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Physics - Energy (Work, Power, Kinetic and Potential Energy)

Grade 8ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Work is defined as the product of the force applied on an object and the displacement produced in the direction of the force. It is measured in Joules (JJ).

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Work is a scalar quantity. 1 Joule1 \text{ Joule} is the work done when a force of 1 Newton1 \text{ Newton} moves a body through a distance of 1 metre1 \text{ metre} in its own direction.

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Power is the rate of doing work or the rate of transfer of energy. Its SI unit is the Watt (WW), where 1 W=1 J/s1 \text{ W} = 1 \text{ J/s}.

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Energy is the capacity to do work. Like work, its SI unit is the Joule (JJ).

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Mechanical Energy exists in two forms: Kinetic Energy (due to motion) and Potential Energy (due to position or configuration).

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Kinetic Energy (KK) depends on the mass (mm) and the square of the velocity (vv) of the object.

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Gravitational Potential Energy (UU) depends on the mass (mm), acceleration due to gravity (gg), and the height (hh) above the reference level.

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The Law of Conservation of Energy states that energy can neither be created nor destroyed, but only transformed from one form to another. The total energy of an isolated system remains constant (K+U=constantK + U = \text{constant}).

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Commercial unit of energy is the kilowatt-hour (kWhkWh). 1 kWh=3.6×106 J1 \text{ kWh} = 3.6 \times 10^6 \text{ J}.

📐Formulae

W=F×sW = F \times s

P=WtP = \frac{W}{t}

K=12mv2K = \frac{1}{2}mv^2

U=mghU = mgh

1 HP=746 W1 \text{ HP} = 746 \text{ W}

P=F×v (where v is constant velocity)P = F \times v \text{ (where } v \text{ is constant velocity)}

💡Examples

Problem 1:

A coolie lifts a luggage of 15 kg15 \text{ kg} from the ground and puts it on his head 1.5 m1.5 \text{ m} above the ground. Calculate the work done by him. (Take g=10 m/s2g = 10 \text{ m/s}^2)

Solution:

Given: m=15 kgm = 15 \text{ kg}, s=h=1.5 ms = h = 1.5 \text{ m}, g=10 m/s2g = 10 \text{ m/s}^2. \nWork done W=F×s=mghW = F \times s = mgh \nW=15×10×1.5=225 JW = 15 \times 10 \times 1.5 = 225 \text{ J}.

Explanation:

Since the force applied is against gravity, the work done is equal to the potential energy gained by the luggage.

Problem 2:

An electric motor of power 500 W500 \text{ W} takes 20 seconds20 \text{ seconds} to lift a water bucket. How much work is done by the motor?

Solution:

Given: P=500 WP = 500 \text{ W}, t=20 st = 20 \text{ s}. \nUsing P=WtP = \frac{W}{t}, we get W=P×tW = P \times t \nW=500×20=10,000 JW = 500 \times 20 = 10,000 \text{ J} or 10 kJ10 \text{ kJ}.

Explanation:

Work done is the product of power and the time interval for which the power is exerted.

Problem 3:

Calculate the kinetic energy of a ball of mass 200 g200 \text{ g} moving with a velocity of 5 m/s5 \text{ m/s}.

Solution:

Given: m=200 g=0.2 kgm = 200 \text{ g} = 0.2 \text{ kg}, v=5 m/sv = 5 \text{ m/s}. \nK=12mv2K = \frac{1}{2}mv^2 \nK=12×0.2×(5)2K = \frac{1}{2} \times 0.2 \times (5)^2 \nK=0.1×25=2.5 JK = 0.1 \times 25 = 2.5 \text{ J}.

Explanation:

Mass must be converted to the SI unit (kg) before substituting into the kinetic energy formula.

Energy (Work, Power, Kinetic and Potential Energy) Revision - Class 8 Science ICSE