Work Energy And Power
Class 11th Physics NCERT Exemplar Solution
Mcq I
Question 1.An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system during motion, one ignores the magnetic force of one on another. This is because,
A. the two magnetic forces are equal and opposite, so they produce no net effect.
B. the magnetic forces do no work on each particle.
C. the magnetic forces do equal and opposite (but non-zero) work on each particle.
D. the magnetic forces are necessarily negligible.
Answer:
From work energy theorem, we know that change in kinetic energy of the system is equal to work done by net force acting on the system. According to the question, the magnetic forces due to motion of an electron and a proton act in perpendicular direction to the direction of motion which results in the centripetal force for the particle and hence particle performs uniform circular motion (i.e.; speed of the particle remains constant)
∴ no work is done by the forces and so no change in K.E. of the particle.
Hence, magnetic forces do no work on each particle.
Option (b) is correct.
Question 2.
A proton is kept at rest. A positively charged particle is released from rest at a distance d in its field. Consider two experiments; one in which the charged particle is also a proton and in another, a positron. In the same time t, the work done on the two moving charged particles is
A. same as the same force law is involved in the two experiments.
B. less for the case of a positron, as the positron moves away more rapidly and the force on it weakens.
C. more for the case of a positron, as the positron moves away a larger distance.
D. same as the work done by charged particle on the stationary proton.
Answer:
We know that, force between two protons is equal to the force
between proton and a positron (∵ their charges are same)
now,
mass of positron =
times mass of the proton
greater the mass lesser the distance a particle travels
∴ positron will travel the largest distance
And ,work done=
Force is same in both the cases only distance varies (positron’s
Distance is larger than proton) and
∴work done will be more in case of positron.
Option (c) is correct.
Question 3.
A man squatting on the ground gets straight up and stand. The force of reaction of ground on the man during the process is
A. constant and equal to mg in magnitude.
B. constant and greater than mg in magnitude.
C. variable but always greater than mg.
D. at first greater than mg, and later becomes equal to mg.
Answer:
According to the question more force is required by the man
during the process of standing than when he stands up i.e.;
during standing up, normal force = friction + mg ⇒ N>mg
while standing N=mg
option (d) is correct.
Question 4.
A bicyclist comes to a skidding stop in 10 m. During this process, the force on the bicycle due to the road is 200N and is directly opposed to the motion. The work done by the cycle on the road is
A. + 2000J
B. – 200J
C. zero
D. – 20,000J
Answer:
We know that, Work done=F× d
But here road is not moving ∴ work done by cycle on
road is equal to zero.
Option (c) is correct.
Question 5.
A body is falling freely under the action of gravity alone in vacuum. Which of the following quantities remain constant during the fall?
A. Kinetic energy.
B. Potential energy.
C. Total mechanical energy.
D. Total linear momentum.
Answer:
During free fall,
K.E. increases, momentum increases and P.E. decreases
Total mechanical energy=K.E.+P.E.
As P.E. decreases and K.E. increases ∴ M.E. remains constant.
Option (c) is correct.
Question 6.
During inelastic collision between two bodies, which of the following quantities always remain conserved?
A. Total kinetic energy.
B. Total mechanical energy.
C. Total linear momentum.
D. Speed of each body.
Answer:
In inelastic collision no external force is present.
∴ total linear momentum of the system remains conserved.
Option (c) is correct.
kinetic energy appears in various forms i.e.; energy
May be lost in various forms of heat, sound etc.
∴ either (initial K.E. or final K.E. will be greater)
M.E.=K.E.+P.E. (∵ K.E. Is not constant ∴ M.E. will also be
not constant)
Option (a), (b), (d) is incorrect.
Question 7.
Two inclined frictionless tracks, one gradual and the other steep meet at A form where two stones are allowed to slide down from rest, one on each track as shown in Fig. 6.1. Which of the following statement is correct?
A. Both the stones reach the bottom at the same time but not with the same speed.
B. Both the stones reach the bottom with the same speed and stone I reaches the bottom earlier than stone II.
C. Both the stones reach the bottom with the same speed and stone II reaches the bottom earlier than stone I.
D. Both the stones reach the bottom at different times and with different speeds.
Answer:
Considering the ball in incline AB has mass 
,acceleration
and the ball in incline AC has mass 
,acceleration 
.
from the above fig.
As,
We know that , 
From above equation we see that t is inversely proportional to a.
∴ 
Now,
From law of conservation of energy, we get to see in the above
Fig. that
K.E. at B=K.E. at C =P.E. at A
∴ 
….(1)
…. (2)
Option (c) is correct.
Question 8.
The potential energy function for a particle executing linear SHM is given by V (x) = 
kx2where k is the force constant of the oscillator (Fig. 6.2). For k = 0.5N/m, the graph of V(x) versus x is shown in the figure. A particle of total energy E turns back when it reaches x = ±xm. If V and K indicate the P.E. and K.E., respectively of the particle at x = +xm, then which of the following is correct?
A. V = O, K = E
B. V = E, K = O
C. V < E, K = O
D. V = O, K < E
Answer:
Given, force constant k=0.5N/m.
V(x) =
k.E. =
from law of conservation of energy,
total energy=K.E. + P.E.
At the moment of turns back then velocity=0⇒ K.E.=0
∴ total energy =P.E.
option (b) is correct.
Question 9.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V as shown in Fig. 6.3.
If the collision is elastic, which of the following (Fig. 6.4) is a possible result after collision?
A. 
B. 
C. 
D. 
Answer:
for elastic collision initial K.E. = final K.E.
let m be the mass of each ball bearings
K.E. before collision =
Now, K.E after collision in different cases
Case(a): 
Case (b): 
Case (c): 
Case (d): 
From the above cases we see that in case(b) K.E. is conserved.
Option (b) is correct.
Question 10.
A body of mass 0.5 kg travels in a straight line with velocity v = a x3/2 where a = 5 m-1/2 s-1. The work done by the net force during its displacement from x = 0 to x = 2 m is
A. 1.5 J
B. 50 J
C. 10 J
D. 100 J
Answer:
Given, 
Initial velocity, u (at x=0) =0
Final velocity ,v(at x=2m)=
Work done=change in kinetic velocity
=
=
=
Option (b) is correct.
Question 11.
A body is moving unidirectionally under the influence of a source of constant power supplying energy. Which of the diagrams shown in Fig. 6.5 correctly shows the displacement-time curve for its motion?
A. 
B. 
C. 
D. 
Answer:
According to the question, body is moving unidirectionally
under the influence of a source of constant power supplying
energy. We know that,
power=force× velocity
P=F× v
P=ma× v…… (1)
We also know that, v=u+at…. (2)
V=at (∵ u=0 body starts initially from rest)
From eq (1) and (2)
P=ma× at
P =
… (3)
We also know that, 
(∵ u=o)
….(4)
From eq (3) and (4)
⇒ 
From the above equation we see that s is directly proportional to 
Displacement-time graph is correctly shown by Option (b).
Question 12.
Which of the diagrams shown in Fig. 6.6 most closely shows the variation in kinetic energy of the earth as it moves once around the sun in its elliptical orbit?
A. 
B. 
C. 
D. 
Answer:
when earth is near the sun then due to gravitational pull
K.E. decreases and when earth is away from sun gravitational
pull is less as a result K.E. increases.
option(d) best represents it.
Question 13.
Which of the diagrams shown in Fig. 6.7 represents variation of total mechanical energy of a pendulum oscillating in air as function of time?
A. 
B. 
C. 
D. 
Answer:
When a pendulum oscillates in air its total mechanical energy
decreases continuously in overcoming resistance due to air.
We know that, total mechanical energy of pendulum decreases
exponentially with time
∴ option (c) is correct representation
of it.
Question 14.
A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be
A. 250π2
B. 100π2
C. 5π2
D. 0
Answer:
Given, m=5kg, r=1m, frequency of revolution, f =300re/min
Kinetic energy=
∴ angular velocity =
=
= (300× 2× 3.14) rad/60s
=
rad/s
=
We know that v=wR
=
=
Kinetic energy=
=
Question 15.
A raindrop falling from a height h above ground, attains a near terminal velocity when it has fallen through a height (3/4) h. Which of the diagrams shown in Fig. 6.8 correctly shows the change in kinetic and potential energy of the drop during its fall up to the ground?
A. 
B. 
C. 
D. 
Answer:
we know that at height h, P.E. is maximum and K.E.=0(∵ v=0)
Now, as the raindrop falls its P.E. starts decreasing and K.E.
starts increasing. Hence, total mechanical energy remains
conserved if air resistance is neglected.
If there is some air resistance, then there exists a force called
upthrust. So, as velocity increases upthrust also increases.
∴ during rainfall first its velocity increases then become
constant after sometime. (this velocity is termed as terminal
velocity.) Hence, K.E. also becomes constant and P.E.
decreases continuously as drop is falling continuously.
option (b) represents the correct diagram.
Question 16.
In a shotput event an athlete throws the shotput of mass 10 kg with an initial speed of 1m s-1 at 45° from a height 1.5 m above ground. Assuming air resistance to be negligible and acceleration due to gravity to be 10 m s -2, the kinetic energy of the shotput when it just reaches the ground will be
A. 2.5 J
B. 5.0 J
C. 52.5 J
D. 155.0 J
Answer:
Given, h=1.5m, v=1m/s , a=
,m=10kg
If air resistance is negligible then total mechanical energy
remains constant (P.E. at ground is zero)
initial energy= 
=
=
=155J
Now,
from law of conservation of mechanical energy,
⇒ 
∴ final K.E.=155J.
option (d) is correct.
Question 17.
Which of the diagrams in Fig. 6.9 correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity?
A. 
B. 
C. 
D. 
Answer:
When an iron sphere is falling freely in the lake, it will
accelerate first then its velocity increases and resistance
due to water causes a viscous force which oppose its
motion. Hence, during fall of the sphere, first its velocity
increases and then become constant after sometime.
option (b) is best representable graph of it.
Question 18.
A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be
A. 10.5 N
B. 21 N
C. 1.05 ×104 N
D. 2.1 × 104 N
Answer:
Given, m=150g=0.15kg, u=126km/hr=35m/s, ∆t=0.001s
v=-35m/s
it is an elastic collision.
we know that 
(∵ u=v,v=-v)
option (c) is correct.
Mcq Ii
Question 1.A man, of mass m, standing at the bottom of the staircase, of height L climbs it and stands at its top.
A. Work done by all forces on man is equal to the rise in potential energy mgL.
B. Work done by all forces on man is zero.
C. Work done by the gravitational force on man is mgL.
D. The reaction force from a step does not do work because the point of application of the force does not move while the force exists
Answer:
according to question,
When a man of mass m ,climbs up the staircase of height L
work done by gravitational force on man =-mgL
work done by muscular forces=- (work done against gravitational
force) =mgL
∴ work done by all the forces =mgL-mgL=0
as the point of application of contact forces does not move
so, work done is zero.
option (b, d) is correct.
Question 2.
A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?
A. The velocity of the bullet will be reduced to half its initial value.
B. The velocity of the bullet will be more than half of its earlier velocity.
C. The bullet will continue to move along the same parabolic path.
D. The bullet will move in a different parabolic path.
E. The bullet will fall vertically downward after hitting the target.
F. The internal energy of the particles of the target will increase.
Answer:
From law of conservation of energy,
⇒ 
⇒ 
⇒ 
…..(1)
it is given that while it is moving downward it emerges out with
half the kinetic energy
⇒ 
⇒ 
(substituting the value of v’ from eq (1))…..(2)
from eq (1) and (2)
⇒ 
∴ velocity of the bullet is more than half of its earlier velocity.
Hence, the bullet will move in a different parabolic path and internal
energy of the particle will increase.
option (b,d,f) is correct
Question 3.
Two blocks M1 and M2 having equal mass are free to move on a horizontal frictionless surface. M2 is attached to a massless spring as shown in Fig. 6.10. Initially M2 is at rest and M1 is moving toward M2 with speed v and collides head-on with M2.
A. While spring is fully compressed all the KE of M1 is stored as PE of spring.
B. While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.
C. If spring is massless, the final state of the M1 is state of rest.
D. If the surface on which blocks are moving has friction, then collision cannot be elastic.
Answer:
Collision is elastic ∴ velocities gets interchanged and K.E. and
linear momentum remains conserved.
when comes in contact with spring, it is retarded by the
spring force and is accelerated by the spring force.
now let’s see the cases,
(a) the spring will compress until two blocks acquire common
velocities .some of the K.E. of is stored as P.E. and
some into K.E. of ∴ (a) is incorrect.
(b)momentum is conserved. option (b) is incorrect.
(c) In head on elastic collision two bodies exchange their
velocities and so finally will come to rest.
option (c) is correct.
(d) ∵ there is loss in K.E. when the blocks collide on rough
surface therefore, collision is inelastic. Option (d) is correct.
Vsa
Question 1.A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline? Is there then a dissipation of energy?
Answer:
There is no dissipation of energy as the block rests on the
incline, the force of friction acting between block and incline
opposes the block to slide and as a result the block is not
in motion with respect to incline therefore, work done by
force of friction between block and incline plane is zero.
Question 2.
Why is electrical power required at all when the elevator is descending? Why should there be a limit on the number of passengers in this case?
Answer:
Electrical power is required when an elevator is descending
to prevent it from free fall under gravity.
The cable of the elevator has the capacity of some limiting
value of tension developed in it. therefore, there is a limit
on the number of passengers.
Question 3.
A body is being raised to a height h from the surface of earth. What is the sign of work done by
(a) applied force
(b) gravitational force?
Answer:
(a) When force is applied to raise the body then the angle
between force and displacement is 0°.
∴ Work done =Fscosθ =Fscos0=Fs
sign of work done is positive.
(b)As gravitational force acts in downward direction then the
angle between force and displacement is 180°.
∴ work done =Fscosθ =Fscos180°=-Fs.
sign of work done is negative.
Question 4.
Calculate the work done by a car against gravity in moving along a straight horizontal road. The mass of the car is 400 kg and the distance moved is 2m.
Answer:
Given m=400kg, d=2m
car is displaced horizontally and force (weight of the car) is
acting downward, angle between force and displacement is 90 �
∴ Work done=FscosƟ=Fscos90 �=0
Thus, work done is zero.
Question 5.
A body falls towards earth in air. Will its total mechanical energy be conserved during the fall? Justify.
Answer:
As we know that gravitation is conservative force;
Therefore, mechanical energy of the body falling towards
earth in air will be conserved.
Question 6.
A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.
Answer:
No, the work done in moving the body is not necessarily zero until the body reaches its initial position. If the body does not reach the
Question 7.
In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact).
(a) Kinetic energy.
(b) Total linear momentum? Give reason for your answer in each case.
Answer:
During the collision the there might occur deformation in the billiard balls due to which some of the kinetic energy may get lost in the potential energy. While linear momentum is always conserved so only linear momentum will be conserved at the time of collision.
Question 8.
Calculate the power of a crane in watts, which lifts a mass of 100 kg to a height of 10 m in 20s.
Answer:
Given: mass = 100kg
Initial Height 
= 0 m
Final height 
= 10 m
Time taken t = 20 s
Work done in lifting the mass W
= change potential energy
=
∴ Power = Work done per unit time =
Question 9.
The average work done by a human heart while it beats once is 0.5 J. Calculate the power used by heart if it beats 72 times in a minute.
Answer:
Work done by heart in one beat = 0.5 J
∵ it beats 72 times in a minute
∴ Work done in minute = 
∴ Power used by heart = work done per unit time = 
Question 10.
Give example of a situation in which an applied force does not result in a change in kinetic energy.
Answer:
Imagine a pendulum doing its oscillatory motion. Now the tension due to the rod is always perpendicular to the displacement. ∴ the work done by the tension force will always be the zero and hence it will never bring about any change in kinetic energy.
Question 11.
Two bodies of unequal mass are moving in the same direction with equal kinetic energy. The two bodies are brought to rest by applying retarding force of same magnitude. How would the distance moved by them before coming to rest compare?
Answer:
For particle a,
Change in kinetic energy = work done by the force
For particle b,
Change in kinetic energy = work done by the force
Now,
∵ 
and the magnitude of the force is same
∴ displacement will also be the same
Question 12.
A bob of mass m suspended by a light string of length L is whirled into a vertical circle as shown in Fig. 6.11. What will be the trajectory of the particle if the string is cut at
(a) Point B?
(b) Point C?
(c) Point X?
Answer:
(a) Vertically downward as the tangential velocity at the point B will be vertically downwards
(b) Parabolic path with vertex at C as the tangential velocity will be in the horizontal direction but due to downward gravitational force it will follow parabolic path
(c)It will also be a parabolic path with vertex at x, due to the tangential velocity and the gravitational force
Sa
Question 1.A graph of potential energy V (x) verses x is shown in Fig. 6.12. A particle of energy E0 is executing motion in it. Draw graph of velocity and kinetic energy versus x for one complete cycle AFA.
Answer:
i.) Kinetic energy vs x
We know that,
At C, P.E =0,
∴ 
Till point D there is no change in P.E, ∴ the K.E remains constant
At F, 
∴ 
From B to C the potential decreases so the kinetic energy increases ∴ we get inverted curve of potential.
From C to D potential remains constant so the kinetic energy will also remain same and thus we get a straight line parallel to x- axis
From D to F the potential energy increases to Eo linearly so the kinetic energy must fall linearly to Eo.
∴ we get the following plot,
ii.) Velocity vs X
Now, 
-----(2)
From B to C, from above equation, velocity increases parabolically symmetric from x-axis.
From C to D there is no change in P.E energy so the velocity remains constant.
From D to F, the P.E increases linearly and thus the velocity decreases parabolically from equation (2)
Now since we get two values of velocities from equation (2), we will get a mirror image of the graph in the negative side of the fourth quadrant too.
∴ we obtain following the graph,
Question 2.
A ball of mass m, moving with a speed 2v0, collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than 90°.
Answer:
a) In this problem if we prove that both the velocities after collision is positive will be enough to prove part a)
Conserving momentum,
---(1)
Where 
are the velocity of both the ball
Putting equation 1 in above equation we get,
from this we get,
∵ 
For first inequality, 
------(2)
From second inequality, 
-------(3)
Combining (2) and (3),
We get,
∵ after collision both the velocities are positive
∴ both the balls move in same forward direction.
b)
Let po be initial momentum and p1and p2be the momentum of ball 1 and 2 respectively after collision. (here bold text means vector)
In inelastic collision,
Conserving momentum,
By vector addition of momentums, we get,
-----(1)
But ∵ there is a loss in kinetic energy
∴ 
----- (2)
From equation (1) we can write equation (2) as,
-------- (3)
∴ the angle between the two velocities of scattered balls is less than 90°.
Question 3.
Consider a one-dimensional motion of a particle with total energy E. There are four regions A, B, C and D in which the relation between potential energy V, kinetic energy (K) and total energy E is as given below:
Region A: V > E
Region B: V < E
Region C: K > E
Region D: V > K
State with reason in each case whether a particle can be found in the given region or not.
Answer:
(Note: this problem is based the concept that potential energy can be negative also and E = V+K)
Region A: No, it is not possible
∵ 
and 
∴ 
which is not possible
Region B: Yes, it is possible.
∵ 
and
∴ 
which is true
Region C: Yes, it is possible.
∵ 
and
∴ 
which is possible
Region D: Yes, it is possible.
∵ 
which is possible
Question 4.
The bob A of a pendulum released from horizontal to the vertical hits another bob B of the same mass at rest on a table as shown in Fig. 6.13. If the length of the pendulum is 1m, calculate
(a) the height to which bob A will rise after collision.
(b) the speed with which bob B starts moving. Neglect the size of the bobs and assume the collision to be elastic.
Answer:
Let 
be the mass of the bob and 
be the initial height of the bob.
Potential energy of a particle raised to the height h is = 
And kinetic energy of the particle with velocity v is given as
=
Just before collision, bob A reaches ground and its potential energy gets converted to kinetic energy,
∴ conserving energy we get,
-----(1)
(a) Using the result that when two bodies of same mass collide, their velocities are mutually interchanged,
Bob A must come to rest and bob B must gain the velocity in equation (1).
a) ∴ from above we conclude the bob A will not rise to any height since its velocity will become zero after collision.
b) ∵ the velocities are interchanged,
∴ velocity of bob B =
Question 5.
A raindrop of mass 1.00 g falling from a height of 1 km hits the ground with a speed of 50 m s–1. Calculate
(a) the loss of P.E. of the drop.
(b) the gain in K.E. of the drop.
(c) Is the gain in K.E. equal to loss of P.E.? If not why.
Take g = 10 m s-2
Answer:
Mass m of drop = 1g = 0.001 kg
Change in height Δ h = 1 km = 1000 m
Final velocity = 50 
∵ the drop starts from rest, initial velocity = 0
∴ initial K.E = 0 J
And final K.E = 
a) Loss in P.E. = 
b) Gain in K.E. 
c) No, the gain in K.E. equal to loss of P.E. is not the same because of resistance like drag and viscous force against the drop.
Question 6.
Two pendulums with identical bobs and lengths are suspended from a common support such that in rest position the two bobs are in contact (Fig. 6.14). One of the bobs is released after being displaced by 100 so that it collides elastically head-on with the other bob.
(a) Describe the motion of two bobs.
(b) Draw a graph showing variation in energy of either pendulum with time, for 0 ≤ t ≤ 2T, where T is the period of each pendulum.
Answer:
a) At time period t=0
Bob A has maximum potential energy and bob B is at rest with zero energy.
At t=T/4
Bob A gains maximum velocity and collides with the bob B. bob A comes to rest and bob B gains the velocity of bob A because the collision is elastic and both have the same mass ∴ their velocity gets interchanged.
Now bob B has maximum kinetic energy.
At t=T/2
Bob B reaches the maximum height and its kinetic energy is converted to maximum potential energy. While bob A remains at rest.
At =3T/4
Bob B reaches to bob A with its maximum potential energy getting converted to max kinetic energy and collides with bob A. During the collision the interchanging of velocity takes place and bob A gains the velocity of bob B and bob B comes to rest.
At t=T
Bob A reaches the maximum height and its kinetic energy is converted to maximum potential energy. While bob B remains at rest.
And after t=T the whole process between 0<t<T repeats itself in the time period of T<t<2T
b) Here E is total energy of bob A and B ∴ we get straight lines, since they E is constant.
Now from the above discussion in part a) we get the following graphs,
For bob A,
For bob B,
Question 7.
Suppose the average mass of raindrops is 3.0 × 10-5 kg and their average terminal velocity 9 m
s-1. Calculate the energy transferred by rain to each square metre of the surface at a place which receives 100 cm of rain in a year.
Answer:
Average mass of the raindrop = 
Terminal velocity v 
Kinetic energy before hitting the ground 
∴ energy transferred by 1 drop 
Volume of water received = 
∴ total mass of the water received 
∴ no. of droplets = 
∴ total energy transferred by all the droplets
Question 8.
An engine is attached to a wagon through a shock absorber of length 1.5m. The system with a total mass of 50,000 kg is moving with a speed of 36 km h-1 when the brakes are applied to bring it to rest. In the process of the system being brought to rest, the spring of the shock absorber gets compressed by 1.0 m. If 90% of energy of the wagon is lost due to friction, calculate the spring constant.
Answer:
Mass of car m = 50,000 kg
Velocity v 
∴ kinetic energy 
∵ 90% of the energy is lost due to friction
∴ only 10% of the energy is stored by the spring
∴ 
where k is spring constant and x=compression
Question 9.
An adult weighing 600N raises the centre of gravity of his body by 0.25 m while taking each step of 1 m length in jogging. If he jogs for 6 km, calculate the energy utilised by him in jogging assuming that there is no energy loss due to friction of ground and air. Assuming that the body of the adult is capable of converting 10% of energy intake in the form of food, calculate the energy equivalents of food that would be required to compensate energy utilised for jogging.
Answer:
1 step is of 1 m
∴ in 6km there are 6000 steps
Person raises the body by 0.25 m in each step
Work done in each step = 
∴ total work in 6000 steps
W= 
W = 10% of food energy
Food energy = 10× W=
Question 10.
On complete combustion a litre of petrol gives off heat equivalent to 3×107 J. In a test drive a car weighing 1200 kg. including the mass of driver, runs 15 km per litre while moving with a uniform speed on a straight track. Assuming that friction offered by the road surface and air to be uniform, calculate the force of friction acting on the car during the test drive, if the efficiency of the car engine were 0.5.
Answer:
In 1litre of the petrol car moves distance d=15km=15000 m
With 0.5 efficiency heat energy to drive the car in one litre
Since the car moves with the uniform velocity,
Work done by the force of friction = heat energy produced by 1litre of petrol
∴ the force of friction = 1000 N
La
Question 1.A block of mass 1 kg is pushed up a surface inclined to horizontal at an angle of 30° by a force of 10 N parallel to the inclined surface (Fig. 6.15). The coefficient of friction between block and the incline is 0.1. If the block is pushed up by 10 m along the incline, calculate
(a) work done against gravity
(b) work done against force of friction
(c) increase in potential energy
(d) increase in kinetic energy
(e) work done by applied force.
Answer:
Mass of the block m = 1kg
Angle of inclination = 
Coefficient of friction = 0.1
Displacement = 10 m
a) Work done against gravity
= 
b) Work done against friction
= 
c) Vertical raise in height h 
∴ change in P.E.
d) Applying work energy theorem,
-(1)
Putting these in equation (1),
We get,
∴ there is a decrease in the kinetic energy by 41.3 J
e) Applied force F = 
Displacement d = 
Work done W by the force F is given as
= 
=
Question 2.
A curved surface is shown in Fig. 6.16. The portion BCD is free of friction. There are three spherical balls of identical radii and masses. Balls are released from rest one by one from A which is at a slightly greater height than C.
With the surface AB, ball 1 has large enough friction to cause rolling down without slipping; ball 2 has a small friction and ball 3 has a negligible friction.
(a) For which balls is total mechanical energy conserved?
(b) Which ball(s) can reach D?
(c) For balls which do not reach D, which of the balls can reach back A?
Answer:
(a) ball 1 and 3 the total mechanical energy is conserved.
For ball 1, friction is so large that starts rolling without slipping and gains rotational kinetic energy.
For ball 3, friction is negligible so there is no external force responsible for energy loss ∴ the energy is conserved.
(b) ball 1 acquires some rotational kinetic energy ∴ it will not be
able to reach C and thus will not reach point D
ball 2 loses its energy due to friction, thus it cannot reach point D
ball 3 has exactly the amount of energy to reach C and D because it does not roll and does not lose energy due to friction.
∴ only ball 3 will reach point D
(c) ball 1 and 2 starts moving in the back direction before
reaching point C.
for ball 1, it loses its energy while slipping back to point B, thus it will not reach back to point A
for ball 2, it will have clockwise rotation as it moves backs to point B due to which there will be kinetic friction coming to play due to which it will lose energy and won’t be able to reach point A.
Question 3.
A rocket accelerates straight up by ejecting gas downwards. In a small time interval ∆t, it ejects a gas of mass ∆m at a relative speed u. Calculate KE of the entire system at t + ∆t and t and show that the device that ejects gas does work = 
∆m u2 in this time interval (neglect gravity).
Answer:
At time t,
Kinetic energy 
Now at time t+Δt,
Rocket loses mass Δm and gains velocity Δv and releases gas with relative velocity of u
Velocity of rocket w.r.t ground
Velocity of gas w.r.t ground
∴ total kinetic energy of the system w.r.t ground
[neglecting terms with 
as both of them are very small]
Kinetic gained in Δt interval,
----(1)
∵ there is no external force ∴ internal forces must cancel out,
Putting the above in equation (1), we get,
By work energy theorem,
Work done = change in kinetic energy
W
∴ the device that ejects gas does work = ∆m u2 in this time interval
Question 4.
Two identical steel cubes (masses 50g, side 1cm) collide head-on face to face with a speed of 10cm/s each. Find the maximum compression of each. Young’s modulus for steel = Y= 2 × 1011 N/m2.
Answer:
From Hooke’s law, 
,
Where F = Force,
A= Area of cross section
Y= young’s modulus
L=initial length of the cube
Δ L=change in length
Kinetic energy of both boxes K.E.
Now, the kinetic energy gets converted to potential energy due to compression and,
P.E.
Now, P.E = K.E
Question 5.
A balloon filled with helium rises against gravity increasing its potential energy. The speed of the balloon also increases as it rises. How do you reconcile this with the law of conservation of mechanical energy? You can neglect viscous drag of air and assume that density of air is constant.
Answer:
According to Archimedes principle when a body is partially or fully dipped into a fluid at rest, the fluid exerts an upward force of buoyancy equal to the weight of the displaced fluid.
The balloon similarly experiences a buoyant force which provides it an acceleration 
.
If
are the mass , density and volume of the balloon and
is the density of
.
Then,
Now writing acceleration a, in its differential form,
Integrating both the sides, we obtain,
Now for the height by which it raises till time t, we use equation of motion
Putting the value of a and initial velocity to be zero,
now the kinetic energy of the balloon at time t,
K.E
Rearranging the terms, we get,
But 
are terms which represent potential energy of air and helium balloon respectively.
∴ 
So, we can say the balloon raises by pushing the air downward, thus its potential energy increases as the air’s decreases.