krit.club logo

Motion, Forces and Energy - Energy, work and power

Grade 11IGCSEPhysics

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

🔑Concepts

•

Work is done whenever a force FF moves an object through a distance dd in the direction of the force. It is measured in Joules (JJ).

•

Energy is the capacity to do work. According to the Law of Conservation of Energy, energy cannot be created or destroyed, only transferred between stores (e.g., kinetic, gravitational potential, chemical, thermal).

•

Kinetic Energy (EkE_k) is the energy possessed by a body due to its motion. It is proportional to the mass mm and the square of the velocity vv.

•

Gravitational Potential Energy (EpE_p) is the energy stored in an object due to its position in a gravitational field, determined by mass mm, gravitational field strength gg, and height hh.

•

Power is the rate at which work is done or the rate at which energy is transferred. Its unit is the Watt (WW), where 1W=1J/s1 W = 1 J/s.

•

Efficiency is a measure of how much of the total energy input to a system is converted into useful energy output. No system is 100%100\% efficient due to energy 'losses' (usually as heat).

📐Formulae

W=FΔdW = F \Delta 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 car of mass 1200 kg1200 \text{ kg} is traveling at a constant speed of 20 m/s20 \text{ m/s}. The driver applies the brakes, and the car comes to a stop over a distance of 40 m40 \text{ m}. Calculate the average braking force applied.

Solution:

First, calculate the initial kinetic energy: Ek=12×1200×202=240,000 JE_k = \frac{1}{2} \times 1200 \times 20^2 = 240,000 \text{ J}. Since the car stops, all this energy is dissipated as work done by the brakes: W=F×d  ⟹  240,000=F×40W = F \times d \implies 240,000 = F \times 40. Therefore, F=240,00040=6000 NF = \frac{240,000}{40} = 6000 \text{ N}.

Explanation:

This problem uses the Work-Energy Theorem, where the work done by the friction (braking force) equals the change in kinetic energy of the car.

Problem 2:

An electric motor lifts a 50 kg50 \text{ kg} mass through a vertical height of 12 m12 \text{ m} in 15 s15 \text{ s}. The motor is supplied with an electrical power of 500 W500 \text{ W}. Calculate the efficiency of the motor (use g=9.8 m/s2g = 9.8 \text{ m/s}^2).

Solution:

Useful work done (Gain in EpE_p) =mgh=50×9.8×12=5880 J= mgh = 50 \times 9.8 \times 12 = 5880 \text{ J}. Useful power output =Wt=588015=392 W= \frac{W}{t} = \frac{5880}{15} = 392 \text{ W}. Efficiency =Useful PowerTotal Power Input×100=392500×100=78.4%= \frac{\text{Useful Power}}{\text{Total Power Input}} \times 100 = \frac{392}{500} \times 100 = 78.4\%.

Explanation:

Efficiency is calculated by comparing the rate of useful work (power output) to the electrical power supplied to the motor.

Problem 3:

A ball of mass 0.5 kg0.5 \text{ kg} is dropped from a height of 20 m20 \text{ m}. Calculate its velocity just before it hits the ground, assuming air resistance is negligible (g=10 m/s2g = 10 \text{ m/s}^2).

Solution:

Loss in Ep=E_p = Gain in EkE_k. mgh=12mv2mgh = \frac{1}{2}mv^2. The mass mm cancels out: gh=12v2  ⟹  10×20=12v2gh = \frac{1}{2}v^2 \implies 10 \times 20 = \frac{1}{2}v^2. 200=0.5v2  ⟹  v2=400  ⟹  v=20 m/s200 = 0.5v^2 \implies v^2 = 400 \implies v = 20 \text{ m/s}.

Explanation:

By the Principle of Conservation of Energy, potential energy at the maximum height is fully converted into kinetic energy at the bottom if friction is ignored.

Energy, work and power - Revision Notes & Key Formulas | IGCSE Grade 11 Physics