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Motion, Forces and Energy - Energy, work and power

Grade 12IGCSEPhysics

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

🔑Concepts

Energy is defined as the capacity to do work and is measured in Joules (JJ).

Work Done (WW) occurs when a force FF results in a displacement dd in the direction of the force.

Kinetic Energy (EkE_k) is the energy of an object due to its motion, depending on mass mm and velocity vv.

Gravitational Potential Energy (EpE_p) is the energy stored by an object due to its position in a gravitational field.

The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another; thus, Etotal,initial=Etotal,finalE_{total, initial} = E_{total, final}.

Power (PP) is the rate at which work is done or the rate of energy transfer, measured in Watts (WW), where 1W=1J/s1 W = 1 J/s.

Efficiency is a ratio (often expressed as a percentage) of the useful energy output to the total energy input, accounting for energy dissipated as heat or sound.

📐Formulae

W=Fdcos(θ)W = F d \cos(\theta) balance

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

ΔEp=mgh\Delta E_p = mgh

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

P=FvP = Fv

Efficiency=Useful Energy OutputTotal Energy Input×100%\text{Efficiency} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%

Eelastic=12kx2E_{elastic} = \frac{1}{2}kx^2

💡Examples

Problem 1:

A roller coaster car of mass 500 kg500 \text{ kg} starts from rest at the top of a hill 40 m40 \text{ m} high. Calculate its speed at the bottom of the hill, assuming no friction. Use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution:

mgh=12mv2mgh = \frac{1}{2}mv^2 (500)(9.8)(40)=12(500)v2(500)(9.8)(40) = \frac{1}{2}(500)v^2 196,000=250v2196,000 = 250v^2 v2=784v^2 = 784 v=28 m/sv = 28 \text{ m/s}

Explanation:

According to the Law of Conservation of Energy, the initial gravitational potential energy at the top is converted entirely into kinetic energy at the bottom in the absence of resistive forces.

Problem 2:

An electric crane lifts a 1200 kg1200 \text{ kg} load to a height of 15 m15 \text{ m} in 20 s20 \text{ s}. If the motor has an efficiency of 75%75\%, calculate the total electrical power input required.

Solution:

Useful Work Done=mgh=1200×9.8×15=176,400 J\text{Useful Work Done} = mgh = 1200 \times 9.8 \times 15 = 176,400 \text{ J} Useful Power Output=176,40020=8,820 W\text{Useful Power Output} = \frac{176,400}{20} = 8,820 \text{ W} Total Power Input=PoutEfficiency=8,8200.75=11,760 W=11.76 kW\text{Total Power Input} = \frac{P_{out}}{\text{Efficiency}} = \frac{8,820}{0.75} = 11,760 \text{ W} = 11.76 \text{ kW}

Explanation:

First, calculate the useful work (potential energy gain) and power output. Then, adjust for efficiency to find the total power the motor must consume from the electrical source.

Energy, work and power - Revision Notes & Key Formulas | IGCSE Grade 12 Physics