Work, Energy and Power

This lesson covers AQA Spec 3.4.1.7–3.4.1.8 (Work, energy and power) in full depth, including work done at angles, energy transformations, efficiency, and the relationship P = Fv.

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1 Work Done

Key Definition: Work done by a force is the product of the force and the displacement in the direction of the force.

W = Fs cos θ

where θ is the angle between the force and the displacement.

• If θ = 0° (force parallel to displacement): W = Fs
• If θ = 90° (force perpendicular to displacement): W = 0
• If θ > 90° (force has a component opposing displacement): W is negative

SI unit of work: joule (J). 1 J = 1 N m.

Worked Example 1 — Work Done at an Angle

A person pulls a suitcase 50 m along a horizontal floor using a force of 40 N at 30° above the horizontal. Find the work done.

Solution:

W = Fs cos θ = 40 × 50 × cos 30° = 40 × 50 × 0.866 = 1732 J ≈ 1.73 kJ

Exam Tip: Only the component of force in the direction of motion does work. A force perpendicular to the motion (like the normal contact force on a horizontal surface) does zero work.

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2 Work Done Against Gravity

When an object of mass m is raised through a vertical height h:

W = mgh

This equals the gain in gravitational potential energy.

Worked Example 2

A crane lifts a 500 kg girder through a vertical height of 20 m in 15 s. Find (a) the work done against gravity and (b) the power of the crane.

Solution:

(a) W = mgh = 500 × 9.81 × 20 = 98 100 J ≈ 98.1 kJ

(b) P = W/t = 98 100/15 = 6540 W ≈ 6.54 kW

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3 Energy Forms and Transfers

Kinetic Energy

KE = ½mv²

This is the energy an object has due to its motion. Derived from the work-energy theorem: the net work done on an object equals its change in kinetic energy.

Derivation: From F = ma and v² = u² + 2as:

W = Fs = mas = m × (v² − u²)/(2s) × s = ½mv² − ½mu²

Therefore: W_net = ΔKE = ½mv² − ½mu² (the work-energy theorem)

Gravitational Potential Energy

GPE = mgh

This is the energy stored due to an object's position in a gravitational field, where h is the height above a chosen reference level.

Worked Example 3 — Energy Conservation (Roller Coaster)

A 600 kg roller coaster car starts from rest at the top of a 35 m hill. Assuming no friction, find its speed at the bottom.

Solution: By conservation of energy:

Loss in GPE = Gain in KE

mgh = ½mv²

v² = 2gh = 2 × 9.81 × 35 = 686.7

v = √686.7 = 26.2 m s⁻¹

Key Point: Mass cancels — the speed at the bottom is independent of mass (when friction is negligible).

Worked Example 4 — With Friction

The same roller coaster car loses 50 000 J to friction. Find the speed at the bottom.

Solution:

GPE = KE + Energy lost to friction

mgh = ½mv² + 50 000

600 × 9.81 × 35 = ½ × 600 × v² + 50 000

206 010 = 300v² + 50 000

300v² = 156 010

v² = 520.0

v = 22.8 m s⁻¹

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4 Power

Key Definition: Power is the rate of doing work (or the rate of energy transfer).

P = W/t = E/t

SI unit: watt (W). 1 W = 1 J s⁻¹.

The Relationship P = Fv

If a constant force F moves an object at constant velocity v:

P = W/t = Fs/t = F × (s/t) = Fv

This is extremely useful for problems involving vehicles at constant velocity.