Recall and use the equation for mechanical working W = F d = ΔE

1.7.2 Work, Energy & Power

Learning Objective

Students will be able to:

• Define mechanical work and state the equation W = F d = ΔE.
• Explain the vector nature of force and displacement and why work is a scalar.
• State the power definition P = W/t and the electrical‑work formula W = V I t = P t.
• Identify the correct SI units (J for work/energy, W for power) and the sign convention for work.
• Apply the above formulae to simple, inclined and frictional situations.

Core Checklist (Cambridge IGCSE 0625)

What is Work?

Work is done when a force causes an object to move through a distance in the direction of the force. Because the dot product  \(\vec F\!\cdot\!\vec d = Fd\cos\theta\) is a scalar, work itself is a scalar quantity that measures the transfer of energy to or from a system.

Mathematical Formulation

• General (vector) form: \(\displaystyle W = \vec F\cdot\vec d = Fd\cos\theta\)
• Force parallel to displacement (θ = 0°): \(W = Fd\)
• Force opposite to displacement (θ = 180°): \(W = -Fd\) (negative work)
• Force perpendicular to displacement (θ = 90°): \(W = 0\)
• Variable force: Work equals the area under a force–displacement graph,

  \(\displaystyle W = \int F\,\mathrm{d}x\).

Key Equations (Cambridge IGCSE)

Special Cases of Mechanical Work

• Work against gravity: \(W = mgh\)
• Work against kinetic friction: \(W = -Ff d\) (where \(Ff = μ_k N\))
• Inclined force (constant magnitude): \(W = Fd\cos\theta\)

Work‑Energy Theorem (Supplementary)

The net work done on an object equals the change in its kinetic energy:

\(\displaystyle W{\text{net}} = ΔKE = \frac12 m vf^{\,2} - \frac12 m v_i^{\,2}\)

This links mechanical work directly to the energy transferred to the object's motion.

Power and Everyday Energy

• Power: rate of doing work, \(P = W/t\) (unit = watt, W).
• Kilowatt‑hour (kWh): common unit for household electricity.

  \(1\;\text{kWh}=1000\;\text{W}\times3600\;\text{s}=3.6\;\text{MJ}=3.6\times10^{6}\;\text{J}\).

• Cost example: A 1500 W heater running 4 h uses

  \(E = 1.5\;\text{kW}\times4\;\text{h}=6\;\text{kWh}\).

  At £0.20 kWh⁻¹ the cost is £1.20.

Common Situations & Corresponding Formulae

Suggested Diagram

Worked Examples

Example 1 – Horizontal push

A 5.0 kg crate is pushed with a constant horizontal force of 20 N over 3.0 m. Find the work done.

1. Force component in direction of motion: \(F = 20\;\text{N}\).
2. Displacement: \(d = 3.0\;\text{m}\).
3. Work: \(W = Fd = 20\times3.0 = 60\;\text{J}\) (positive).

Example 2 – Lifting a box

Lift a 12 kg box vertically through 2.5 m (take \(g = 9.8\;\text{m s}^{-2}\)).

1. Weight: \(mg = 12\times9.8 = 117.6\;\text{N}\).
2. Work against gravity: \(W = mg h = 117.6\times2.5 = 294\;\text{J}\).

Example 3 – Work against friction

A 4 kg block is pulled 5 m across a rough surface (μ_k = 0.20). Find the work done by friction.

1. Normal reaction: \(N = mg = 4\times9.8 = 39.2\;\text{N}\).
2. Friction force: \(Ff = μk N = 0.20\times39.2 = 7.84\;\text{N}\).
3. Work by friction: \(Wf = -Ff d = -7.84\times5 = -39.2\;\text{J}\).

Example 4 – Electrical work

A 60 W lamp operates for 3 h. Determine the energy used and the cost if electricity costs £0.15 kWh⁻¹.

1. Energy: \(E = Pt = 60\;\text{W}\times3\;\text{h}=0.18\;\text{kWh}\).
2. Convert to joules (optional): \(0.18\;\text{kWh}=0.18\times3.6\;\text{MJ}=6.48\times10^{5}\;\text{J}\).
3. Cost: \(0.18\;\text{kWh}\times£0.15 = £0.027\) (≈ 2.7 p).

Practice Questions

1. A 10 kg block is lifted vertically 2.5 m. Calculate the work done against gravity (use \(g=9.8\;\text{m s}^{-2}\)).
2. A force of 15 N acts at 45° to the horizontal on a cart that moves 4.0 m horizontally. Find the work done by the force.
3. A person holds a 30 N weight stationary for 10 s. How much work is done on the weight?
4. An 800 kg car accelerates from rest to 20 m s⁻¹ in 5 s. Assuming constant net force, calculate the work done on the car.
5. A 1200 W heater runs for 6 h.
   

   (a) Express the energy used in kWh and in joules.
   

   (b) If electricity costs £0.22 per kWh, what is the cost?

6. A block is pulled across a rough surface with a variable force shown in the graph below. (The shaded area under the curve is 250 N·m.) Determine the work done.

Summary Table – Sign of Work

Key Take‑away

Work quantifies the transfer of energy by a force acting through a distance. Use the simple form W = Fd when force and displacement are aligned; include cos θ** for an angle and the sign to indicate whether energy is added (+) or removed (–). Power is the rate of doing work (P = W/t). The same symbols apply to electrical work (W = VIt), and everyday energy use is expressed in kilowatt‑hours.