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Energy - Work, Power, and Efficiency

Grade 9IB

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

🔑Concepts

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Work (WW) is defined as the product of the force (FF) applied to an object and the displacement (ss) moved in the direction of the force. It is measured in Joules (JJ).

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Energy is the capacity to do work. According to the Law of Conservation of Energy, energy cannot be created or destroyed, only transformed from one form to another (e.g., from EpE_p to EkE_k).

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Kinetic Energy (EkE_k) is the energy an object possesses due to its motion, depending on its mass (mm) and the square of its velocity (v2v^2).

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Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its vertical position (height hh) relative to a reference point and the acceleration due to gravity (g≈9.8 m/s2g \approx 9.8 \text{ m/s}^2).

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Power (PP) is the rate at which work is done or energy is transferred over time (tt). The SI unit for power 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. No real machine is 100%100\% efficient due to energy 'losses' (usually as thermal energy due to friction).

📐Formulae

W=F⋅sW = F \cdot s

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

Ep=mghE_p = mgh

P=WtP = \frac{W}{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 crane lifts a 500 kg500 \text{ kg} container to a height of 20 m20 \text{ m} in 10 s10 \text{ s}. Calculate the work done by the crane and the power generated. (Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2)

Solution:

W=F⋅s=(m⋅g)⋅h=500 kg×9.8 m/s2×20 m=98,000 JW = F \cdot s = (m \cdot g) \cdot h = 500 \text{ kg} \times 9.8 \text{ m/s}^2 \times 20 \text{ m} = 98,000 \text{ J}. \ P=Wt=98,000 J10 s=9,800 WP = \frac{W}{t} = \frac{98,000 \text{ J}}{10 \text{ s}} = 9,800 \text{ W}.

Explanation:

First, we find the force required to lift the mass, which is its weight (mgmg). Then we multiply by the vertical displacement to find work. Finally, divide work by time to find power in Watts.

Problem 2:

An electric motor has a total power input of 250 W250 \text{ W}. It is used to lift a weight, providing a useful power output of 200 W200 \text{ W}. Calculate the efficiency of the motor.

Solution:

Efficiency=200 W250 W×100%=0.8×100%=80%\text{Efficiency} = \frac{200 \text{ W}}{250 \text{ W}} \times 100\% = 0.8 \times 100\% = 80\%.

Explanation:

Efficiency is the ratio of useful power to total power. Here, 50 W50 \text{ W} of power is wasted (likely as heat or sound), resulting in an 80%80\% efficiency rating.

Problem 3:

A 0.5 kg0.5 \text{ kg} ball is rolling at a velocity of 4 m/s4 \text{ m/s}. Calculate its kinetic energy (EkE_k).

Solution:

Ek=12mv2=12(0.5 kg)(4 m/s)2=0.25×16=4 JE_k = \frac{1}{2}mv^2 = \frac{1}{2}(0.5 \text{ kg})(4 \text{ m/s})^2 = 0.25 \times 16 = 4 \text{ J}.

Explanation:

Using the kinetic energy formula, we square the velocity before multiplying by half the mass.