Law of Conservation of Energy Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. State the law of conservation of energy.

Ans. According to the law of conservation of energy, energy can neither be created nor destroyed but it can be transferred from one body to another.

Ques. Define mechanical energy.

Ans. The sum of potential energy and kinetic energy possessed by a body due to its motion or position is called mechanical energy.

Ques. In which device mechanical energy is converted into electrical energy?

Ans. One common device where mechanical energy is converted into electrical energy is a generator, also known as an electric generator or dynamo.

Ques. What is the unit of energy?

1. Radian
2. Coulomb
3. Joule
4. Watt

Ans. The correct option is c. Joule

Explanation: The SI unit of energy is Joule (J). One joule is defined as the work done by a force of 1 N in displacing a body through a distance of 1 m.

Ques. Light bulbs transform _____ into light energy.

1. Nuclear energy
2. Mechanical energy
3. Sound energy
4. Electrical energy

Ans. The correct option is d. Electrical energy.

Explanation: A light bulb converts electrical energy into light energy.

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Short Answers Questions [2 Marks Questions]

Ques. Give an example of the law of conservation of energy.

Ans. A swinging pendulum is the best example of the law of conservation of energy, where the potential energy of the bob of the pendulum is continuously converted into kinetic energy and vice versa, while the total energy of the system remains conserved.

Ques. Give a common example where you can witness electrical energy transform into sound energy.

Ans. A common example where electrical energy is converted into sound energy is a loudspeaker. Musical instruments such as guitar, and piano is also an example in which electrical energy is converted into sound energy.

Ques. When a bomb explodes, which form of the energy is converted to which form?

Ans. When a bomb explodes, the chemical energy of the bomb is converted into various forms of energy such as

• Kinetic energy
• Light energy
• Sound energy
• Thermal energy

Ques. What is conservative force?

Ans. When a body moves under the action of a conservative force, the work done done by the conservative force depends on the initial and final position of the body and it is independent of the path followed by the body.

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Long Answers Questions [3 Marks Questions]

Ques. A body is thrown vertically upward from the surface of the earth with 10 J of kinetic energy. What is the potential energy at the highest point of its journey?

Ans. Let x be the potential energy of the particle at the highest point.

At bottom

• K.E. = 10 J
• P.E. = 0

At highest point

• K.E. = 0
• P.E. = x

Applying law of conservation of energy

(K.E. + P.E.)_bottom = (K.E. + P.E.)_{highest point}

⇒ (10 + 0)_bottom = (0 + x)_{highest point}

⇒ x = 10 J

Ques. The string in the figure is 50 cm long. When the ball is released from rest, it will swing along the arc shown. How fast in m/s, will it be going at the lowest point in its swing?

Ans. Let m be the mass of the ball. At the lowest point, the velocity of the ball is maximum i.e. v, and at the highest point, the velocity of the ball will be zero.

Applying the law of conservation of energy to the highest and lowest point, we get

mgh + 0 = 0 + 1/2 mv²

⇒ v = √(2gh)

⇒ v = √(2 x 9.8 x 50 x 10^-2) = 3.1 m/s

Ques. Prove the law of conservation of energy in case of a freely falling body.

Ans. Let us consider a ball of mass m is dropped from height H.

At point A:

• Kinetic energy, KE = 0
• Potential energy, PE = mgH

Total mechanical energy, E_A = 0 + mgH = mgH

At point B: Potential energy, PE = mg (H - x)

Using the equation of motion, v² = u² + 2as, we get

v² = 0 + 2(-g)(-x) = 2gx

Now, kinetic energy, KE = 1/2 mv² = 1/2 m(2gx) = mgx

Total mechanical energy, E_B = mg (H - x) + mgx = mgH

At point C: Potential energy, PE = 0

Using the equation of motion, v² = u² + 2as, we get

v² = 0 + 2(-g)(-H) = 2gH

Now, kinetic energy, KE = 1/2 mv² = 1/2 m(2gH) = mgH

Total mechanical energy, E_C = 0 + mgH = mgH

It is clear that total mechanical energy during the whole path is conserved. Hence the law of conservation of energy proved.

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Very Long Answers Questions [5 Marks Questions]

Ques. State and prove the work-energy theorem.

Ans. The work-energy theorem states that work done by all the forces acting on a particle is equal to the change in the kinetic energy of the particle.

W_total = Change in kinetic energy (ΔK.E.)

Let a particle of mass m is moving with a velocity v, then its kinetic energy is given by

K.E. = 1/2 mv²

Differentiating both sides with respect to time t, we get

\(\frac{d(K.E.)}{dt} = \frac{d}{dt}\) (\(\frac{1}{2}\) mv²)

⇒ \(\frac{d(K.E.)}{dt} = \frac{1}{2} m \frac{d}{dt}\)v²

⇒ \(\frac{d(K.E.)}{dt} = \frac{1}{2} m (2v \frac{dv}{dt})\) = mva = Fv

Where

• acceleration, a = dv/dt
• Force, F = ma

Also, we have velocity, v = dx/dt, therefore

⇒ \(\frac{d(K.E.)}{dt}\)= F\(\frac{dx}{dt}\)

⇒ d (K.E.) = Fdx

Integrating from initial position x_i to final position x_f we get

\(\int _{K.E_i}^{K.E_f} d(K.E.) = \int_{x_i}^{x_f}Fdx\)

(K.E.)_f – (K.E.)_i = \(\int_{x_i}^{x_f}Fdx\)

But \(\int_{x_i}^{x_f}Fdx\) = W (work done)

⇒ (K.E.)_f – (K.E.)_i = W

Where (K.E.)_f - (K.E.)_i = ΔK.E, change in kinetic energy

⇒ Work done, W = Change in kinetic energy (ΔK.E.)

Ques. What is the relationship between kinetic energy and linear momentum?

Ans. Consider a particle of mass m is moving with a velocity v, the linear momentum p of the particle is given by

p = mv …(i)

Also, the kinetic energy of the particle is given by

KE = 1/2 mv²

Multiplying and dividing the above equation by m, we get

KE = m²v² / 2m

Using equation (i)

KE = p²/2m

Or p = \(\sqrt{2m (KE)}\)

Ques. Derive Einstein’s mass-energy equivalence relationship.

Ans. Let a particle of rest mass m₀ is moving with velocity v, then according to Einstein the moving mass of the particle is given by

m = \(\frac{m_0}{(1 – \frac{v^2}{c^2})^\frac{1}{2}}\)

⇒ m₀ = m \((1 – \frac{v^2}{c^2})^\frac{1}{2}\)

⇒ m₀² = m \((1 – \frac{v^2}{c^2})\)

⇒ m₀²c² = m²c² – m²v²

Differentiating both sides, assuming m₀ and c are constant, we get

0 = c²(2m dm) - [v²(2m dm) + m²(2v dv)]

⇒ 2m (c² dm – v² dm – mv dm) = 0

Since 2m is not equal to zero, therefore

c² dm – v² dm – mv dm = 0 ….(i)

Now from work-energy theorem, change in kinetic is equal to work done.

ΔK = W ⇒ dK = F dx

But, F = d/dt (mv) = m dv/dt + v dm/dt

⇒ dK = (m dv/dt + v dm/dt) dx

⇒ dK = m (dx/dt) dv + v (dx/dt) dm

⇒ dk = mv dv + v² dm

On substituting above equation in equation (i), we get

c² dm - dk = 0

⇒ dk = c² dm

Let the mass increases from m₀ to m as the kinetic energy increases from 0 to E

\(\int _0^E\) dE = c² \(\int _{m_0}^m\) dm

E – 0 = c² (m – m₀)

⇒ E = mc² – m₀c²

Where m₀c² is the rest mass energy.

If m₀c² = 0, then

E = mc²

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