.

SINGLE ANSWER TYPE

LEVEL – I

1. Two springs A and B(KA = 2KB) are stretched by applying forces of equal magnitudes at the

four ends. If the energy stored in A is E, that in B is

a) E/2 b) 2E c) E d) E/4

2. Two equal masses are attached to the two ends of a spring constant K. The masses are pulled out symmetrically to stretch the spring by a length x over its natural length. The work done by the spring on each mass is

a) ½ Kx2 _{b) -1/2 Kx}2 _{c) ¼ Kx}2 _{d) -1/4 Kx}2

3. The negative of the work done by the conservative internal forces on a system equals the change in

a) total energy b) kinetic energy c) potential energy d) none of these

4. The work done by the external forces on a system equals the change in

a) total energy b) kinetic energy c) potential energy d) none of these

5. The work done by all the forces (external and internal) on a system equals the change in a) total energy b) kinetic energy c) potential energy d) none of these

6. ________ of a two particle system depends only on the separation between the two particles. The most appropriate choice for the blank space in the above sentence is

a) kinetic energy b) total mechanical energy c) potential energy d) total energy

7. A small block of mass ‘m’ is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity ‘v’ and the block does not slide on the wedge. The work done by the force of friction on the block in time ‘t’ will be

a) zero b) mgvt cos2θ _{c) mgvt sin}2θ _{d) mgvt sin2}θ

8. A block of mass ‘m’ slides down a smooth vertical circular track. During the motion, the block is in

a) vertical equilibrium b) horizontal equilibrium c) radial equilibrium d) none of these

9. A particle is rotated in a vertical circle by connecting it to a string of length ‘l’ and keeping the other end of the string fixed. The minimum speed of the particle when the string is horizontal for which the particle will complete the circle is

a) gl b) 2gl c) 3gl d) 5gl

10. Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass m moving a distance ‘l’ on a rough surface. The following quantities will be same as observed by the two observers

a) kinetic energy of the block at time t b) work done by friction c) total work done on the block d) acceleration of the block

11. A particle of mass ‘m’ is attached to a light string of length ‘l’, the other end of which is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle

a) the string becomes slack when the particle reaches its highest point b) the velocity of the particle becomes zero at the highest point c) the kinetic energy of the ball in initial position was ½ mv2 _{= mgl}

d) the particle again passes through the initial position

12. The kinetic energy of a particle continuously increases with time

a) the resultant force on the particle must be parallel to the velocity at all instants b) the resultant force on the particle must be at an angle less than 900 _{all the time}

c) its height above the ground level must continuously decrease d) the magnitude of its linear momentum is increasing continuously

13. One end of a light spring of spring constant k is fixed to a wall and the other end is tied to a block placed on a smooth horizontal surface. In a displacement, the work done by the spring is ½ kx2_{. The possible cases are}

a) the spring was initially compressed by a distance x and was finally in its natural length b) it was initially stretched by a distance x and finally was in its natural length

c) it was initially in its natural length and finally in a compressed position d) it was initially in its natural length and finally in a stretched position

14. A block of mass ‘M’ is hanging over a smooth and light pulley through a light string. The other

end of the string is pulled by a constant force F. The kinetic energy of the block increases by 20J in

1s

a) the tension in the string is Mg b) the tension in the string is F c) the work done by the tension on the block is 20J in the above is 1s

d) the work done by the force of gravity is –20J in the above 1s

15. A particle of mass 0.25kg moves under the influence of a force F = (2x-1). If the velocity of the particle at x = 0 is 4m/s. its velocity at x = 2m will be

A) 4 2 m/s B) 2 2 m/s C) 8m/s D) 6m/s

16. Work done to accelerate a car from 10 to 20m/s compared with that required to accelerate it from 0 to 10m/s is

A) twice B) three times C) four times D) same

17. Two springs have their force constant as K1 and K2 (K1 > K2). When they are stretched by

the same force :

A) no work is done in case of both the springs B) equal work is done in case of both the springs

C) more work is done in case of second spring D) more work is done in case of first spring

18. The kinetic energy K of a particle moving in a straight line depends upon the distance s as K = as2 _{where a is a constant. The force acting on the particle is}

A) 2as B) 2mas C) 2a D) as2

19. A particle moves in a straight line with a retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

A) x B) x2 _{C) ln x D) e}x

20. 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) B) C)

D) 21. 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

2 2 dx y d will be : A)

[

M−1L−2T4

]

B)

[

M2L−3T−2

]

C)

[

M−2L−4T−4

]

D)

[

ML3T−6

]

24. 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 25. Identify the correct statement for a non-conservative force

A) A force which is not conservative is called a non-conservative force B) The work done by this force depends on the path followed

C) The word done by this force along a closed path is zero D) The work done by this force is always negative

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

A) x = x1 B) x = x2

C) both x1 and x2 D) neither x1 and x2

29. 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

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/r

2

_{. The total energy of the particle is}

a)

r 2 C −

b)

r 2 C

c) C x 2r

d) Zero

Sol:

Fcentipetal F =

₂ 2 r C r mv =

; v =

r C dr r C Fdr= 2 = −

_∫

_∫

− α

;

∴

E

1

= E

K

+ v = C/2r – C/r

= -C/2r

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 = 10

5

_w

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) t

3/2

_{c) t}

3/4

_{d) t}

2

Sol:

P = Fv = constant or ma . at = constant or a

2

_{t = constant}

Q

S = 1/2at

2

_{or S}

α

_at

2

_{But a}

α

_1/

_t

∴

S

α

t

2

_/

_t

_{or S}

α

_t

3/2

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

b) 8m

c) 4m

d) 6m

Sol:

∴

₂ 2 1 2 1 h 10 60 100 or h h u u = =

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)

2 CX₁2

b)

2 1 CX

c)

3 1 CX

d) 0

Sol:

W =

2 2 x 0 2 x 1 x Cx 2 1 dx cx Fdx=

∫

=

∫

6. A uniform chain of mass M and length L lies on a horizontal table such that one

third of its length hangs from the edge of the table. The work done is pulling the

hanging part on the table will be

a)

3 MgL

b) MgL

c)

9 MgL

d)

18 MgL

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 = t

3

_{/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 mv

22

– ½ mv

12

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

b) L/4

c) L/3

d) 2L/3

Sol:

      −x 2 L 4 w 3

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 10

3

_HP

Sol:

P = Fv = 4500 x 2 = 9 kw

10. A body of mass m is accelerated to velocity v in time et

1

. The work done by the

force as a function of time t will be

a)

₂ 2 2 e 2 t mv

b)

2 2 t t mv 2 1      

_c)

_t2 t 2 mv

d)

t 2 mvt2

Sol:

Acceleration produced in a body a =

1 t v

; W =

2 1

ma

2

_t

2

₌

2 2 1 2 t t mv 2 1

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 10

3

_N

_{b) 3.72 x 10}

2

_N

_{c) 3.73 x 10}

1

_N

_{d) None of}

the above

Sol:

F = P/v

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

c) 3/4

d) 4/3

Sol:

According to question

      = 2 2 _U 2 M 2 1 x 2 1 Mv 2 1

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

Sol:

m

1

v

1

= m

2

v

2

;

_K2 m₂v2₂

2 1 E =

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

b) mgRe/2

c) mgRe/4

d) 2mgRe

Sol:

2 mgR R 2 GMm =

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%

b) 30%

c) 20%

d) 10%

Sol:

M =

x100 m M m +

16. A force

F

= (3x

2

_{+ 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

b) 70 Joule

c) 115 Joule

d) 270 Joule

Sol:

W =

Fdx

(

3x 2x 7

)

dx s 0 2 2 x 1 x

∫

∫

− + =

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%

b) 50%

c) 100%

d) 95%

Sol:

x100 m M m 100 x E E 2 1 2 + = ∆

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 mv

2

_{= ½ kx}

2

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

b) 0 m

c) 3 m

d) 2 m

Sol:

t =

x

+ 3 or x = (t – 3)

2

_;

∴

_{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)

100 10 x 108

Joule

c) 0 Joule

d)

108 10 x 100

Joule

Sol:

W = Fd cos

θ

= Fd cos 90

0

_{= 0}

22. A particle moves in a potential region given by u = 8x

2

_{– 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

b) 11/15

c) 11/9

d) 9/11

Sol:

P =

t w t wh t mgh α =

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

Sol:

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

9 8 E 9 E E E E E i K i K i K 1 K F K 1 K − ₌ − ₌

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

_{12 6} r

b r

a ₋

,

where a and b are positive constants and r is inter atomic distance. The equilibrium

between two atoms is

a)

6 / 1 a b      

_b)

1/6 b a      

_c)

1/6 a 2 b      

_d)

1/6 b a 2      

Sol:

0 r b 6 r a 12 dr du 7 13 + = − =

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/s

2

₎

a) 2.6 kw

b) 2.6 w

c) 0.6 w

d) 0.6 kw

Sol:

t mv 2 / 1 mgh t w P 2 + = =

27. A body weighing 80N is moved up a slope of angle 60

0

_{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

θ

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:

W

PR

= W

PQ

+ W

QR

29. Two particles each of mass m and traveling with velocities u

1

and u

2

collide

perfectly inelastically. The loss of energy will be

a) ½ m(u

1

– u

2

)

2

b) ¼ m(u

1

– u

2

)

2

c) m(u

1

– u

2

)

2

d) 2m(u

1

–

u

2

)

2

Sol:

∆

E =

(

1 2

)

2 2 1 2 1 _{u u} m M m m 2 1 ₋ +

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 10

3

_{c) 1.44 x 10}

2

_{d) 1.44 x 10}

4

Sol:

U = kq

2

_/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)

2

times

c) 1/2 times

d) 2 times

Sol:

E = ½ mv

2

_;

∴

_{v =}

_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

_{b) f =}

(

)

2 1 A A +

c) f =

(

)

2 1 A uA +

d)

f

=

(

)

2 1 A A 4 +

Sol:

f =

(

) (

2 _{1 A}

)

2 A 4 m M Mm 4 + = +

33. Two masses m

1

= 2kg and m

2

= 5kg are moving on a frictionless surface with

velocities 10 m/s and 3 m/s respectively. m

2

is ahead of m

1

. An ideal spring of

spring constant k = 1120 N/m is attached on the backside of m

2

. The maximum

a) 0.51 m

b) 0.062 m

c) 0.25 m

d) 0.72 m

Sol:

∆

E =

(

)

2 2 2 1 2 1 2 1 _kx 2 1 u u m m m m 2 1 _{− =} +

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 k

1

and k

2

. If

P3

and k

3

are the

momentum and kinetic energy respectively of the third part then the ratio k

2

/k

3

will

be

a) 2/5

b) 3/5

c) 4/5

d) 1/5

Sol:

Conceptual

35. A block falls down from a table 0.5m high. It falls on an ideal vertical spring of

constant 4 x 10

2

_{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/s

2

₎

a) 0.5 kg

b) 2 kg

c) 1.2 kg

d) 0.9 kg

Sol:

mgh = ½ kx

2

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

b) 1500 joule

c) 1800 joule

d) 2100 joule

Sol:

W = mgh =

12kg 2 9 15 = +

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 w

1

and in the second case w

2

, then

a) w

2

= w

1

b) w

2

= 2w

1

c) w

2

= 3w

1

d) w

2

= 4w

1

Sol:

w

1

= ½ kx

2

, w

3

= ½ k(2x

2

)

38. A 2k body is projected, at an angle of 30

0

_{with the horizontal, with a velocity of}

10m/s. The kinetic energy of the body after 1 second will be

a) 10 joule

b) 50 joule

c) 100 joule

d) 200 joule

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 mx

2

_{= 200 cos 30}

0

_x

_{5 3}

40. A block of mass m is pushed towards a movable wedge of mass

β

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)

β + 1 u

c) u(1 +

β

)

d) 0

Sol:

mu = (m +

β

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)

    β +1 1 gh 2

b)

β 2gh

c)

2gh

d)

    β −1 1 gh 2

Sol:

½ mu

2

_{= mgh + ½ (m +}

β

_m)V

2

_cm

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) dgh

2

_a

c) 2gdh

2

_a

d) 4dgh

2

_a

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

b) 3.29 w

c) 1.19 w

d) 9.11 w

Sol:

P = hdg

dt dv

44. A block of mass 1kg is pulled up on an incline of angle 30

0

_{with the horizontal. The}

block moves with an acceleration of 1 m/s

2

_{. 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

θ

= ma or F = mg sin

θ

+ ma

45. A particle of mass m is moving in a circular path of constant radius r. The

centripetal acceleration of the particle (a

c

) is varying with time t according to

following relation a

c

= k

2

n

2

where k is a constant. The power delivered to the

particle by the forces acting on it will be

a) mk

2

_r

2

_t

2

_{b) m}

2

_k

2

_r

2

_t

2

_{c) m}

2

_k

2

_rt

_{d) mk}

2

_r

2

_t

Sol:

a

c

= v

2

/r = k

2

n

2

; w = ½ mv 2/2 – ½ mv

12

= ½ m k

2

r

2

t

2

= 0;

∴

P = dw/dt

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

Sol:

W =

_ − _+_ − 2C_ 2 B 2 B mv 2 1 mv 2 1 mv 2 1 mgh

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

a)

m gh 2 M     ₊

b)

gh 2 m M     +

c)

gh 2 m M     +

d) (M + m)

gh

Sol:

gh w 2 m M =      +

48. A particle has shifted along some trajectory in the x – y plane from point

r₁=iˆ−2jˆ

to

another point

r₂ =2iˆ+3jˆ

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

forces

F_i =3iˆ+4jˆ and F₂ =2iˆ−7jˆ+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 mv

2

_{+ F}

_fr

_h

50. A shell is fired from a cannon with a velocity v and at an angle

θ

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

θ

b) 2 v cos

θ

c) 3 v cos

θ

d)

3

/2 v

cos

θ

Sol:

mv cos

θ

= m/2 v cos

θ

+ m/2 v

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

c) 700 J

d) J

Sol:

mv = m

1

v

1

+ m

2

v

2

; E = ½ m

1

v

12

+ ½ m

2

v

22

– ½ mv

2

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

b) 10 J

c) 20 J

d) none of

these

Sol: As F

ext

= 0;

∴

W

ext

=

F_ext . S =0

53. 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: W

int

= ½ mv

2

+ ½ mv

2

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 B) 4.5J C) 2.25J D) 9J

55. A bullet, moving with a speed of 150m/s, strikes a wooden plank. After passing through the plank its speed becomes 125m/s. Another bullet of the same mass and size strikes the plank with a speed of 90m/s. It speed after passing through the plank would be

A) 25m/s B) 35m/s C) 50m/s D) 70m/s

56. A man of mass 60kg climbs a staircase inclined at 450 _{and having 10steps. Each step is}

20cm high. He takes 2 seconds for the first five steps and 3 seconds for the remaining five steps. The average power of the man is

A) 245W B) 245 2 W C) 235 2 W D) 235W

57. The potential energy of a particle moving in x-y plane is given by U = x2 _{+ 2y. The force}

acting on the particle at (2, 1) is

A) 6N B) 20 N C) 12 N D) 0

58. 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

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}

60. 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 B) 6J C) 0J D) 3J

61. A body of mass m was slowly pulled up the hill by a force F which at each point was directed along the tangent of the trajectory. All surfaces are smooth. Find the work performed by this force

A) mg l B) -mg l C) mgh D) zero

62. A rope ladder with a length l carrying a man of mass m at its end, is attached to the basket of a balloon of mass M. The entire system is in equilibrium in air. As the man climbs up the ladder into the balloon, the balloon descends by height h. Then the potential energy of man

A) increases by mg l B) increases by mg (l -h) C) increases by mgh D) increases by mg (2 l -h)

63. 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 B) l-mg/k C) l + mg/k D) l - k/mg

65. 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

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 l = πR/2, then the potential energy of the chain with reference level at the centre of sphere is give by

A) m R g B) 2m R g C) 2/π m R g D) 1/π m R g 67. If the force acting on a particle is given by F = 2i + xyj + xz2_{k, how much work is done}

when the particle moves parallel to Z-axis from the point (2, 3, 1) to (2, 3, 4) ?

A) 42J B) 48J C) 84J D) 36J

68. A uniform chain of length ' l ' and mass m is placed on a smooth table with one-fourth of its length hanging over the edge. The work that has to be done to pull the whole chain back onto the table is

A) 4 1 mgl B) 8 1 mgl C) 16 1 mgl D) 32 1 mgl

69. A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is W1 and in the

second case is W2

A) W2 = W1 B) W2 = 2W1 C) W2 = 3W1 D) W2 = 4W1

70. 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

1) zero B) gl C) 1.5gl D) 2gl

71. A force F acting on a body depends on its displacement x as Fα xn. The power delivered by F will be independent of x if n is

A) 1/3 B) -1/3 C) 1/2 D) -1/2

72. A particle is moving in a conservative force field from point A to B. UA and UB are the

potential energies of the particle at points A and B and Wc is the work done in the

process of taking the particle from A to B.

A) Wc = UB - UA B) Wc = UA - UB C) UA > UB D) UB > UA

73. 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

74. 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) kx2 2 1 B) 2 1 (kx) (s+x) C) kxs 2 1 D) kx

(

s x s

)

2 1 + + 75. Force acting on a block moving along x-axis is given by :

F = N 2 x 4 2     + −

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

A) positive B) negative

C) zero D) may be positive or negative

75. The system is released from rest with both the springs in unstretched positions. Mass of each block is 1 kg and force constant of each springs is 10 N/m. Extension of horizontal spring in equilibrium is:

A) 0.2m B) 0.4m C) 0.6m D) 0.8m

77. In a projectile motion, if we plot a graph between power of the force acting on the projectile and time then it would be like :

A) B) C) D)

78. A golfer rolls a small ball with speed u along the floor from point A. If x = 3R, determine the required

speed u so that the ball returns to A after rolling on the circular surface in the vertical plane from B to C and becoming a projectile at C. (Neglect friction)

A) gR 5 2 B) gR 2 5 C) gR 7 5 D) none of these

79. 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

A) v B) ν2 _C) ν3 _D) ν4 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

LEVEL – III

1. A block m is pulled by applying a force F as shown in fig. If the block has moved up through a distance 'h', the work done by the force F is

A) 0 2) Fh

C) 2Fh D)

2 1

Fh

2. A body of mass m, having momentum p is moving on a rough horizontal surface. If it is stopped in a distance x, the coefficient of friction between the body and the surface is given by

A) µ = p/(2mg x) B) µ = p2 _{/ (2mg s) C)} µ _{= p}2 _{/ (2g m}2_{s) D)} µ _{= p}2 _{(2g m}2_s2₎

3. A body of mass m moves from rest, along a straight line, by an engine delivering constant power P. the velocity of the body after time t will be

A) m Pt 2 B) m Pt 2 C) m 2 Pt D) m 2 Pt

4. The spring shown in fig has a force constant k and the mass of block is m. Initially, the spring is unstretched when the block is released. The maximum elongation of the spring on the releasing the mass will be

A) k mg B) 2 1 k mg C) 2 k mg D) 4 k mg

5. A skier starts from rest at point A and slides down the hill, without turning or braking. The friction coefficient is µ. When he stops at point B, his horizontal displacement in S. The height difference h between points A and B is

A) h = S/µ B) h = µS C) h = µS2 _{D) h = S/}µ2

6. A small block of mass m is kept on a rough inclined surface of inclination θ fixed in an elevator. The elevator goes up with a uniform velocity v and the block does not slide on the wedge. The work done by the force of friction on the block in time t will be

A) zero B) mg vt cos2θ _{C) mg vt sin}2θ _{D) mg vt sin2}θ

7. 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 µk and compresses a spring with force constant

k, a distance x before momentarily coming to rest. The spring then extends and the block travels back across the rough surface, sliding up the plane. The maximum height h' that the block reaches on its return is

A) h' = h - 2µd B) h' = h - 2µd - 2 1

kx2

C) h' = h - 2µd + kx2 _{D) h' = h - 2}µ_{d - kx}2

8. A chain of length 3 l and mass m lies at the top of smooth prism such that its length l is one side and 2 l 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) 2gl B) gl 3 2 C) gl 3 1 D) gl 2 1

9. If a constant power P is applied in a vehicle, then its acceleration increases with time according to the relation

A) a = t m 2 P         B) a = t3/2 m 2 P         C) a = 1/t m 2 P         D) a = mt 2 P

10. A body of mass m slides downward along a plane inclined at an angle α. The coefficient of friction is µ. The rate at which kinetic energy plus gravitational potential energy dissipates expressed as a function of time is

A) µmtg2 _cos α _B) µ_mtg2 _cos α _(sin α _- µ _cos α₎

C) µmtg2 _sin α _D) µ_mtg2 _sin α _(sin α _- µ _cos α₎

11. 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, π/4) is

A) 1 B)

2

C) 1/

2

D) 0

12. A uniform rope of length ' l ' 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 µ. The work done by the friction during the period the chain slips completely off the table is A) 2/9 µmgl B) 2/3 µmgl C) 1/3 µmgl D) 1/9 µmgl

13. A particle is moving in a force field given by potential U = - λ(x + y + z) from the point (1, 1, 1) to (2, 3, 4). The work done in the process is

A) 3λ B) 1.5λ C) 6λ D) 12λ

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 A) x mg kh B) mg xkh C) x mg kh 2 D) kh x mg

15. 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 µ = 1/5. Find maximum value of vo so

that after pressing the spring the block does not return back but stops there permanently.