Work, Energy and Power

(i) When the displacement of the body is in the direction of force then the work done is said to be positive.

Therefore, W = F × S

Example - When a child pushes a toy table then the force applied by the child and the displacement of the table are in the same direction. So, work done is said to be positive.

(ii) When the displacement of the body is in opposite direction to the force applied then the work done is negative.

Therefore, W = – F × S

Example - When a ball is thrown upwards with a force, the balls moves to a height (h).

However, the displacement is opposite to the direction of force of gravity.

W = -mgh

When body of mass m falls through a height (h) either vertically or on an inclined plane work is said to be done.

We know, work done is

W = F x S

However, the force of gravity on the body is F = mg acting vertically downwards and the vertical displacement in the direction of force is S = h.

Applying the values of F and S we get,

Work done by the force of gravity is

W = F x S
= m x g x h

The two factors on which the power spent by a source depends are :

1. The amount of work done by the source
2. The time taken by the source to do the said work.

For example:

Let us suppose a man X takes 5 minute to lift a load to the roof of a house and a man Y takes 10 minutes to lift the same load to the roof of the same house.

The work done by both the persons remain the same, but the power spent by the man X is twice the power spent by the man Y because man X does the work faster than man Y.

${W} = \vec{F}.\vec{S}$

Reason — As force F and displacement S are vector quantities and work W is a scalar quantity so work is expressed as a dot product of force and displacement vectors. Hence, we get the expression : ${W} = \vec{F}.\vec{S}$

The dot product of two vectors is a scalar.

0°

Reason — If the displacement is in the direction of force i.e., θ = 0°, then cos 0° = 1 and work done is maximum.

Zero

Reason — When a body revolves in a circular path under the influence of a force, on completing one round the displacement becomes zero, so the work done by the force is W = 0 because work done is a product of force and displacement.

-mgh

Reason — When a ball m is thrown upwards to a height h, the displacement h (upwards) is opposite to the direction of force of gravity mg (downwards), so the work done by the force of gravity mg in displacement h is W = -mgh i.e., negative.

equal to work done by A

Reason — From formula: ${W} = \vec{F}.\vec{S}$

Given that the distance lifted and the force applied are the same for both coolies, as they are lifting the same load to the same height (the roof of the bus) hence, the work done by both coolies is the same.

The time taken to do the work doesn't affect the amount of work done.

746 W

Reason — 1 horse power = 746 W

10⁷

Reason — 1 Joule is equivalent to 10⁷ erg

4.0 x 10^-19 J

Reason — Given, energy of electron = 2.5 eV

1 eV = 1.6 x 10 ^-19 J

Hence,

2.5 eV = 2.5 x 1.6 x 10 ^-19 = 4 x 10 ^-19 J

600 J

Reason — Given,

Force = 200 N

Displacement = S = 3 m

As we know,

Work done = force × displacement in the direction of force

$W = F × S \\[0.5em] W = 200 \times 3 \\[0.5em] W = 600 J \\[0.5em]$

Therefore, work done is 600J.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true.

As,

Work done (𝑊) =$\vec𝐹$⋅$\vec𝑑$= Fdcosθ .............(i)

Where

F = magnitude of force

d = displacement

θ = angle between force and displacement

If displacement is normal (perpendicular) to the force, then 𝜃 = $90^o$

$\Rightarrow$ cos $90^o$ = 0 $\Rightarrow$ W = 0

Reason (R) is true.

From (i) it is clearly visible that work done depends upon angle between force vector and displacement vector.

Hence, R is the correct explanation of A.

Given,

Force = 20 kgf and g = 10N per kg

So,

Force = 20 x 10 = 200 N

Displacement = S = 1 m

As we know,

Work done = force × displacement in the direction of force

(i) When displacement is in the direction of force

$W = F × S \\[0.5em] W = 200 \times 1 \\[0.5em] W = 200 J \\[0.5em]$

Therefore, work done is 200J.

(ii) When displacement is at an angle of 60° with the force

$W = F × S \space cos \space \theta \\[0.5em] W = 200 \times 1 \space Cos \space 60 \\[0.5em] W = 200 \times 1 \times 0.5 \\[0.5em] W = 100 J$

Therefore, work done is 100J.

(iii) When displacement is normal to the force

$W = F × S \space cos \space \theta \\[0.5em] W = 200 \times 1 \space Cos \space 90 \degree\\[0.5em] W = 200 \times 1 \times 0 \\[0.5em] W = 0$

Therefore, work done is 0.

Given,

m = 40 kg

h = 8 m

t = 5 s

(i) Force of gravity acting on the boy

$F = mg \\[0.5em] F = 40 \times 10 = 400N \\[0.5em]$

(ii) Work done against gravity

$W = F × S \\[0.5em] W = 400 \times 8 = 3200J \\[0.5em]$

(iii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] W = \dfrac{3200}{5} = 640 W$

Given,

Work done = 7.4 kJ

S = 74 m

t = 2.5 s

(i) W = F x S

$\text{Force} = \dfrac{\text{work done}}{\text{distance}} \\[0.5em] F = \dfrac{7.4 \times 10^3}{74} \\[0.5em] \Rightarrow F = 100 N \\[0.5em]$

(ii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] P = \dfrac{7.4 \times 10^3}{2.5} \\[0.5em] \Rightarrow P = 2960 W \\[0.5em]$

We know,

$\text{1 H.P} = \text{746 W} \\[0.5em] \text{1 W}= \dfrac{1}{746} \text{ H.P} \\[0.5em] \therefore \text {2960 W} = \dfrac {2960}{746} \text{ H.P} \\[0.5em] \Rightarrow \text {2960 W} = \text{3.97 H.P}$

Given,

mass = 200kgf

height = 2.5m

time = 5s

Force = mg
= 200 x 10 = 2000N

So, we get force is equal to 2000N

(i) Work done = F x h

$W = 2000 \times 2.5 = 5000 J\\[0.5em]$

Therefore, work done is 5000 J.

(ii)

$\text{Power developed} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] P = \dfrac{5000}{5} = 1000W$

Therefore, power developed by the boy 1000 W.

Given,

Force = 750N

height or distance moved = 16 m

time = 5 s

(i) Energy spent or the work done = F x S

$W = 750 \times 16 = 12000 J \\[0.5em]$

Therefore, energy spent is equal to 12000 J

(ii)

$\text{Power spent} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] W = \dfrac{12000}{5} = 2400 W \\[0.5em]$

Therefore, power spent is equal to 2400 W.

Given,

Power of the electric heater = 3kW

Time for which the electric heater is used = 10 hours

We know,

Energy consumed = power × time

(i) Energy consumed in kWh

$\text{Energy} = 3 \times 10 \\[0.5em] \Rightarrow \text{Energy} = 30 kWh \\[0.5em]$

(ii) Energy consumed in joule.

$1 kWh = 3.6 \times 10^6 J \\[0.5em] 30 kWh = 30 \times 3.6 \times 10^6 \\[0.5em] \Rightarrow 30 kWh = 1.08 \times 10^8 J \\[0.5em]$

Given,

Volume of water raised = 50 L and
50 L = 50 × 10^-3 m³

height = 25m

time spent = 5s

As we know,

Mass of water = Volume of water × density of water

$Mass =\dfrac{50\times 1000}{1000} = 50 Kg$

Work done in raising 50 kg water to height of 25 m is

W = mgh

Power = work done / time taken

Substituting the values we get

$\text{Power} = \dfrac{50 \times 10 \times 25}{5} = 2500W\\[0.5em]$

Therefore, power of the pump is 2500W.

Given,

Mass of water = 600kg

Height to which the water has to be raised = 75 m

Time = 10 s

(a) Work done by the pump = mgh

Substituting the values we get,

$W = 600 \times 10 \times 75 \\[0.5em] \Rightarrow W = 4.5 \times 10^5 J \\[0.5em]$

(b) $\text{Power of the pump} = \dfrac{\text{work done}}{\text{time taken}}$

$\phantom{(b)\space}\Rightarrow\text{Power} = \dfrac{\text{mgh}}{\text{t}}$

Substituting the values we get,

$\text{Power} = \dfrac{600 \times 10 \times 75 }{10} = 45000 \text{ W}\\[1em] \Rightarrow \text{Power} = 45 \text{ kW} \\[1em]$

(c) $\text{Efficiency} = \dfrac{\text{useful power}}{\text{power input}}$

Given,

Efficiency = 40 %

$0.4 = \dfrac{45}{\text{input power}} \\[0.5em] \text{input power} = \dfrac{45 \times 10^3}{0.4} \\[0.5em] = 112.5 \times 10^3 \text{ W} \\[0.5em] = 112.5 \text{ kW}$

∴ Power rating of pump = 112.5 kW.

Given,

Force = 1000 N

Velocity = 30 m/s

Power = force × velocity

Power = 1000 x 30 = 30,000 W

Therefore, power developed by the ox is 30 kWh

Given,

Power = 40 kW

Force = 20,000 N

Power = force × velocity

Hence,

$\text{velocity} = \dfrac{\text{power}}{\text{force}} \\[0.5em] \Rightarrow \text{velocity} = \dfrac {40,000}{20,000} \\[0.5em] \Rightarrow \text{velocity} = 2 \space ms^{-1} \\[0.5em]$

Power exerted by Rajan = force x velocity

Power = 150 x 10 = 1500 W

We know,
work done = Force × distance moved in direction of force
i.e. W = F × S

Given,

Force = 350 N

Distance covered in 30 steps (S)

S = 30 x 20 cm = 600cm
⇒ S = 6m

Substituting the values of F and S

$W = 350 \times 6 \\[0.5em] = 2100J \\[0.5em]$

∴ Work done by boy = 2100J

(ii)

$\text{Power developed} = \dfrac{\text{work done}}{\text{time taken}} \\[0.5em] = \dfrac{2100}{60} \\[0.5em] = 35 W \\[0.5em]$

∴ Power spent by boy = 35W

(i) We know that work done is a product of force and displacement and is independent of time. Hence both persons A and B will do the same amount of work.

Therefore, ratio of work done by A and B = 1:1

(ii) The power developed by the persons A and B :

We know that Power = work done / time

Work done is same by A and B

Time A took = 20 s

Time B took = 15 s

Therefore, more power is spent by B as B does the work at a faster rate.

Substituting the values in formula we get :

Power = 1 / time

${\text{Power spent by A}} : {\text{Power spent by B}} = \dfrac{15}{20} \\[0.5em] \Rightarrow {\text {Power spent by A}} : {\text {Power spent by B}} = 3 : 4 \\[0.5em]$

(i) We know that work done is a product of force and displacement

W = F x S = mgh

Given,

m₁ = 40 kgf
m₂ = 30 kgf
t₁ = 4 min = 4 x 60 =240 sec
t₂ = 3 min = 3 x 60 = 180 sec
h₁ = h₂ = 30 x 20 = 600 cm = 6 m

$\dfrac{\text {Work done by boy}}{\text {Work done by girl}} = \dfrac{m_1gh_1}{m_2gh_2} \\[0.5em] = \dfrac{m_1}{m_2} = \dfrac{40}{30} \\[0.5em] = \dfrac{4}{3} \\[0.5em]$

Therefore,

Work done by boy : Work done by girl = 4 : 3

(ii) The power developed —

We know that,

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] \dfrac{\text{Power developed by boy}}{\text{Power developed by girl}} \\[1em] = \dfrac{\dfrac{\text{work done by boy}}{\text{time taken by boy}}}{\dfrac{\text{work done by girl}}{\text{time taken by girl}}} \\[1em] = \dfrac{\dfrac{40 \times 10 \times 6}{240}}{\dfrac{30 \times 10 \times 6}{180}} \\[1em] = \dfrac{\dfrac{2400}{240}}{\dfrac{1800}{180}} \\[1em] = \dfrac{10}{10} \\[1em] = \dfrac{1}{1}$

Therefore,

Power developed by boy : Power developed by girl = 1:1

(a)

(i) Work done = force x displacement

Hence, work done is same for both men as both carry 50 kg weight to a height of 2m.

Therefore,

Work done by first man : Work done by second man = 1 : 1

(ii)

Let the work done by both men be W.

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[1em] \dfrac{\text{Power developed by Man A}}{\text{Power developed by Man B}} \\[1em] = \dfrac{\dfrac{\text{work done by Man A}}{\text{time taken by Man A}}}{\dfrac{\text{work done by Man B}}{\text{time taken by Man B}}} \\[1em] = \dfrac{\dfrac{W}{20}}{\dfrac{W}{50}} \\[1em] = \dfrac{5}{2}$

Power developed by Man A : Power developed by Man B = 5:2

(b)

(i) As we know,

W = F x S = mgh

$W = 50 \times 10 \times 2 = 1000 J \\[0.5em] \Rightarrow W = 1000 J \\[0.5em]$

(ii) Power developed by Man A

$P = \dfrac{1000}{20} \\[0.5em] \Rightarrow P = 50 W\\[0.5em]$

Power developed by Man B

$P = \dfrac{1000}{50} \\[0.5em] \Rightarrow P = 20 W\\[0.5em]$

(a)

(i) Work done = force x displacement

Work done is same for boy and father as both lift 20 litre water from 20 m deep well.

Work done by boy : Work done by father = 1:1

(ii) Comparing Power developed:

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] {\text {Power of boy}} : {\text {Power of father }} = \dfrac{\dfrac{W}{3}}{\dfrac{W}{2}} \\[0.5em] \Rightarrow {\text {Power developed by boy}} : {\text {Power developed by father}} = \dfrac{2}{3} \\[0.5em]$

Power developed by boy : Power developed by father = 2:3

(b) Work done = mgh

Given,

density of water = 10³kg m ^-3

To convert, density of water to mass per litre

We know,

1 litre = 1000 cm³
and
1 m³ = 10⁶ cm³

So,

$10^3\text{ kgm}^{-3} = \dfrac{1000\text{ kg}}{1\text{ m}^3} \\[1em] = \dfrac{1000\text{ kg}}{10^6 \text{ cm}^3} \\[1em] = \dfrac{1\text{ kg}}{1000\text{ cm}^3} \\[1em] = \dfrac{1\text{ kg}}{1 \text{ litre}}$

Hence,

20L of water = 20 Kg by mass

height = 20m

Now, substituting the values in the formula for work done we get

W = 20 x 9.8 x 20
= 3.92 kJ

The definition of work states that when force is applied on a body and the body moves then work is said to be done.

The condition for work done is displacement of the body when force is applied.

(i) When force is in direction of displacement then work done is given as —

Work done = Force x displacement of the point of application of force in the direction of force.

When,

W = work done

F = force applied

S = displacement of the body then we get,

W = F × S

(ii) When force acts at an angle to the direction of displacement then work done is given as —

Work done = Force x component of displacement in the direction of force.

When,

W = work done

F = force applied

S = displacement of the body then we get,

W = F × S Cos θ

The two conditions for the work done to be zero are,

(i) When there is no displacement of the body on application of force

i.e. S = 0 and

(ii) When displacement is normal to the direction of force

i.e. θ = 90° as cos 90° = 0

When a body moves in a circular path, no work is said to be done by the body as the force is directed towards the centre of circular path (the body is acted upon by the centripetal force).

At all points the displacement is along the tangent to the circular path, (i.e normal to the direction of the force).

When a satellite revolves around the earth in a circular orbit the work done is zero as force of gravity acting on satellite is perpendicular to its displacement.

Work done is given by the formula:

W = F × S Cos θ

Coolie Y is moving the load on a frictionless horizontal platform. The force acting on the load is the force of gravity. In case of Coolie Y as the platform is horizontal so the angle between force and displacement is 90°.

Work done by Coolie Y = F × S Cos 90° = F x S x 0 = 0.

Therefore Coolie Y does zero work.

Coolie X is moving the load on a slope. Assuming the angle of slope as θ:

Work done by Coolie X = F × S Cos θ

Therefore, work done by coolie X is more than coolie Y.

The work done by a fielder when he takes a catch is said to be negative as the force applied by him is in the direction opposite to the displacement of the ball.

The S.I unit of work is Joule.

C.G.S unit of work is erg.

Relation between joule and erg:

1 joule = 1N × 1m

As we know, 1 N = 10⁵ dyne and 1 m = 10² cm

Therefore,

    1 joule = 10⁵ dyne x 10² cm
⇒ 1 joule = 10⁷ dyne cm
So, 1 joule = 10⁷ erg.

The S.I unit of work is Joule.

1 joule of work is said to be done when a force of 1 newton displaces a body through a distance of 1 metre in its own direction.

1 joule = 1N × 1m

As we know, 1 N = 10⁵ dyne and 1 m = 10² cm

Therefore,

    1 joule = 10⁵ dyne x 10² cm
⇒ 1 joule = 10⁷ dyne cm
So, 1 joule = 10⁷ erg.

Energy is defined as the capacity to do work. The S.I unit of energy is Joule.

kilowatt-hour is the energy spent (or work done) by a source of power 1 kW in 1 hour.

1 kWh = 3.6 × 10⁶ joule.

Power is defined as the rate of doing work. The S.I unit of power is watt (W)

The S.I unit of power is watt (W).
The power spent is said to be 1 watt, if 1 joule of work is done in 1 second.

The unit of power is watt (W) while the unit of work is watt-hour (Wh), since power × time = work or energy.

(a) When force acts at an angle θ to the direction of displacement then work done is

W = F × S Cos θ

(b)

(i) For work done to be zero the angle between force and displacement should be equal to 90° as cos 90° = 0

Therefore, product will also be zero.

W = F × S Cos 90° = 0

(ii) For work done to be maximum the angle between force and displacement should be equal to 0° as cos 0° = 1

Therefore, product will also be maximum.

W = F × S Cos 0° = F × S

When a body is moved in a direction opposite to the direction of force then work is done against the force acting on the body.

When the displacement of the body due to application of force is in a direction opposite to the force then,

θ = 180°, and

cos 180° = -1

Therefore, Work done = (-) Force x displacement of the body.

(a) No, work is not done by the man.

(b) No, work is not done by the coolie.

(c) Yes, work is done by the boy in climbing up the 20 stairs.

Explanation

(a) Work is not done by the man as on pushing the wall, it doesn't move hence, there is no displacement of the wall. Work is defined as the product of force and displacement in the direction of the force. Since there is no displacement, no work is done.

(b) Standing with a box on the head doesn't involve any displacement in the direction of the force of gravity acting on the box. The coolie is not moving the box, nor is there any vertical displacement. Therefore, no work is said to be done.

(c) When the boy climbs stairs, he exerts a force to lift his body against gravity. As he climbs, there is a vertical displacement as he moves from one step to the next. Since there is both force and displacement in the direction of the force (upward against gravity), work is done in this scenario.

When a coolie carrying a load moves on a horizontal platform, the force of gravity acting on him is normal to the displacement of the load i.e. the angle between force and displacement of load is 90°.

Work done by Coolie = F × S Cos 90° = F x S x 0 = 0.

Therefore, in this case work done by the force of gravity is zero even though the load gets displaced from its initial position.

When a boy of mass m climb up stairs of vertical height h then

(a) Work done by the boy against the force of gravity is

W = FS = mgh

(b) The work done when the boy uses a lift in climbing the same vertical height

W = FS = mgh

The energy of atomic particles is measured in electron volt (eV)

    1 eV = charge on an electron × 1 volt
⇒ 1 eV = 1.6 x 10^-19 coulomb x 1 volt
⇒ 1 eV = 1.6 x 10^-19 J

(a) 1 J = 0.24 calorie.

(b) 1 kWh = 3.6 x 10⁶ J.

The physical quantity which is measured in calorie is heat energy.

1 calorie = 4.18 joule

(a) Power is the physical quantity measured in terms of horsepower.

(b) The relation between horsepower and S.I. unit of power is 1 horsepower = 746 watt.

(a) Energy is measured in kWh

(b) Power is measured in kW

(c) Energy is measured in Wh

(d) Energy is measured in eV

Yes there are cases when we will see no transfer of energy even on application of force.

Example - When a body moves in circular motion then force acts normal to the displacement of the body then there is no transfer of energy and work done is zero.

The potential energy possessed by a body due to the force of attraction of the earth on it, is called the gravitational potential energy.

When a body is placed at a height above the ground then the amount of work done in lifting the body to that height against the force of gravity is called the gravitational potential energy.

When we lift a body of mass m from the ground to a height h.Then, force of gravity (mg) acts on the body acting vertically downward which must be equal to the least upward force F required to lift the body.

Work done = force of gravity (mg) × displacement (h)

⇒ W = mgh

So, the work done on the body is stored in the form of its gravitational potential energy when it is at a height h.

Hence,

Gravitational potential energy U = mgh.

When the string of a bow is pulled, the work done is stored in the form of elastic potential energy.

When the string is released to shoot the arrow, the potential energy of the bow changes into the kinetic energy.

The kinetic energy is the same potential energy stored when the string was pulled.

When the ball is placed on a compressed spring, the compressed spring has elastic potential energy.

The potential energy of the spring changes into kinetic energy when it is released and as a result it flies.

On being thrown upwards, the kinetic energy of the pebble changes to potential energy.

Its kinetic energy is completely converted into potential energy when the pebble reaches its maximum height.

The potential energy is then converted into kinetic energy while coming down and the potential energy is completely converted to kinetic energy when the pebble reaches the ground.

When the water falls from the height, the potential energy stored in water at a height changes into the kinetic energy.

On reaching the ground, a part of the kinetic energy of water changes into the heat energy which increases the temperature of the water.

Gravitational potential energy

Reason — Gravitational potential energy is the energy an object possesses because of its position in a gravitational field. It arises from the gravitational force between the object and the Earth (or another massive body).

Elastic potential energy

Reason — When an object is deformed (stretched, compressed, or bent), it stores energy within itself due to the deformation. This energy is called elastic potential energy. It arises from the forces within the object's material that act to restore it to its original shape and size when the deforming force is removed.

work done

Reason — According to the work energy theorem, the increase in the kinetic energy of a moving body is equal to work done by a force acting in the direction of the moving body.

vibrational kinetic energy 

Reason — When a wire clamped at both ends is struck in the middle, it will vibrate back and forth. This motion results in vibrational kinetic energy, as the wire undergoes oscillations about its equilibrium position.

both rotational and translational kinetic energy

Reason — When a vehicle is in motion, its wheels are both rotating and translating (moving in a straight line). The motion of the vehicle results in translational kinetic energy, as the vehicle moves forward. Additionally, each wheel is rotating about its axis, so it possesses rotational kinetic energy due to its spinning motion.

$\dfrac{1}{9}$

Reason — Let initial velocity be v.

Kinetic Energy at v = $\dfrac{1}{2}$ x m x v² = $\dfrac{\text{mv}^2}{2}$

Kinetic Energy at $\dfrac{\text{v}}{3}$ = $\dfrac{1}{2}$ x m x $(\dfrac{\text{v}}{3})^2$ = $\dfrac{1}{9}\Big(\dfrac{\text{mv}^2}{2}\Big)$

Hence, the kinetic energy is reduced by $\dfrac{1}{9}$

2000 J

Reason — Given,

Initial velocity = 20 m/s.

Mass = 10 kg

Kinetic Energy = $\dfrac{\text{mv}^2}{2}$

Substituting we get,

Kinetic Energy = $\dfrac{\text{10 x (20) }^2}{2}$ = $\dfrac{\text{10 x 400}}{2}$ = 2000 J

K = $\dfrac{p^2}{2m}$

Reason — The kinetic energy K and momentum p are related as K = $\dfrac{p^2}{2m}$.

momentum

Reason — The momentum of an object is given by the product of its mass (m) and velocity (v) represented by :

p = mv

If the velocity is doubled, then the new momentum (p') becomes:

p' = m x 2v = 2(mv) = 2p

So, the momentum gets doubled when the velocity is doubled.

light energy to electrical energy

Reason — A photoelectric cell, also known as solar cell, converts light energy (photons) into electrical energy (electrons). When light strikes the surface of the cell, it excites electrons, causing them to be ejected from the material (the photoelectric effect). These ejected electrons can then be collected and used as electric current.

chemical energy to kinetic energy

Reason — In a petrol vehicle, the chemical energy stored in the fuel (petrol) is converted into kinetic energy as the vehicle moves. This conversion of chemical energy into kinetic energy occurs through the process of combustion within the engine, where the chemical bonds in the fuel are broken, releasing energy that propels the vehicle forward.

nuclear energy changes to electrical energy

Reason — Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into smaller parts, releasing a significant amount of energy. This energy is in the form of nuclear energy and is used for constructive purposes to produce electrical energy.

chemical to electrical

Reason — An electric cell, such as a battery, converts chemical energy stored within it into electrical energy when in use. The chemical reactions occurring within the battery cause a flow of electrons, generating an electric current.

Both A and R are true and R is the correct explanation of A.

Explanation

Assertion (A) is true. When the momentum of a body increases by 50%, its kinetic energy will increase by 125%.

Let initial momentum be p and kinetic energy be K

If p increases by 50%, the new momentum p' = p + $\dfrac{p}{2}$ = $\dfrac{3p}{2}$

Relation between kinetic energy and momentum is given by; K = $\dfrac{p^2}{2m}$

So, $\dfrac{K'}{K}$ = $\dfrac{\dfrac{p'^2}{2m}}{\dfrac{p^2}{2m}}$ = $\dfrac{p'^2}{p^2}$ = $\dfrac{\Big(\dfrac{3p}{2}\Big)^2}{\dfrac{p^2}{1}}$ = $\dfrac{9}{4}$

So,

Percentage increase in kinetic energy = $\dfrac{K'-K}{K}$ x 100

= $\dfrac{ \dfrac{9K}{4} -K}{K}$ x 100

= $\dfrac{ \dfrac{9K-4K}{4}}{K}$ x 100

= $\dfrac{ \dfrac{5K}{4}}{K}$ x 100

= $\dfrac{5}{4}$ x 100 = 125%

Hence, kinetic energy will increase by 125%.

Reason (R) is true. Kinetic energy is proportional to the square of velocity, according to the formula: KE = $\dfrac{1}{2}$mv².

Given,

Mass of body A and body B = m each

Height H_A = h

Height H_B = 2h

We know, Gravitational potential energy = mgh

Substituting the values we get,

$\text {Ratio of gravitational P.E.} = \dfrac{mgh}{mg2h} \\[0.5em] \Rightarrow \text {Ratio of gravitational P.E.} = \dfrac{1}{2} \\[0.5em]$

Hence, ratio of gravitational potential energies of the two bodies = 1 : 2

Given,

Mass, m = 1 kg

Height, h = 5 m

Gravitational potential energy = mgh
= 1 × 10 × 5
= 50 J

When the mass hits the ground, its entire Potential energy is converted to Kinetic energy.

Hence, Kinetic energy of mass on hitting the ground = 50 J.

Given,

Mass of box = 300 Kg

Gravitational potential energy = 29400 J

We know,

Gravitational potential energy = mgh

Therefore,

$h = \dfrac{\text{Gravitational P.E.}}{\text{mass} \times \text{gravity}}$

Substituting the values in formula we get,

$\text{height} = \dfrac{29400}{300 \times 9.8} \\[1em] \Rightarrow \text{height} = \dfrac{29400}{2940} \\[1em] \Rightarrow \text{height} = 10\text{ m}$

Given,

Mass = 10Kg

Fall in height = 20 - 8 = 12m

(i) Loss in potential energy = mg(h₁ - h₂)

Substituting the vales in formula we get,

Loss in potential energy = 10 x 10 x 12 = 1200 J

(ii) Total energy remains constant for free fall.

Total energy = P.E + K.E

So, at height 20 m, K.E = 0

Therefore total energy = P.E + K.E

Total energy = mgh + 0
= 10 x 10 x 20 = 2000 J

Given,

Mass = 0.5 kg

Energy = 1 J

We know that,

$\text{Gravitational P.E.} = \text{mgh} \\[0.75em] \therefore h = \dfrac{\text{Gravitational P.E.}}{\text{mg}} \\[0.5em]$

Substituting the values we get,

$h = \dfrac{1}{0.5 \times 10} \\[0.5em] \Rightarrow h = 0.2m \\[0.5em]$

Mass = 50 Kgf

Increase in gravitational potential energy = mg(h₂ – h₁)

= 50 x 10 x (9 - 3)

= 50 x 10 x 6

= 3000 J

Given,

Mass = 50 kg

Height = 15 m

Gravitational potential energy = mgh

Substituting the values we get,

$\text {Gravitational P.E.} = 05 \times 10 \times 15 \\[0.5em] \Rightarrow \text {Increase in gravitational P.E.} = 7500J \\[0.5em]$

Given,

Mass = 50 kg

h₁ = 0 m

h₂ = 15 m

(i) Work done by man = mgh₂

Substituting the values we get,

W = 50 x 9.8 x 15 = 7350 J

(ii) Increase in his potential energy = Mg (h₂ – h₁)

= 50 x 9.8 x (15 - 0)

= 7350 J

Given,

F = 100 N

h = 3m

displacement = 5m

We know,

(a) Work done = Force × displacement in the direction of the force

$W = 100 \times 5 \\[0.5em] \Rightarrow W = 500 J \\[0.5em]$

(b) The potential energy gained by the block U = mgh

$\text{P.E.} = 100 \times 3 \\[0.5em] \Rightarrow \text { P.E. gained by the block}= 300 J \\[0.5em]$

(c) The difference (200 J) energy is used in doing work against the force of friction between the block and the slope and it will appear as heat energy.

Given,

m = 5 kg

Velocity, v = 10 m / s

Kinetic energy = $\dfrac{1}{2}$ × mass × (velocity)²

Substituting the values we get,

KE = $\dfrac{1}{2}$ x 5 x 10² = 250 J

Let speed of car be v.

Kinetic Energy at speed v = $\dfrac{1}{2}$ x m x v² = $\dfrac{\text{mv}^2}{2}$

Kinetic Energy at speed $\dfrac{\text{v}}{2}$ = $\dfrac{1}{2}$ x m x $(\dfrac{\text{v}}{2})^2$ = $\dfrac{\text{mv}^2}{8}$ = $\dfrac{1}{4}\Big(\dfrac{\text{mv}^2}{2}\Big)$

Hence, when the speed of the car is halved the kinetic energy becomes one-fourth.

Given,

Velocity of first body v₁ = v

Velocity of second body, v₂ = 2v

As the mass of the two bodies are same and as kinetic energy is directly proportional to the square of the velocity (K α v²)

Hence, ratio of their kinetic energies is

$\dfrac{K_1}{K_2} = \dfrac{(v_1)^2}{(v_2)^2} \\[0.5em] \dfrac{K_1}{K_2} = \dfrac{v^2}{(2v)^2} \\[0.5em] \dfrac{K_1}{K_2} = \dfrac{v^2}{4v^2} \\[0.5em] \dfrac{K_1}{K_2} = \dfrac{1}{4} \\[0.5em]$

We know that,

$\text{K.E.} = \dfrac{1}{2} \times m \times v^2 \\[0.5em]$

Let the two bodies be A and B.

Let mass of body A be m_A and body B be m_B. Let velocity of body A be v_A and body B be v_B.

Given,

Ratio of masses:

$\dfrac {m_A}{m_B} = \dfrac {5}{1} \\[0.5em]$

Ratio of K.E.:

$\dfrac {KE_A}{KE_B} = \dfrac {125}{9} \\[1em] \dfrac {KE_A}{KE_B} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{\dfrac{1}{2} \times m_A \times v_A^2}{\dfrac{1}{2} \times m_B \times v_B^2} \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{m_A}{m_B} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \\[0.5em] \Rightarrow \dfrac {125}{9} = \dfrac{5}{1} \times \Big(\dfrac{v_A}{v_B}\Big)^2 \\[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{125}{9} \times \dfrac{1}{5} \\[0.5em] \Rightarrow \Big(\dfrac{v_A}{v_B}\Big)^2 = \dfrac{25}{9} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \sqrt{\dfrac{25}{9}} \\[0.5em] \Rightarrow \dfrac{v_A}{v_B} = \dfrac{5}{3} \\[0.5em]$

∴ Ratio of their velocities = 5:3

We know that,

$KE =\dfrac{1}{2} \times m \times v^2 \\[0.5em]$

Hence, KE α v²

So we get,

$\dfrac{KE_1}{KE_2} = \dfrac{v_1^2}{v_2^2} \\[0.5em] \dfrac{KE_1}{KE_2} = (\dfrac{15}{45})^2 \\[0.5em] \dfrac{KE_1}{KE_2} = (\dfrac{1}{3})^2 \\[0.5em] \Rightarrow \dfrac{KE_1}{KE_2} = (\dfrac{1}{9}) \\[0.5em]$

Given,

Mass = 0.5 kg

Initial velocity = 10 m / s

Final velocity of the ball = 6 m / s

Kinetic energy = $\dfrac{1}{2}$ × mass × (velocity)²

Substituting the values in the equation we get,

Initial KE = $\dfrac{1}{2}$ x 0.5 x 10²

= $\dfrac{1}{2}$ x 0.5 x 100

= 0.5 x 50

= 25 J

Final KE = $\dfrac{1}{2}$ x 0.5 x 6²

= $\dfrac{1}{2}$ x 0.5 x 36

= 0.5 x 18

= 9 J

So, change in the kinetic energy of the ball = 9 J - 25 J = -16J

Hence, there is a decrease in the kinetic energy of the ball by 16J.

Given,

mass = 500 g = 0.5 kg

Speed = v = 15 m / s

(i) We know,

Kinetic energy = 1 / 2 × mass × (velocity)²

Substituting the values in the equation we get,

$KE = \dfrac{1}{2} \times 0.5 \times 15^2 \\[0.5em] KE = \dfrac{1}{2} \times 0.5 \times 225 \\[0.5em] \Rightarrow KE = 56.25J \\[0.5em]$\

(ii) We know,

Momentum = mass × velocity

Substituting the values in the equation we get,

Momentum = 0.5 x 15
= 7.5 kgm/s

Given,

Initial Mass = 10 kg and

Final mass = 2 × 10 kg = 20kg

Initial Velocity = 20 m / s

Changed velocity v2 = 20 / 2 = 10 m / s

(i) Initial kinetic energy = 1 / 2 × mass × (velocity)²

$\text {Initial Kinetic Energy} = \dfrac{1}{2} \times 10 \times (20)^2 \\[0.5em] \Rightarrow \text{Initial kinetic energy} = 2000J \\[0.5em]$

(ii) Final kinetic energy

$\text {Final Kinetic Energy} = \dfrac{1}{2} \times 20 \times (10)^2 \\[0.5em] \Rightarrow \text{Final kinetic energy} = 1000J \\[0.5em]$

Given,

Mass = 1000 kg

$\text {Initial velocity (u) } = 36 km / h \\[0.5em] = \dfrac {36 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Initial velocity (u) } = 10ms^{-1} \\[0.5em]$

$\text {Final velocity (v) } = 72 km / h \\[0.5em] = \dfrac {72 \times 1000m}{3600s} \\[0.5em] \Rightarrow \text {Final velocity (v) } = 20ms^{-1} \\[0.5em]$

From, equation of motion we get

$s = ( v^2 - u^2) / 2a$

(i) Work done = force x displacement

$W = F.s = ma.s = \dfrac{1000 \times (20^2 -10^2)}{2} \\[0.5em] W = \dfrac{1000 \times (400 -100)}{2} \\[0.5em] W = 500 \times 300 \\[0.5em] \Rightarrow W = 150000J = 1.5 × 10^5J \\[0.5em]$

(ii) Power = work done / time taken

$Power = \dfrac{1.5 \times 10^5}{120} \\[0.5em] \Rightarrow Power = 1.25 × 10^3 W$

Given,

Mass = 60 kg

Momentum = 3000 kgm / s

(i)

$\text{Kinetic Energy} = \dfrac{\text{momentum}^2}{2 \times \text{mass}} \\[0.5em] \Rightarrow \text{KE} = \dfrac{3000^2}{2 \times 60} \\[0.5em] \Rightarrow \text{KE} = \dfrac{3000 \times 3000}{120} \\[0.5em] \Rightarrow \text{KE} = 75000\text{ J} = 7.5 × 10^4\text{ J}$\

(ii)

$\text{Momentum} = \text{Mass} \times \text{Velocity} \\[0.5em] \Rightarrow \text{Velocity} = \dfrac{\text{Momentum}}{\text{Mass}} \\[0.5em] \Rightarrow \text{Velocity} = \dfrac{3000}{60} \\[0.5em] \Rightarrow \text{Velocity} = 50\text{ ms}^{-1}$

Given,

mass = 50gm = 0.05kg

momentum = 500 gcm / s = 0.005 kgm / s

KE = momentum² / 2 x mass

$\text{Work done} = KE = \dfrac {momentum^2}{2 \times mass} \\[0.5em] \text{Work done} = KE = \dfrac{0.005^2} {2 \times 0.05} \\[0.5em] \Rightarrow KE =250 J \\[0.5em]$

Given,

Mass = 20 kg

(a) As there is no displacement of the man, so work done is Zero.

(b) We know,

Work done = Kinetic energy of man = 1 / 2 × mass × (velocity)²

$\text{Work done} = \text{Kinetic energy of man} \\[0.5em] = \dfrac{1}{2} \times \text{mass} \times \text{velocity}^2 \\[0.5em] = \dfrac{1}{2} \times 20 \times 3^2 \\[0.5em] = \dfrac{1}{2} \times 20 \times 9 \\[0.5em] = 90 J \\[0.5em]$

∴ Energy Gained = 90 J

(c) We know,

Potential energy = mgh

$\text{PE} = 20 \times 10 \times 0.5 \\[0.5em] \Rightarrow U = 100 J \\[0.5em]$

Given,

Mass = 50 g = 0.05 kg

Velocity = 500 m / s

Distance (S) = 10 cm = 0.1 m

(a) The kinetic energy possessed by the bullet = $\dfrac{1}{2}$ × mass × (velocity)²

Substituting the values in equation we get,

$KE = \dfrac {1}{2} \times 0.05 \times 500^2 \\[0.5em] KE = \dfrac {1}{2} \times 0.05 \times 250000 \\[0.5em] \Rightarrow KE = 6250J \\[0.5em]$

(b)

$\text{Retarding Force} = \dfrac{\text{Work done by bullet}}{\text{distance}} \\[0.5em] = \dfrac{6250}{0.1} \\[0.5em] = 62500N \\[0.5em]$

∴ Average retarding force offered by target = 62500N

Given,

Mass = 0.5 kg

Velocity = 2 m / s

Potential Energy possessed by the spring is equal to the kinetic energy of moving trolley.

Kinetic energy of trolley

$= \dfrac{1}{2} \times \text{mass} \times \text{velocity}^2 \\[0.5em] = \dfrac{1}{2} \times 0.5 \times 2^2 \\[0.5em] = \dfrac{1}{2} \times 0.5 \times 2^2 \\[0.5em] = 1 \text{ J} \\[0.5em]$

∴ Potential energy = Kinetic energy = 1.0 J

The energy possessed by a body at rest due to its position or size and shape is called the potential energy.

Different forms of potential energy are as follows:

(i) Gravitational potential energy — It is the energy possessed by a body due to the force of attraction of the earth.

Example — A stone at a height has gravitational potential energy.

(ii) Elastic potential energy — It is the energy possessed by a body due to the deformed state because of change in its shape and size.

Example — When a spring is compressed it has elastic potential energy.

A body possesses potential energy even when it is not in motion.

For example — a stone raised at a height has gravitational potential energy.

The work done in lifting a body to a height h is

W = force of gravity (mg) × displacement (h) = mgh

Therefore, this work is stored in the body in the form of its gravitational potential energy.

Hence, Gravitational potential energy U = mgh

Kinetic Energy of a body is the energy possessed by a body in motion.

According to the work theorem,

The work done by a force in the same direction on a moving body is equal to the increase in its kinetic energy.

Mass = m

Change in velocity from u to v, so acceleration acts (a)

Force = f

Work done by the force = force × displacement

From equation of motion we know,

$v^2 = u^2 + 2aS \\[0.5em] \therefore S = \dfrac{v^2 -u^2}{2a}$

Substituting it in the formula for work done:

$W = m \times a \times (\dfrac{v^2-u^2}{2a}) \\[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \\[0.5em] W = (\dfrac{1}{2} \times m \times v^2) - (\dfrac{1}{2} \times m \times u^2) \\[0.5em] W = \dfrac{1}{2} m (v^2 - u^2) \\[0.5em] W = KE_f - KE_i$

Kinetic energy (KE) and momentum (p) are related as

$KE = \dfrac{p^2}{2m} \\[0.5em]$

As both masses have the same momentum p. The kinetic energy k is inversely proportional to the mass of the body.

Hence, a body of light mass has more kinetic energy as smaller the mass, larger is its kinetic energy.

The three forms of kinetic energy are:

(i) Vibrational kinetic energy

Example — A wire clamped at both the ends vibrates when struck.

(ii) Rotational kinetic energy

Example — A spinning top.

(iii) Translational kinetic energy

Example — A car moving in a straight path.

The six different forms of energy are

(i) Heat energy

(ii) Solar energy

(iii) Nuclear energy

(iv) Hydro energy

(v) Light energy

(vi) Mechanical energy

While transforming energy from one from to another desired form, the entire energy does not change into the desired form. A part of energy changes either to some other undesirable form (usually heat due to friction) or a part is lost to the surroundings due to radiation and becomes useless. This conversion of energy to the undesirable form is called dissipation of energy.

Since, this part of energy is not available to us for any productive purpose so we call this part of energy as degraded energy.

While transforming energy from one from to another desired form, the entire energy does not change into the desired form. A part of energy changes either to some other undesirable form (usually heat due to friction) or a part is lost to the surroundings due to radiation and becomes useless. This conversion of energy to the undesirable form is called dissipation of energy or degradation of energy.

Since, this part of energy is not available to us for any productive purpose so we call this part of energy as degraded energy.

Example — When a vehicle is run by using the chemical energy of its fuel a major part is wasted as heat and sound. Only a part of it is changed into useful mechanical energy.

The two forms of mechanical energy are

1. Kinetic energy
2. Potential energy

Wound up watch spring possesses an elastic potential energy.

(a) A moving cricket ball posses Kinetic energy — K.

(b) A compressed spring posses Potential energy — U.

(c) A moving bus posses Kinetic energy — K.

(d) A stretched wire posses Potential energy — U.

(e) An arrow shot out of a bow posses Kinetic energy — K.

(f) A piece of stone placed on the roof posses Potential energy — U.

(a) The expression for a body of mass m moving with a velocity v is given by

${\text {Kinetic energy} = \dfrac{1}{2} \times m \times velocity^2} \\[0.5em] {\text {Kinetic energy} = \dfrac{1}{2} \times m \times v^2} \\[0.5em]$

(b) 2K / v² = Joules / (ms^-1)² = kg

So, Kg is the unit of mass

Kinetic energy and momentum are related as

$p = \sqrt{2mk}$

Since,

the kinetic energy of both bodies are same, momentum is directly proportional to the square root of mass.

As we can see that mass of body B is greater than that a body A.

Therefore, body B will have more momentum than body A.

(a) The kinetic energy of a body is the energy by virtue of its motion.

(b) The potential energy of a body is the energy by virtue of its position.

(c) The conversion of part of energy into an undesirable form is called degradation of energy.

(a) Potential energy of a watch spring converts into kinetic energy

(b) Chemical energy of diesel or petrol converts into mechanical energy

(c) Kinetic energy converts into potential energy

(d) Light energy converts into chemical energy

(e) Electrical energy converts into chemical energy

(f) Chemical energy converts into heat energy

(g) Chemical energy converts into heat and light energy

(h) Chemical energy converts into heat, light and sound energy

(a) Electrical energy changes into sound energy

(b) Heat energy changes into mechanical energy

(c) Sound energy changes into electrical energy

(d) Electrical energy changes into mechanical energy

(e) Electrical energy changes into light energy

(f) Chemical energy changes into heat energy

(g) Light energy changes into electrical energy

(h) Chemical energy changes into heat energy

(i) Chemical energy changes into electrical energy

(j) Chemical energy changes into mechanical energy

(k) Electrical energy changes into heat energy

(l) Electrical energy changes into mechanical energy

(m) Electrical energy changes into magnetic energy

Nuclear fission is the process used for producing electricity from nuclear energy.

It is not possible to completely convert one form of energy into other as a part of the energy is dissipated in the form of heat which is lost to the surroundings.

We know that,

Kinetic energy + potential energy = constant

So, when a body falls from a height h under free fall

(i) At the position A — height h

Initial velocity = 0

Kinetic energy K = 0

Potential energy U = mgh

As, Total energy = KE + PE

Total energy = 0 + mgh

Total energy = mgh       (1)

(ii) At the position B — when it has fallen a distance x.

Then, velocity at B = v₁

Then u = 0, s = x, a = g

From equation of motion:
v² = u² + 2aS
v₁² = 0 + 2gx = 2gx

$\therefore \text{Kinetic energy K}= \dfrac{1}{2} \times m \times v_1^2 \\[0.5em] = \dfrac{1}{2} \times m \times 2gx \\[0.5em] = mgx$

∴ Potential energy U = mg (h – x)

Hence, total energy = K + U = mgx + mg (h – x) = mgh

Total energy = mgh       (2)

(iii) At position C (on the ground) —

Let the velocity acquired by the body on reaching the ground be v₂.

Then u = 0, s = h, a = g

We know,
v²= u² + 2aS
v₂² = 0 + 2gh
v₂² = 2gh

$\therefore \text{Kinetic energy K}= \dfrac{1}{2} \times m \times v_2^2 \\[0.5em] = \dfrac{1}{2} \times m \times 2gh \\[0.5em] = mgh$

And potential energy U = 0 (at the ground when h = 0)

So, total energy = K + U = mgh + 0

Total energy = mgh       (3)

Thus, from equation (1), (2) and (3) we note that the total mechanical energy i.e. the sum of kinetic energy and potential energy always remain constant at each point of motion and it is equal to the initial potential energy at height h.

The kinetic energy decreases and the potential energy becomes maximum at B.

After a moment the the to and fro movement starts again.

So, from B to A, again the potential energy changes into kinetic energy and this process repeat again and again.

So, when the bob in its state of to and from movement it has potential energy at the extreme position B or C and kinetic energy at resting position A.

It has both the kinetic energy and potential energy at an intermediate position.

However, the sum of kinetic and potential energy remain same at every point of movement.

Total mechanical energy of a system remains constant if no external force is worked on it.

Reason — The conservation of mechanical energy states that the total mechanical energy (sum of kinetic energy and potential energy) of a system remains constant if no external forces, such as friction or air resistance, are acting on it.

It remains constant

Reason — As the skater glides across the ice, he possess kinetic energy due to his movement. The potential energy remains constant as there is no significant change in height. Assuming a smooth ice surface with no friction or air resistance, the external forces are absent, hence, according to the principle of conservation of mechanical energy, the total mechanical energy (KE + PE) of the skater remains constant.

It remains constant

Reason — When the spring is compressed, it stores potential energy. Upon release, this potential energy is converted into kinetic energy as the spring propels an object forward. Assuming there are no significant external forces acting on the system (such as friction), the total mechanical energy of the system remains constant.

Potential energy of the ball at the highest point is mgh.

Reason — At the highest point, the ball has maximum potential energy. The potential energy at a height h, above the ground is given by the formula mgh.

mechanical energy

Reason — Mechanical energy, which is the sum of kinetic and potential energy, remains constant during the motion of the body. Initially, when the body is released from a height, it has potential energy. As it falls, this potential energy decreases while its kinetic energy increases. However, the sum of kinetic and potential energy (mechanical energy) remains constant in the absence of external forces like air resistance.

The sum of its kinetic and potential energy remains constant throughout the motion.

Reason — The sum of kinetic and potential energy does not remain constant throughout the motion of a pendulum. It varies as the pendulum swings due to the interconversion of kinetic and potential energy. However, the sum of kinetic and potential energy is constant if we consider the entire system (kinetic energy + potential energy = mechanical energy), as long as there are no external forces like friction acting on the pendulum.

0.28 J

Reason — Given,

m = 400 g = 0.4 kg

g = 10 m/s²

h = 7 cm = 0.07 m

The potential energy of the pendulum at its extreme position (maximum height) is : mgh

and kinetic energy at this point = 0

Substituting we get,

Total energy = PE + KE = mgh + 0 = (0.4 x 10 x 0.07) + 0 = 0.28 J

assertion is true but reason is false.

Explanation

Assertion (A) is true. When a hammer is made to fall on a nail fixed upright on a wooden piece, the nail begins to penetrate the wood due to the force applied by the falling hammer.

Reason (R) is false. The statement about the conversion of kinetic energy into potential energy as the hammer falls is incorrect. Potential energy is the energy that is stored in an object due to its position above the earth's surface. Hence, when a hammer is lifted it stores potential energy in it and when the hammer starts falling, this potential energy begins to change into kinetic energy and is used in driving a nail into the wood.

Given,

Mass = 0.20 kg

Initial velocity = 20ms^-1

We know,

$KE = \dfrac{1}{2} \times m \times v^2$

Maximum potential energy at the maximum height = initial kinetic energy

Substituting the values in the equation

$KE = \dfrac{1}{2} \times 0.20 \times 20^2 \\[0.5em] KE = \dfrac{1}{2} \times 0.20 \times 400$

∴ Maximum potential energy at the maximum height = Initial kinetic energy = 40 J

Mass = 500g

Velocity = 15ms^-1

(a) We know,

Potential energy at maximum height = initial kinetic energy

mgh = $\dfrac{1}{2}$ mv²

Substituting the values in the equation we get,

$KE = \dfrac{1}{2} \times 0.500 \times 15^2 \\[0.5em] KE = \dfrac{1}{2} \times 0.500\times 225 \\[0.5em] = 56.25J$

∴ Potential energy at maximum height = Initial kinetic energy = 56.25J

(b) As we know that,

Kinetic energy on reaching the ground = potential energy at the greatest height

= 56.25J

(c) Total energy at its halfway point = $\dfrac{1}{2}$ (K + U) = 56.25J

Given,

Mass = 2 kg

Height = 5m

(a)

(i) the potential energy possessed by the ball when it is initially at rest = mgh = 2 x 10 x 5 = 100J

(ii) As we know that,

The kinetic energy of the ball just before hitting the ground = initial potential energy = mgh = 2 x 10 x 5 = 100J

(b) Mechanical energy of the ball gets converted into heat and sound energy after the ball hits the ground and comes to rest.

As the person swings on a rope from a cliff of height h = 20 m then potential energy at the top gets converted into kinetic energy at the lowest point.

From conservation of energy,

Potential energy at top = Kinetic energy at lowest point

mgh = $\dfrac{1}{2}$mv²

gh = $\dfrac{1}{2}$v²

10 x 20 = $\dfrac{1}{2}$v²

400 = v²

$v=\sqrt{400}=20$ m/s

∴ The person is moving at a speed of 20 m/s at the lowest point of the swing.

Given,

Mass = 60 kg

(a)

$\text {Loss in potential energy} = \text{mg (h1 – h2)} \\[0.5em] = 60 \times 10\times (75-15) \\[0.5em] = 60 \times10 \times 60 \\[0.5em] = 3.6 \times 10^4 J \\[0.5em]$

(b) When kinetic energy at B is 75% of (3.6 × 10⁴)

$\text{Kinetic energy at B} = \dfrac{75}{100} \times 3.6 \times 10^4 \\[0.5em] = 27000J \\[0.5em] = 2.7 \times 10^4 J \\[0.5em]$

Since,

Kinetic energy = $\dfrac{1}{2}$ mv²

Substituting the values in equation we get,

$27000 = \dfrac{1}{2} \times 60 \times v^2 \\[0.5em] \Rightarrow 27000 = 1 \times 30 \times v^2 \\[0.5em] \Rightarrow v^2 = \dfrac{27000}{30} \\[0.5em] \Rightarrow v^2 = 900 \\[0.5em] \Rightarrow v = \sqrt{900} \\[0.5em] \Rightarrow v = 30 \\[0.5em]$

∴ The speed at which the skier arrives at B = 30ms^-1

Given,

Efficiency = 40%

Work done = 40% of potential energy

$\text{Work done} = 0.4 \times mgh \\[0.5em] = 0.4 \times m \times 10 \times 50 \\[0.5em] = 0.4 \times m \times 500 \\[0.5em] = 200 \times m$

$\text{Power} = \dfrac{\text{work done}}{\text{time}} \\[0.5em] 1MW = 200 \times \text {mass of water flowing per second} \\[0.5em] \Rightarrow 1 \times 10^6 = 200 \times m \\[0.5em] \Rightarrow \text m = \dfrac{1 \times 10^6 }{200} \\[0.5em] = 5000$

Hence,

Mass of water which must flow through the turbine = 5000Kg

Given,

Energy lost = 0.6 x KE

KE = $\dfrac{1}{2}$ x mass x velocity²

Substituting the values we get,

$KE = \dfrac{1}{2} \times mass \times 5^2 \\[0.5em]$

Amount of energy lost,

$\text{Energy lost} = \dfrac{1}{2} \times mass \times 5^2 \times 0.6 \\[0.5em]$

Amount of energy available,

$\text{Energy available} = \dfrac{1}{2} \times mass \times 5^2 \times 0.4\\[0.5em]$

Applying the rule for the conservation of energy we get,

Kinetic energy available = potential energy

$\dfrac{1}{2} \times mass \times 5^2 \times 0.4 = mass \times gravity \times height\\[0.5em] \Rightarrow \dfrac{1}{2} \times 5 \times 5 \times 0.4 = 10 \times height\\[0.5em] \Rightarrow \dfrac{1}{2} \times 5 \times 2 = 10 \times height\\[0.5em] \Rightarrow 5 = 10 \times height\\[0.5em] \Rightarrow height = 0.5m \\[0.5em]$

∴ Maximum vertical height reached = 0.5m

Given,

h = 5 m, m = 200 g = 0.2 kg, g = 10 ms^-2

(a) Potential energy U_B at B is given by

U_B = m x g x h

Substituting the values we get,

$U_B = 0.2 \times 10 \times 5 \\[0.5em] \Rightarrow U_B = 10 J$

Hence, the potential energy of the pendulum at the position B = 10 J

(b) Total mechanical energy at point C = 10 J

The total mechanical energy is same at all points of the path due to conservation of mechanical energy.

(c) At A, bob has only kinetic energy which is equal to potential energy at B,

Therefore,

$\dfrac{1}{2} \times m \times (v_A)^{2} = U_B \\[0.5em] 0.5 \times 0.2 \times (v_A)^{2} = 10 \\[0.5em] (v_A)^{2} = \dfrac{10}{0.1} \\[0.5em] \Rightarrow v_A = \sqrt{100} \\[0.5em] \Rightarrow v_A = 10 \text{ m s}^{-1}$