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Physics - Work, Energy, and Power

Grade 9IGCSE

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

🔑Concepts

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Work is done when a force FF moves an object through a distance dd in the direction of the force. The unit of work is the Joule (JJ), where 1 J=1 Nm1 \text{ J} = 1 \text{ Nm}.

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Energy is the capacity to do work. Like work, it is measured in Joules (JJ).

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Kinetic Energy (EkE_k) is the energy possessed by an object due to its motion. It depends on the mass mm and the square of the velocity vv.

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Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its vertical position (height hh) above a reference point in a gravitational field gg.

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The Law of Conservation of Energy states that energy cannot be created or destroyed, only transferred from one form to another. Total energy in a closed system remains constant.

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Power (PP) is the rate at which work is done or the rate at which energy is transferred. The unit is the Watt (WW), where 1 W=1 J/s1 \text{ W} = 1 \text{ J/s}.

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Efficiency is a measure of how much of the total energy input is converted into useful energy output, often expressed as a percentage.

📐Formulae

W=F×dW = F \times d

Ek=12mv2E_k = \frac{1}{2}mv^2

ΔEp=mgh\Delta E_p = mgh

P=Wt=ΔEtP = \frac{W}{t} = \frac{\Delta E}{t}

Efficiency=Useful energy outputTotal energy input×100%\text{Efficiency} = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100\%

Efficiency=Useful power outputTotal power input×100%\text{Efficiency} = \frac{\text{Useful power output}}{\text{Total power input}} \times 100\%

💡Examples

Problem 1:

A crate of mass 20 kg20 \text{ kg} is lifted vertically through a height of 5 m5 \text{ m}. Calculate the work done on the crate. (Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

W=mgh=20×9.8×5=980 JW = mgh = 20 \times 9.8 \times 5 = 980 \text{ J}

Explanation:

To lift the crate, a force equal to its weight (mgmg) must be applied. The work done is the force multiplied by the distance moved in the direction of the force.

Problem 2:

A car of mass 1000 kg1000 \text{ kg} is traveling at a speed of 20 m/s20 \text{ m/s}. Calculate its kinetic energy.

Solution:

Ek=12mv2=12×1000×(20)2=0.5×1000×400=200,000 J (or 200 kJ)E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times (20)^2 = 0.5 \times 1000 \times 400 = 200,000 \text{ J (or } 200 \text{ kJ)}

Explanation:

The kinetic energy is calculated using the formula Ek=12mv2E_k = \frac{1}{2}mv^2. Ensure the velocity is squared before multiplying by mass and 0.50.5.

Problem 3:

An electric motor uses 600 J600 \text{ J} of energy to lift a load in 12 seconds12 \text{ seconds}. Calculate the power of the motor.

Solution:

P=Et=60012=50 WP = \frac{E}{t} = \frac{600}{12} = 50 \text{ W}

Explanation:

Power is defined as the rate of energy transfer. Dividing the total energy used by the time taken gives the power in Watts.

Problem 4:

A motor has a total power input of 200 W200 \text{ W}. If it provides 150 W150 \text{ W} of useful mechanical power, what is its efficiency?

Solution:

Efficiency=(150200)×100=75%\text{Efficiency} = \left( \frac{150}{200} \right) \times 100 = 75\%

Explanation:

Efficiency is the ratio of useful output to total input. Here, 50 W50 \text{ W} is wasted (likely as heat and sound).