# Question-1

**State and explain the work-energy principle.**

**Solution:**

The work done by a force in displacing a body measures the change in its kinetic energy.

Consider a body of mass m moving with velocity u. A force F acts on it and the body acquires a velocity v.

âˆ´ Work done by the force = m

=

= final K.E. â€“ Initial K.E.

= change in K.E.

This proves the work-energy principle.

# Question-2

**One end of an elastic spring of force constant K is fixed to the wall and the other end is attached to a block lying on a horizontal frictionless surface. The mass is pulled to the right (along the +x direction) by a distance x. Find an expression for the work done.**

**Solution:**

When we stretch an elastic spring through a distance x, the spring exerts a storing force on us, such that F = -kx. where k is the force constant of the spring. This force depends on the displacement x and is opposite to x.

In order to stretch a spring we must exert a force Fâ€² on it, equal but opposite to us. Hence, the applied force is given by Fâ€² = kx.

The work done by FÏ† in stretching the spring through x is given by W, such that

W =

The work done can also be calculated using the force-displacement curve from x = 0 to x = x. It is shown by the shaded portion of the straight-line graph below.

Work done by the applied force Fâ€² = Area of Î” OMN

= Ã— Base Ã— Altitude

= x(kx) = kx^{2}

This work done is stored in the spring in the form of potential spring.

# Question-3

**A ball falls under gravity from a height of 10 m with an initial downward velocity v**

_{0}. If collides with the ground, loses 50% of its energy in collision and then rises back to the same height. Find the initial velocity.**Solution:**

Height h = 10 m

Initial velocity = v

_{0}

Let the mass of the ball be m.

The ball is projected downward with v

_{0}from height h.

âˆ´ Total energy = mgh +

Total energy after the collision =

The ball rises back to the same height h after the collision, hence

v

_{0}= = 14 m/s.

# Question-4

**A body of mass 1 kg, initially at rest explodes and breaks into three fragments of masses in the ratio 1:1:3. The two pieces of equal mass fly off perpendicular to each other with a speed of 30 m/s each. What is the velocity of the heavier fragment?**

**Solution:**

A and B be two pieces of equal mass fly off perpendicular to each other with equal velocity (30 m/s).

Momentum of A and B = mv = = 6 kg ms^{-1}

Resultant momentum of A and

B = 6 kg ms^{-1 }

It will act along the bisector of âˆ AOB

Using the law of conservation of momentum, we have

V = Ã— 6

V = 10 kg ms^{-1 }

Thus, C will have momentum of 6 kg ms^{-1} along OC.

# Question-5

**When a constant force is applied to a body moving with a constant acceleration, is power of the force a constant? If not, how will force vary with speed for the power to be constant?**

**Solution:**

Power = force Ã— velocity

Since the body has acceleration, therefore, velocity changes and power also changes. In order to keep the power constant, the force should vary inversely as velocity varies with time. In order words, if velocity increases with time then force should decrease with the same rate as the velocity is increasing.

# Question-6

**Solution:**

Let us find the resultant of the two momenta acting at right angles, along OP, the bisector of âˆ YOX. The resultant is given by,

= kg m/s

Using the conservation of momentum, the remaining fragment will act along OQ such that

MV = 20 or M = ( V = 40 m/s) = 0.5 kg.

# Question-7

**A running man has half the kinetic energy that a boy of half his mass has. The man speeds up by 1 m/s and then has the same kinetic energy as the boy. What were the original speeds of the man and the boy?**

**Solution:**

Let m

_{1}and m

_{2}be the masses of the man and the boy respectively. If u

_{1}and u

_{2}be their respective speeds, we have, m

_{1}= 2 m

_{2}.

âˆ´

u

_{1}=

Again

or m

_{1}(u

_{1}+ 1)

^{2}=

or

or

âˆ´ u

_{1}=

Taking +ve sign, u

_{1}= + 1 = 2.4 m/s

âˆ´ u

_{2}= 4.8 m/s.

# Question-8

**What is the minimum stopping distance for a car of mass m, moving with speed**

**Î½ along a level road, if the coefficient of static friction between the tyres and the road is Î¼**

**?**

**Solution:**

Minimum value of frictional force = Î¼ mg

Let y be the minimum stopping distance, then the work done against the frictional force is W = Î¼ mg y

= Kinetic energy of the car

= m Î½ ^{2 }

âˆ´ Y =^{ .}

# Question-9

**What is the power output of the sun if 4**

**Ã—**

**10**

^{9 }kg of matter per second^{ }is j converted into energy in the sun?**Solution:**

E = mc

^{2 }= 4 Ã— 10

^{9}(3 Ã— 10

^{8})

^{2}= 3.6 Ã— 10

^{26 }J

âˆ´ Power output of the sun = 3.6 Ã— 10

^{26}w.

# Question-10

**(c) What kind of energy transformation take place at a hydro-electric power house?**

**Solution:**

(a) Zero.

(b) Zero.

(c) The potential energy of water changes into K. E. and then into electrical energy.

# Question-11

**(ii) they are stretched by the same force?**

**Solution:**

(i) W = Kx

^{2 }

F = Kx, where K is force constant of spring and x is distance through which the spring is stretched

âˆ´ W_{A }= K_{A }x^{2 }

W_{B }= K_{B }x^{2} (Since K_{A} > K_{B, }W_{A} > W_{B})

(ii) Putting x = in (i) we have,

W_{A }= K_{A }= and

W_{B} =

_{ }

_{}

âˆ´ W_{A }< W_{B.}

# Question-12

**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. What is the total energy of the particle?**Solution:**

^{âˆ´ }K.E.=

Potential energy, U =

Total energy, E = U + K.E.

=

# Question-13

**A porter moving vertically up the stairs with a suitcase on his head does work. Explain.**

**Solution:**

The porter lifts the suitcase vertically to the upstairs. He has vertical displacement against weight. Hence he does work.

# Question-14

**A lorry and a car with the same kinetic energy are bought to rest by the application of brakes which provide equal retarding forces. Which of them will come to rest in a shorter distance.**

**Solution:**

Loss of kinetic energy E = Work done = FS or S =

Since the retarding force F and kinetic energy E are the same in both the cases, therefore, both car and lorry will come to rest after traveling equal distances.

# Question-15

**The velocity of an aeroplane is doubled:**

(a) What will happen to the mountain? Will the momentum remain conserved?

(b) What will happen to the kinetic energy? Will the energy remain conserved?

(a) What will happen to the mountain? Will the momentum remain conserved?

(b) What will happen to the kinetic energy? Will the energy remain conserved?

**Solution:**

(a) When the velocity of aeroplane is doubled, its momentum gets doubled but the momentum of air also gets increased by the same amount in opposite direction, so that total momentum of plane. In fact, the total energy will remain conserved.

(b) The kinetic energy will become four times. Additional energy is obtained by burning of the fuel of the plane. In fact, the total energy will remain conserved.

# Question-16

**A crane lifts a car up to a certain height in 1 min. A second crane lifts the same car up to the same height in 2 min. Which crane supplies the greater power? Which crane consumes more fuel?**

**Solution:**

W = mgh, hence the two cranes will do the same amount of work and will therefore consume the same amount of fuel. The first crane does the same amount of work in half the time than the second one. Hence the first crane supplies two times power than the second one.

# Question-17

**Why does a coolie does no work when he moves on a level road while carrying a box on his head?**

**Solution:**

We know that work done W = FS cos Î¸

where Î¸ is the angle between the force and the displacement.

In this case, coolie applies a force at right angle to the displacement, i.e. Î¸ = 90Â°Therefore W = FS cos 90 = 0.

# Question-18

**If a block attached to a string is pulled up to x**

_{0}and released the amplitude of its motion cannot exceed**Â±**

**x**

_{0}. Why?**Solution:**

If the amplitude of motion exceed Â± x

_{0}, then the potential energy would be greater than kx

^{2}

_{0}. This would mean that mechanical energy is not conserved. So the block cannot go beyond Â± x

_{0}

# Question-19

**An engine develops 10 kW of power. How much time will it take to lift the mass of 200 kg to a height of 40 m? Take g = 10 ms**

^{-2}.**Solution:**

We know that,

**Power =**

Work done to raise the body through height h = mgh

Therefore P =

t =

=

= 8 s.

# Question-20

**(a) What happens when a light sphere collides head on with a massive sphere initially at rest? Explain.**

(b) What happens when a massive sphere collides head on with a light sphere initially at rest. Explain.

(b) What happens when a massive sphere collides head on with a light sphere initially at rest. Explain.

**Solution:**

(a) The light sphere rebounds almost with the same velocity. There is practically no change in the velocity of the massive sphere.

(b) The velocity of the massive sphere remains unchanged by the light sphere, moves with a velocity almost twice that of massive sphere.

# Question-21

**A spring is kept compressed by tying its ends together tightly. It is then placed in a strong acid, and dissolved. What happens to its stored potential energy?**

**Solution:**

The loss in potential energy appears in the form of kinetic energy of the molecules of the acid.

# Question-22

**When an arrow is shot from its bow, it has kinetic energy. From where does it get this kinetic energy?**

**Solution:**

The potential energy of the bow and arrow system is converted into the kinetic energy of the arrow.

# Question-23

**A cake of mud is thrown on a wall where it sticks. What happens to its initial kinetic energy?**

**Solution:**

A part of the kinetic energy is used in deforming the cake and the remaining part is converted into heat and sound energy.

# Question-24

**A particle is moving in a circular path of given radius with**

(i) constant

(ii) increasing

(iii) decreasing speed. What happens to the work done in all three cases?

(i) constant

(ii) increasing

(iii) decreasing speed. What happens to the work done in all three cases?

**Solution:**

We know that work done = change in kinetic energy, hence.

(i) For constant speed, kinetic energy, does not change, and work done is zero.

(ii) For increasing speed, kinetic energy is increasing, and work is done by the external force.

(iii) For decreasing speed, kinetic energy is decreasing, and work is done by the particle itself.

# Question-25

**A spring has a spring constant of 80 Ncm**

^{-1}. The spring is compressed 12 cm by a ball of mass 15 g. How much is the potential energy of the spring? If the trigger is pulled, what will the velocity of the ball be?**Solution:**

k= 80 Ncm^{-1 }= 80Ã—100 Nm^{-1}

x=12 cm= 0.12 m

m=15 g= 15Ã—10^{-3} kg

Potential energy,

Also, kinetic energy of the ball = potential energy of the spring

Or

# Question-26

**An object is attached to a vertical spring and slowly lowered to its equilibrium position. This stretches the spring by a distance d. If the same object is attached to the same vertical spring but permitted to fall freely, through what distance does it stretch the spring?**

**Solution:**

If mg is the weight of the object, then in equilibrium position,

When the object falls freely, its gravitational potential energy is converted into elastic potential energy of the spring.