3.4.1.7MechanicsCore

Work, Energy and Power

Doing $work$ is how energy gets transferred from one store to another. Push a box and you transfer energy to its kinetic store; lift it and you fill its gravitational store. This lesson ties together work done, kinetic and potential energy, and the rate at which energy is transferred — $power$ — along with how efficiently a machine does it.

45 min Video by Science Shorts 3.4.1 Force, energy and momentum

Work Done, Kinetic Energy & GPE — GCSE & A-Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

Calculate the work done by a force, including when it acts at an angle
Recall and use the equation for kinetic energy
Use the change in gravitational potential energy near the Earth’s surface
Relate work done to the transfer of energy between stores
Define power and use both P = W/t and P = Fv
Calculate efficiency as a ratio of useful to total energy or power

Work done by a force

$Work done$ is the energy transferred when a force moves its point of application. It equals the force multiplied by the distance moved $in the direction of the force$ . The unit is the joule (J): one joule is transferred when a force of one newton moves through one metre.

If the force acts at an angle $θ$ to the direction of motion, only the component along the motion does work, so $W = F s cos θ$ . A force at right angles to the motion (like the tension holding a conker in a circle) does no work at all.

W = F s cos θ

Work done = force × distance moved in the direction of the force.

1Only the horizontal component of the force does work:

F cos θ = 50 \times cos 3 0^{\circ}

W = F s cos θ = 50 \times 12 \times cos 3 0^{\circ}

W = 600 \times 0.866 = 520

Answer

W \approx 520

Tip — Watch the angle: it is the angle between the force and the direction of motion. When they are parallel (θ = 0), cos θ = 1 and W = Fs.

Kinetic energy

$Kinetic energy$ is the energy an object has because of its motion. Doing work to accelerate a mass from rest transfers energy into its kinetic store, giving $E_{k} = \frac{1}{2} m v^{2}$ .

Because the speed is squared, doubling the speed quadruples the kinetic energy — which is exactly why stopping distances grow so sharply with speed.

E_{k} = \frac{1}{2} m v^{2}

Kinetic energy of a mass m moving at speed v.

1Final KE

= \frac{1}{2} (1200) (20)^{2} = 240, 000

2Initial KE

= \frac{1}{2} (1200) (8.0)^{2} = 38, 400

Δ E_{k} = 240, 000 - 38, 400 = 2.0 \times 1 0^{5}

Answer

Δ E_{k} \approx 2.0 \times 1 0^{5}

Gravitational potential energy

Lifting an object against gravity transfers energy to its $gravitational potential$ store. Near the Earth’s surface, where $g$ is effectively constant, the change is $Δ E_{p} = m g Δ h$ .

Only the change in $height$ matters, not the path taken — carrying a bag up a ramp or straight up a ladder to the same height needs the same gain in gravitational potential energy.

Δ E_{p} = m g Δ h

Change in gravitational PE = weight × change in height.

Power and efficiency

$Power$ is the rate of transferring energy, or equivalently the rate of doing work, measured in watts (1 W = 1 J s⁻¹). For a force pushing an object along at a steady speed, power can also be written as force × velocity.

$Efficiency$ compares the useful energy (or power) output with the total input. It is always less than 1 (or 100%) because some energy is dissipated, usually as heat through friction and resistance.

P = \frac{W}{t} = \frac{Δ E}{t}

Power = energy transferred per second (W).

P = F v

Power delivered by a force moving at velocity v.

efficiency = \frac{useful energy (or power) out}{total energy (or power) in}

A ratio between 0 and 1; ×100 for a percentage.

1The motor lifts against gravity, so the force needed is the weight

= m g = 250 \times 9.81 = 2453

2Power

= F v = 2453 \times 0.40

P = 981

W ≈

0.98

kW.

Answer

P \approx 980

Equation recap

W = F s cos θ

Work done by a force (θ between force and motion).

E_{k} = \frac{1}{2} m v^{2}

Kinetic energy.

Δ E_{p} = m g Δ h

Change in gravitational PE near the surface.

P = \frac{W}{t} = F v

Power as rate of working, or force × velocity.

η = \frac{useful out}{total in}

Efficiency (×100 for a percentage).

Common mistakes to avoid

Using the full force when it acts at an angle to the motion.

Only the component along the motion does work: use W = Fs cos θ.

Forgetting to square the speed in kinetic energy.

E_k = ½mv², so the speed is squared — doubling v quadruples the kinetic energy.

Using the total distance travelled for gravitational PE.

ΔE_p depends only on the vertical change in height Δh, not the path length.

Quoting an efficiency greater than 100%.

Useful output can never exceed total input, so efficiency is always ≤ 1 (≤ 100%).

Key takeaways

Work done = Fs cos θ transfers energy between stores; the unit is the joule.
Kinetic energy is ½mv²; gravitational PE change near the surface is mgΔh.
Power is the rate of energy transfer: P = W/t = Fv, measured in watts.
Efficiency = useful output ÷ total input, always less than 100% because of dissipation.

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