WORK POWER & ENERGY

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WORK, POWER & ENERGY 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 gets some displacement along the direction of force then work is said to be done by the force. The work done, W by a constant → force → F when its point of application undergoes a displacement S is measured as

W=

→

→ → F. S =|F | |

→

S | 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 = .

→ → $ → $ $ F x i + F y j + F z k and

→

S = ∆x

$ i + ∆y $j + ∆z k$ then W =

→

F

→

S = Fx ∆x+Fy∆y+Fz ∆z Positive and Negative work : The work is said to be positive if the angle θ is acute (θ

0 < 900) and negative if the angle θ is obtuse (θ > 900). If the angle between → F and S is 90 then work done by the force is zero.

→

If the force is variable then the work done by the variable force is given by dW = → F . dS or

→

S2 →

W=

→

∫ F . dS

S1

Work depends on frame of reference. With change of frame of reference displacement may change, so the work done by a force will be different in different frames. Illustration – 1 : A particle of mass 2 kg moves under the action of a constant force → F =

( 5i$− 2 $j ) N. If

its displacement is 6 $j m. What is the work done by the force → F ? 1 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

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Solution : → The work done → F . x

=

( 5i$− 2 $j ) . 6 $j = - 12 Joule

Illustration – 2 : A load of mass m = 3000 kg is lifted by a rope with an acceleration a = 2 m/s 2. Find the work done during the first one and a half seconds from the beginning of motion. Solution : The height to which the body is lifted during the first 't' second is h =

1 2 at tension in 2

the rope T = mg + ma

 1 2 at  2 

∴ Work done = T.h

= m(g + a) 

1



2 = 3000 (10 + 2)  x 2 x ( 1.5)  2  

= 81 KJ

WORK DONE BY A SPRING FORCE : Whenever a spring is stretched or compressed, the spring force always tend to restore it to the equilibrium position. If x be the displacement of the free end of the spring from its equilibrium position then, the magnitude of the spring force is FS = - kx The negative sign indicates that the force is restoring. The work done by the spring force for a displacement from xi to xf is given by Ws = −

x f

∫ kxdx x i

⇒ Ws = −

(

1 k x f2 − x i2 2

)

WORK DONE BY FRICTION : Work done by friction may be zero, positive or negative depending upon the situations:  When a block is pulled by a force F and the block does not move, the work done by friction is zero.  When a block is pulled on a stationary surface, the work done by the kinetic friction is negative.  When one block is placed on another block and is pulled by a force then friction force does negative work on top block and positive work on the lower block

WORK DONE BY GRAVITY : Here the force of gravity is Fg = - mg

$j and the displacement is given by

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S = ∆x

$ i + ∆y $j + ∆z k$ →

→

∴ Work done by gravity is Wg = F g . S = - mg ∆y ∆y =

y f − yi = - h

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

DEPENDENCE OF WORK ON FRAME OF REFERENCE : Work depends upon the frame of reference from where it is calculated. As the displacement as well as force, depends upon the different frames of reference. Therefore, the work also changes. For example :- If you calculate work from a non inertial frame work due to pseudo force has to be included.

CONSERVATIVE AND NON CONSERVATIVE FORCES : In Conservative force field the work done by the force is independent on path followed and depends only on initial and final co-ordinates. Such forces are known as Conservative forces. Examples are gravitational, electrostatic forces. If the work done depends on path followed. Such forces are called nonConservative 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 = Fdcos180 0 = -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 3 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

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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

v m 0 m

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

ap = (a)

µmg = µg m

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

( a = −2µg )

0 = V0 + at ⇒ t=

V0 2µg

Velocity of the plank during this time is Vp = u p + ap t = µg

V0 V = 0 2µg 2

 2  V0 2   V0 −    2   2   = 3V0 Displacement of the block = S =   2×a 8µg      

µmg ma p m

3 V02 ( −1) = − 3 mv20 Work done by friction on the block = F.S. cosπ = µmg. 8 µg 8 (b)

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

=

0 − V02 − 2µg

µg =

⇒

mV02 2

⇒ work done by friction from the Plank frame = − (c)

mV02 2

Final velocity of the block

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= Velocity of the plank =

V0 2

WORK ENERGY THEOREM : Now we have to study which physical quantity changes when work is done on a particle. If a constant force F acts through a displacement x, it does work W = Fx

 vf2 = vi2 + 2 ax ∴W=

(

m vf2 − vi2 2

The quantity k =

)=

1 1 m vf2 m vi2 2 2

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

the energy posses by the particle by virtue of its motion. Thus the equation takes the form W = K f − K i = ∆K The work done by a force changes the kinetic energy of the particle. This is called the work -Energy Theorem. Illustration – 5 : →

→

The velocity of an 800 gm object changes from v 0 = 3 $ i - 4 $j to v f = -6

$j + 2 k$ m/s.

What is the change in K.E of the body? Solution : Here m = 800gm = 0.8 kg →

vo

∴

=

→

32 + ( − 4) 2 = 5

change in K.E =

1 x 0.8 2

vf 2 2 → →   vf −v 0      

=

=

(−6)2 +(2)2 =

40

1 x 0.8 x ( 40− 25) = 6J 2

Illustration – 6 : The coefficient of sliding friction between a 900 kg car and pavement is 0.8. If the car is moving at 25 m/s along level pavement, when it begins to skid to a stop, how far will it go before stopping? Solution : Here m = 900kg µ = 0.8, v = 25 m/s S =? K.E = work done against friction ⇒s=

1 mv 2 = F.s = µ N.s = µmgs 2

( 25) v2 = ~ 39 m 2 x 0.8 x 10 2µg 2

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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/s 2) Solution :

∴

Let upward push of air be F The resultant downward force = mg - F As work done = gain in K.E

1 mv 2 2 1 (10 x 10 - F) x 20 = x 10 x (10)2 ⇒ F = 75 N 2 (mg - F) x S =

∴ ∴

Work done by push of air = 75 x 20 = 15 Joule This work done is negative.

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

du = - F .dr for example : (i) If an object is placed at any point in gravitational field work is to be done against gravitational field force. The magnitude of this work done against the gravitational force gives the measure of gravitational potential energy of the body at that position which is U = mgh. Here h is the height of the object from the reference level. (ii) The magnitude of work done against the spring force to compress it gives the measure of elastic potential energy, which is U = (iii)

1 k x2 2

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 2

U2-U1= - W= −∫F .

dr

1

CONSERVATION OF MECHANICAL ENERGY : Change in potential energy ∆U = - WC where WC is the work done by conservative forces. From work energy theorem Wnet = ∆k Where Wnet is the sum of work done by all the forces acting on the mass. If the system 6 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

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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 = k i + Ui

∆E = 0, integrating both sides

E = constant.

Illustration – 8 : A projectile is fired from the top of a 40m. high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground. Solution : Taking ground as the reference level we can conserve the mechanical energy between the points A and B v ∆ (K + U) = 0 ⇒ Ki + Ui = Kf + Uf A

∴ ⇒ ⇒

θ

1 1 mv2 + mgH = mv' 2 + 0 2 2 1 1 (50)2 + 40 x 10 = v' 2 2 2

⇒

(1250 + 400) x 2 = v'

⇒

v' 2 = 3300 v' ~ 58 m/s

H

2

B v'

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 =

∆W ∆t

The S.I. unit of power is J/S or watt (W). Thus 1 W = 1 J/S The instantaneous power is the limiting value of Pav as ∆t Instantaneous power may also be written as P =

→

0 that is P =

dW dt

dW → → = F . v Since work and energy are dt 7

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

dE . dt

Illustration – 9 : A car of mass 500 kg moving with a speed 36km/hr in a straight road unidirectionally doubles its speed in 1 minute. Find the average power delivered by the engine. Solution : Its initial speed V1 = 10 m/s then V2 = 20 m/s

∴ ∴

∆k =

1 2 1 2 m v 2 − mv 1 2 2

Power delivered by the engine

(

1 m v 22 − v 12 P = ∆K 2 = ∆t ∆t 1 x 500202 − 102 = 2 60

(

)

)

= 1250 W.

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

O θ

T

P

mg cos θ

mg sin θ

mv 2 T - mg cosθ = r

mv 2 r 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 ⇒

T = mg cosθ +

critical situation T = 0 and θ = π

mv 2min R

∴

mg =

⇒

v min =

⇒

gR v2 min =

gR

Now conserving energy between the lowest and the highest point

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1 1 mu 2 = mv 2 min min + mg ( 2R ) 2 2 u2 min = gR + 4gR = 5gR

⇒

u min = 5gR If u min ≥ 5gR the particle will complete the circle. At u = point is v = If u <

5gR ,velocity at highest

gR and tension in the string is zero. 5gR , the tension in the string become zero before reaching the highest point and

at that point the particle will leave the circular path. After leaving the circle the particle will follow a parabolic path. Above conditions are applicable even if a particle moves inside a smooth spherical shell of radius R. The only difference is that the tension is replaced by the normal reaction N. Illustration – 10 : A heavy particle hanging from a fixed point by a light inextensible string of length l is projected horizontally with speed

g .

Find the speed of the particle and the inclination of

the string to the vertical at the instant of the motion when the tension in the string is equal to the weight of the particle. Solution : Let T = mg at an angle θ as shown in figure h = l (1 - cosθ) Conserving mechanical energy between A and B ⇒

• θ

1 1 mu2 = mv2 + mgh 2 2 u = v + 2gh ⇒ 2

2

T - mg cosθ =

mv 

2

⇒

h 2

2

v = u - 2gh …. (i)

mg = mg cos θ +

⇒

v2 = g l (1- cosθ)

mg sin θ

u = g

mv 2 

T= mg cosθ +

⇒

• A

B • mg cos θ

mv 2  ……. (ii)

From (i) and (ii) u2 - 2gl (1 - cosθ) = gl (1 - cosθ) ⇒

cosθ =

2 3

⇒

 2   3

θ = cos-1 

putting the value of cosθ in equation

 

v2 = gl  1 −

2 g  = 3 3

⇒

…… (ii) v=

g 3

Equilibrium :As we have studied earlier a body is said to be in translational equilibrium if net 9 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

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force acting on the body is Zero. Fnet = 0 ∴

If the forces are Conservative F = -

dU dr

dU =0 dr

⇒

∴ At Equilibrium slope of U and r graph is Zero (or) Potential energy either maximum or minimum or constant at that position. At the stable equilibrium position P.E is minimum At the unstable equilibrium position P.E is maximum Illustration – 11 : The P.E of a Conservative system is given as U = 10 + (x-2) 2. Find the equilibrium position and discuss type of equilibrium. Solution : For Equilibrium F = 0

dU = −2(x − 2) = 0 dx

∴

F=-

⇒

x=2

d 2U

and ∴

dx2

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

B)

2E M

C)

E m

D)

2E is k

E 2m

Solution : Potential energy of particle at x =

2E is zero k

∴ K.E = E

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⇒

1 mv2 = E or v = 2

2E m

Example : 08 A block is suspended by an ideal spring of force constant k. If the block is pulled down by applying a constant Force 'F' and if maximum displacement of block from its initial position of rest is δ then A)

F 2F 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 The kinetic energy K of a particle moving in a straight line depends upon the distance s as K = as2 where a is a constant. The force acting on the particle is A) 2as

19.

20.

B) 2mas

C) 2a

D)

as 2

A particle moves in a straight line with a retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to A) x B) x2 C) ln x D) ex Choose the wrong option 16

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21.

22.

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

23.

24.

25.

26.

 M −1 L−2T 4 

B)

 M 2 L−3T −2 

C)

 M −2 L−4T −4 

D)

d2y dx 2

will be :

 ML3T −6 

Choose the correct statement Work done by a variable force uu r ur A) Is defined as F .S B) Is independent of path C) Is always dependent on the initial and final positions D) None of these Identify the correct statement for a non-conservative force A) A force which is not conservative is called a non-conservative force 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 The figure shows a plot of potential energy function, u(x) = kx 2 where x is the displacement and k is a constant. Identify the correct conservative force function F(x)

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 17

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D) In moving from O to A, work done by the body and is negative

LEVEL - II 1.

A particle of mass m is moving in a circular path of radius r under the influence of centripetal force F = C/r2. The total energy of the particle is a) −

2.

3.

.

C 2r

b)

C 2r

c) C x 2r

d) Zero

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 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 b) t3/2 c) t3/4 d) t2 1 A ball is dropped from a height of 10m. If 40% of its energy is lost on collision with the earth then after collision the ball will rebound to a height of a) 10m b) 8m c) 4m d) 6m 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)

4.

5.

a)

6.

8.

b) CX 12

c) CX 13

d)

0 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)

7.

CX12 2

MgL 3

b) MgL

c)

MgL 9

d)

MgL 18

An electric motor produces a tension of 4500N in a load lifting cable and rolls it at the rate of 2m/s. The power of the motor is a) 9 kw b) 15 kw c) 225 kw d) 9 x 103 HP The relation between time and displacement of a particle moving under the influence of a force F is t =

9.

10.

11.

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 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 Two particles each of mass m and traveling with velocities u 1 and u2 collide perfectly inelastically. The loss of energy will be a) ½ m(u1 – u2)2 b) ¼ m(u1 – u2)2 c) m(u1 – u2)2 d) 2m(u1 2 – u2) 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 18

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12.

area of cross section of the tube is a. The work done in pumping the liquid will be a) dgha b) dgh2a c) 2gdh2a d) 4dgh2a A man pulls a bucket full of water from a h metre deep well. If the mass of the rope is m and mass of bucket full of water is M, then the work done by the man is

M

M + m   gh  2 



+ m  gh a)  2  13.

14.

15.

16.

17.

18.

19.

20.

b) 



m

c) M +  gh 2 

d)

(M + m) gh The force-displacement curve for a body moving on a smooth surface under the influence of force F acting along the direction of displacement s has been shown in fig. If the initial kinetic energy of the body is 2.5J. its kinetic energy at s = 6m is A) 7J B) 4.5J C) 2.25J D) 9J 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 A man of mass 60kg climbs a staircase inclined at 45 0 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

C) 235 2 W D) 235W 2W The potential energy of a particle moving in x-y plane is given by U = x 2 + 2y. The force acting on the particle at (2, 1) is A) 6N

B)

20 N

C)

D) 0 12 N 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 A 5kg brick of dimensions 20cm x 10cm x 8cm is lying on the largest base. It is now made to stand with length vertical. If g = 10m/s2, then the amount of work done is A) 3J B) 5J C) 7J D) 9J A body of mass m was slowly pulled up the hill by a force F which at each point was directed along the tangent of the trajectory. All surfaces are smooth. Find the work performed by this force A) mg  B) -mg  C) mgh D) zero 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) 19

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C) increases by mgh 21.

22.

23.

24.

D) increases by mg (2 l -h)

Two springs s1 and s2 have negligible masses and the spring constant of s 1 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 s 1 to s2 is A) 1/9 B) 1/3 C) 1 D) 3 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 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 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  = 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

25.

26.

If the force acting on a particle is given by F = 2i + xyj + xz2k, how much work is done when the particle moves parallel to Z-axis from the point (2, 3, 1) to (2, 3, 4) ? A) 42J B) 48J C) 84J D) 36J A uniform chain of length '  ' and mass m is placed on a smooth table with one-fourth of its length hanging over the edge. The work that has to be done to pull the whole chain back onto the table is A)

27.

28.

1 mgl 4

B)

1 mgl 8

C)

1 mgl 16

D)

1 mgl 32

A spring, which is initially in its unstretched condition, is first stretched by a length x and then again by a further length x. The work done in the first case is W 1 and in the second case is W2 A) W2 = W1 B) W2 = 2W1 C) W2 = 3W1 D) W2 = 4W1 A particle of mass m is fixed to one end of a light rigid rod of length '  ' and rotated in a vertical circular path about its other end. The minimum speed of the particle at its highest point must be A) zero

B)

gl

C)

1.5gl

D)

2gl

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IIT- P-WPE

29.

30.

31.

A particle is moving in a conservative force field from point A to B. U A and UB are the potential energies of the particle at points A and B and W c 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 A force is given by Mv2/r when the mass moves with speed v in a circle of radius r. The work done by this force in moving the body over upper half circle along the circumference is A) zero B) ∞ C) Mv2 D) Mv2 π/2 A moving railway compartment has a spring of constant 'k' fixed to its front wall. A boy in the compartment stretches this spring by distance x and in the mean time the compartment moves by a distance s. The work done by boy w.r.t earth is A)

32.

33.

34.

1 kx 2 2

1 (kx) (s+x) 2

C)

1 kxs 2

D)



4

1 kx ( s + x + s ) 2



 N Force acting on a block moving along x-axis is given by : F = −   x2 + 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 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 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)

35.

B)

B)

C)

D)

A golfer rolls a small ball with speed u along the floor from point A. If x = 3R, determine the required speed u so that the ball returns to A after rolling on the circular surface in the vertical plane from B to C and becoming a projectile at C. (Neglect friction) A)

2 5

LEVEL - III

gR

B)

5 2

gR

C)

5 7

gR

D) none of these

. 21

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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)

1 Fh 2

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 = p2 / (2g m2s) D) µ = p2 (2g m2s2)

3.

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

2Pt m

A)

B)

2Pt m

C)

Pt 2m

D)

Pt 2m 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)

5.

6.

mg k

B)

1 mg 2 k

C) 2

mg k

D) 4

mg k

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 is 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 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 -

1 2 kx 2

C) h' = h - 2µd + kx2 D) h' = h

- 2µd - kx2 7.

A chain of length 3  and mass m lies at the top of smooth prism such that its length  is one side and 2  is on the other side of the vertex. The angle of prism is 120 0 and the prism is not free to 22

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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) 8.

B)

2gl

2 gl 3

C)

1 gl 3

D)

1 gl 2

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

 P A) a =   2m 

 t  

 P B) a =   2m

 3/ 2 t  

 P C) a =   2m 

 1 / t  

D) a =

P 2mt

A body of mass m slides downward along a plane inclined at an angle α. The coefficient

9.

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) µmtg2 cos α (sin α - µ cos α) C) µmtg2 sin α 10.

The potential energy for a force field F is given by U(x, y) = sin (x + y). The force acting on the particle of mass m at (0, π/4) is A) 1

11.

D) µmtg2 sin α (sin α - µ cos α)

B)

C) 1/ 2 D) 0 2 A uniform rope of length '  ' and mass m hangs over a horizontal table with two third part on the table. The coefficient of friction between the table and the chain is 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

12.

13.

14.

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 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

kh mg

B)

C) x

2kh mg

D)

xkh mg

mg x kh

A block of mass m = 2kg is moving with velocity v o 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. A) 6 m/s B) 12m/s C) 8m/s D) 10m/s

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IIT- P-WPE

15.

16.

Potential energy of a particle moving along x-axis under the action of only conservative forces is given as : U = 10 + 4 sin(4πx). Here U is in Joule and x in meters. Total mechanical energy of the particle is 16J. Choose the correct option. A) At x = 1.25m, particle is at equilibrium position. C) both A and B are correct B) Maximum kinetic energy of the particle is 20J D) both A and B are wrong. A system shown in figure is released from rest. Pulley and spring is massless and friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with ground is (Take force constant of spring k = 40 N/m and g = 10 m/s2) A)

2 m/s`

B) 2

C) 2m/s 17.

18.

2 m/s

D) 4

2 m/s Two blocks of masses m1 = 1 kg and m 2 = 2 kg are connected by a non-deformed light spring. They are lying on a rough horizontal surface. The coefficient of friction between the blocks and the surface is 0.4 what minimum constant force F has to be applied in horizontal direction to the block of mass m1 in order to shift the other block? (g = 10 m/s2) A) 8 N B) 15 N C) 10 N D) 25 N A block of mass m is attached with a massless spring of force constant k. The block is placed over a rough inclined surface for which the coefficient of friction is µ = ¾. The minimum value of M required to move the block up the plane is (Neglect mass of string and pulley and friction in pulley). A) 3/5m B) 4/5m C) 6/5m D) 3/2m

MULTIPLE ANSWER TYPE QUESTIONS . 1.



The potential energy U for a force field F is such that U = - kxy, where k is a constant 

kyˆ i + kxˆ j A) F =



2.



kx ˆ i + kyˆ j B) F =



C) The force F is a conservative force D) The force F is a non-conservative force Two blocks A and B each of mass m are connected by a light spring of natural length L and spring constant k. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A to B and collides with A. Then A) the kinetic energy of the A-B system at maximum compression of the spring is zero B) the kinetic energy of the A-B system at maximum compression of the spring is

24 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

IIT- P-WPE  mv 2   4 

   

 v C) the maximum compression of the spring is   

3.

4.

5.

6.

7.

8.

m k

   

 m   v D) the maximum compression of the spring is   2k    The kinetic energy of a body moving along a straight line varies directly with time t. If the velocity of the body is v at time t, then the force F acting on the body is such that A) F ∝ t1/2 B) F ∝ t-1/2 C) F ∝ v D) F ∝ v-1 A car of mass m is moving on a level road at a constant speed vmax while facing a resistive force R. If the car slows down to vmax/3, then assuming the engine to be working at the same power, what force F is developing and what is the acceleration a of the car ? A) F = 3R B) F = 2R C) a = 3P/m vmax D) a = 2 P/m vmax The potential energy of a particle of mass 1 kg moving in xy plane is given by U = 10 4x - 3y The particle is at rest at (2, 1) at t = 0. Then A) velocity of particle at t = 1 is 5 m/s B) the particle is at (10m, 7m) at t = 2s C) work done during t = 1s to t = 2s is 75J D) the magnitude of force acting on particle is 5N A block of mass m is gently placed on a vertical spring of stiffness k. Choose the correct statement related to the mechanical energy E of the system. A) It remains constant B) It decreases C) It increases D) Nothing can be said A spring of stiffness k is pulled by two forces F A and FB as shown in the figure so that the spring remains in equilibrium. Identify the correct statement (s) : A) The work done by each force contributes into the increase in potential energy of the spring B) The force undergoing larger displacement does positive work and the force undergoing smaller displacement does negative work C) Both the forces perform positive work D) The net work done is equal to the increase in potential energy A particle of mass m is released from a height H on a smooth curved surface which ends into a vertical loop of radius R, as shown in figure. If θ is the instantaneous angle which the line joining the particle and the centre of the loop makes with the vertical, the identify the correct statement(s) related to the normal reaction N between the block and the surface. 25 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

IIT- P-WPE

A) The maximum value N occurs at θ = 0

B) The minimum value of N occurs at N = π

C) The value of N becomes negative for π/2 < θ <

9.

3π 2

D) The value of N becomes zero only when θ > π/2 An engine is pulling a train of mass m on a level track at a uniform speed v. The resistive force offered per unit mass is f A) Power produced by the engine is mfv B) The extra power developed by the engine to maintain a speed v up a gradient of h in

mghv s C) The frictional force exerting on the train is mf on the level track D) None of above is correct A particle of mass 5 kg moving in the x-y plane has its potential energy given by U = (7x + 24y) J, where x and y are in metre. The particle is initially at origin and has a s is

10.



1 velocity u =(14.4ˆi +4.2ˆj) ms −

A) The particle has a speed of 25 ms-1 at t = 4 s B) The particle has an acceleration of 5 ms-2 C) The acceleration of the particle is perpendicular to its initial velocity D) None of the above is correct

*****

MULTIPLE MATCHING TYPE QUESTIONS : . 1. List - I a) Area under F - S b) Work energy theorem c) change in PE d) conservative force

List - II e) Change in KE f) negative of work done to gravitational force g) work done by F h)

 −∫F . dx , where F is conservative

force i) gravitational force 2. 26 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

IIT- P-WPE

List - I

List - II e) depends on frame of reference f) defined for conservative force only g) independent on frame of reference h) same for either compression elongation for same distance

a) KE b) work done c) PE d) spring PE

or

3. List - I a) stable equilibrium b) unstable equilibrium

List - II e) PE in Max f) Fnet = 0 g) PE is Min

dF ≤0 dx dF ≥0 d) dx c)

h) slope of F-x graph is +ve

4. List - I a) work done by frictional force b) work done by electrostatic force c) work done by gravitational force for closed loop d) for slowly moving body, wc + wn.c equal to

List - II e) indepent of path f) non-conservative g) depends on path h) define PE i) zero

*****

ASSERTION AND REASON : . Read the assertion and reason carefully to mark the correct option out of the options given below: (a) If both assertion and reason are true and the reason is the correct explanation of the assertion. (b) If both assertion and reason are true but reason is not the correct explanation of the assertion. (c) If assertion is true but reason is false. (d) If assertion is false but reason is true. 1.

Assertion : A person working on a horizontal road with a load on his head does no work. Reason : No work is said to be done, if directions of force and displacement of load are perpendicular to each other. 27 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

IIT- P-WPE

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. 13. 14. 15.

Assertion : Work done by friction on a body sliding down an inclined plane is positive. Reason : Work done is greater than zero, if angle between force and displacement is acute or both are in same direction. Assertion : When a gas is allowed to expand, work done by gas is positive. Reason : Force due to gaseous pressure and displacement (of piston) are in the same direction. Assertion : The instantaneous power of an agent is measured as the dot product of instantaneous velocity and the force acting on it at that instant. Reason : The unit of instantaneous power is watt. Assertion : The change in kinetic energy of a particle is equal to the work done on it by the net force. Reason : Change in kinetic energy of particle is equal to the work done only in case of a system of one particle. Assertion : A spring has potential energy, both when it is compressed or stretched. Reason : In compressing or stretching, work is done on the spring against the restoring force. Assertion : Comets move around the sun in elliptical orbits. The gravitational force on the comet due to sun is not normal to the comet’s velocity but the work done by the gravitational force over every complete orbit of the comet is zero. Reason : Gravitational force is a non conservative force. Assertion : The rate of change of total momentum of a many particle system is proportional to the sum of the internal forces of the system. Reason : Internal forces can change the kinetic energy but not the momentum of the system. Assertion : Water at the foot of the water fall is always at different temperature from that at the top. Reason : The potential energy of water at the top is converted into heat energy during falling. Assertion : The power of a pump which raises 100 kg of water in 10sec to a height of 100 m is 10 KW. Reason : The practical unit of power is horse power. Assertion : According to law of conservation of mechanical energy change in potential energy is equal and opposite to the change in kinetic energy. Reason : Mechanical energy is not a conserved quantity. Assertion : When the force retards the motion of a body, the work done is zero. Reason : Work done depends on angle between force and displacement. Assertion : Power developed in circular motion is always zero. Reason : Work done in case of circular motion is zero. Assertion : A kinetic energy of a body is quadrupled, when its velocity is doubled. Reason : Kinetic energy is proportional to square of velocity. Assertion : Work done by or against gravitational force in moving a body from one point to another is independent of the actual path followed between the two points. 28

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16.

17. 18.

Reason : Gravitational forces are conservative forces. Assertion : Graph between potential energy of a spring versus the extension or compression of the spring is a straight line. Reason : Potential energy of a stretched or compressed spring, proportional to square of extension or compression. Assertion : Heavy water is used as moderator in nuclear reactor. Reason : Water cool down the fast neutron. Assertion : If two protons are brought near one another, the potential energy of the system will increase. Reason

19.

20.

21.

: The charge on the proton is +1.6 ×10−19 C .

Assertion : In case of bullet fired from gun, the ratio of kinetic energy of gun and bullet is equal to ratio of mass of bullet and gun. Reason : In firing, momentum is conserved. Assertion : Power of machine gun is determined by both, the number of bullet fired per second and kinetic energy of bullets. Reason : Power of any machine is defined as work done (by it) per unit time. Assertion : Mountain roads rarely go straight up the slope. Reason : Slope of mountains are large therefore more chances of vehicle to slip from roads.

*****

IIT TRACK QUESTIONS : . 1.

2.

A ball hits the floor and rebounds after inelastic collision. In this case [IIT 1986] (a) The momentum of the ball just after the collision is the same as that just before the collision (b) The mechanical energy of the ball remains the same in the collision (c) The total momentum of the ball and the earth is conserved (d) The total energy of the ball and the earth is conserved A uniform chain of length L and mass M is lying on a smooth table and one third of its 29 Head Office : Balaji Towers : Z-21, Opp. Vishal Mega Mart, Zone-I, M.P. Nagar, Bhopal – 462 011. Ph. 4285066, 4274749, 4275672.

IIT- P-WPE

3.

length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is [IIT 1985] (a) MgL (b) MgL/3 (c) MgL/9 (d) MgL/18 A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to − K / r 2 , where K is a constant. The total energy of the particle is 1977] (a)

K 2r

(b) −

K 2r

(c) −

K r

[IIT

(d)

4.

K r The displacement x of a particle moving in one dimension under the action of a constant

5.

force is related to the time t by the equation t = x + 3 , where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is [IIT 1979] (a) 9 J (b) 6 J (c) 0 J (d) 3 J A force F = −K (yi + xj) (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is [IIT 1998]

6.

(a) − 2Ka2 (b) 2Ka2 (c) − Ka2 (d) Ka2 If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of earth to a height equal to the radius of the earth R, is [IIT 1983]

1 1 mgR mgR (b) 2 mgR (c) mgR (d) 2 4 A lorry and a car moving with the same K.E. are brought to rest by applying the same retarding force, then [IIT 1973] (a) Lorry will come to rest in a shorter distance (b) Car will come to rest in a shorter distance (c) Both come to rest in a same distance (d) None of the above A particle free to move along the x-axis has potential energy given by (a)

7.

8.

U(x) = k[1 −exp( −x)2] for

9.

−∞ ≤ x ≤ +∞ , where k is a positive constant of appropriate

dimensions. Then [IIT-JEE 1999] (a) At point away from the origin, the particle is in unstable equilibrium (b) For any finite non-zero value of x, there is a force directed away from the origin (c) If its total mechanical energy is k/2, it has its minimum kinetic energy at the origin (d) For small displacements from x = 0, the motion is simple harmonic A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to [IIT 1984] (a) t1 / 2

(b) t3 / 4

(c) t3 / 2

(d)

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IIT- P-WPE

10.

11.

t2 A shell is fired from a cannon with velocity v m/sec at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in m/sec of the other piece immediately after the explosion is [IIT 1984] 3 3 v cosθ (a) 3v cosθ (b) 2v cosθ (c) (d) v cosθ 2 2   Two particles of masses m1 and m2 in projectile motion have velocities v1 and v2  respectively at time t = 0. They collide at time t0 . Their velocities become v1 ' and  2t0 v2 ' at time while still moving in air. The value of | (m1v1 '+m2 v2 ') −(m1v1 +m2 v2 ) | is [IIT-JEE Screening 2001]

(b) (m1 + m2 )gt0

(a) Zero

(c) 2(m1 + m2 )gt0

(d)

1 (m1 + m2 )gt0 2 12.

A particle is placed at the origin and a force F = kx is acting on it (where k is positive constant). If U(0) = 0 , the graph of U(x) versus x will be (where U is the potential energy function) [IIT-JEE (Screening) 2004] U(x)

(a)

U(x) x

(b)

x

(c)

U(x)

U(x) x

(d)

x

*****

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IIT- P-WPE

SINGLE ANSWER TYRE . LEVEL - I 1 B 16 B

2 D 17 C

.

3 C 18 A

LEVEL - II 1 A 16 B 31 A

2 A 17 B 32 B

5 B 20 D

6 C 21 B

7 C 22 A

4 D 19 C 34 B

5 A 20 B 35 B

6 D 21 D

7 A 22 B

8 D 23 C

9 C 24 B

10 D 25 B

11 AD 26 A

12 BD

13 AB

14 B

15 A

11 B 26 D

12 C 27 C

13 A 28 A

14 B 29 B

15 D 30 A

.

3 B 18 A 33 B

LEVEL - III 1 C 12 C

4 A 19 B

8 B 23 B

9 B 24 C

10 B 25 A

.

2 C 13 C

3 A 14 D

4 C 15 A

5 B 16 B

6 A 17 A

7 B 18 A

8 D

6 B

7 ACD

9 B

10 A

11 A

MULTIPLE ANSWER TYPE QUESTIONS . 1 AC

2 BD

3 BD

4 AD

5 ABD

8 ABD

9 ABC

10 ABC

MULTIPLE MATCHING TYPE QUESTIONS . 1 a-eg, b-e, c-fh, d-i

2 a-e, b-e, c-ef, d-gh

3 a-fg, b-efh, c-g, d-eh

4 a-fg, b-eh, c-i, d-i

ASSERTION AND REASON . 1 A 12 D

2 D 13 D

3 A 14 A

4 B 15 A

5 C 16 D

6 A 17 C

5 C

6 A

7 C 18 B

8 D 19 A

9 A 20 A

10 B 21 A

11 C

IIT TRACK QUESTIONS : . 1 C

2 D

3 B

4 C

7 C

8 D

9 C

10 A

11 C

12 A

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