Energy is the capacity to do work, and work is the mechanism by which energy is transferred. In this chapter you will develop a precise mathematical description of work, explore the relationship between work and kinetic energy through the work-energy theorem, and apply the law of conservation of mechanical energy to frictionless and frictional systems. Power measures how quickly energy is transferred.
4.1 Work Done by a Force
Definition
Work (W)
Work is done on an object when a force causes the object to be displaced in the direction of the force. W = FΔx cosθ, where F is the applied force (N), Δx is the displacement (m), and θ is the angle between the force vector and the displacement vector. Work is a scalar quantity. SI unit: joule (J = N·m). No work is done if the force is perpendicular to displacement (θ = 90°).
Work is only done when a force has a component along the direction of motion. The angle θ is measured between the applied force vector and the displacement vector. If θ = 0° the force is parallel to motion and all of it does work (W = FΔx). If θ = 90° the force is perpendicular to motion — no work is done (cos 90° = 0). This is why the normal force and gravity do no work on a horizontal surface. Work is a scalar quantity measured in joules (J). Positive work means the force transfers energy TO the object; negative work means energy is removed FROM the object.
Formula
Work done by a constant force
W = work done (J), F = magnitude of force (N), Δx = displacement (m), θ = angle between force and displacement
SI unit: J
Watch Out
Common mistake: students use the angle between the force and the horizontal instead of the angle between the force and the displacement. Always draw a diagram and identify the displacement direction first.
Positive, Negative and Zero Work
| Property | Situation | Sign of Work |
|---|---|---|
| Push a box in direction of motion (θ = 0°) | W = FΔx (positive) | Energy added to object |
| Friction opposes motion (θ = 180°) | W = −FΔx (negative) | Energy removed from object |
| Normal force on horizontal surface (θ = 90°) | W = 0 | No energy transfer |
Worked Example
A person pulls a 15 kg crate along a horizontal floor with a force of 80 N at an angle of 30° above the horizontal. The crate moves 6 m. Calculate the work done by the applied force.
Given
- F = 80 N
- θ = 30°
- Δx = 6 m
Find
W_applied
Solution
- 1W = FΔx cosθ
- 2W = 80 × 6 × cos 30°
- 3W = 480 × 0,866
- 4W = 415,7 J
Practice Question
A 200 N horizontal force moves a box 5 m along the floor. Calculate the work done by the force.
(2 marks)
Practice Question
A porter carries a 30 kg bag on his head and walks 10 m horizontally at constant velocity. Calculate the work done by the normal (support) force of his head on the bag.
(3 marks)
Exam Tip
Exam tip: in every work calculation, state the formula, substitute with units, and give the sign with the numerical answer. Examiners deduct marks for missing signs.
4.2 Kinetic Energy and the Work-Energy Theorem
Definition
Work-energy theorem
The work-energy theorem states that the net (total) work done on an object equals the change in its kinetic energy: W_net = ΔEk = Ek,f − Ek,i = ½mv_f² − ½mv_i². If the net work is positive, kinetic energy increases (object speeds up). If negative, kinetic energy decreases (object slows down). Individual forces do work; only the NET work equals ΔEk.
Formula
Kinetic energy
Ek = kinetic energy (J), m = mass (kg), v = speed (m·s⁻¹)
SI unit: J
Formula
Work-energy theorem
Wnet = net work done (J), Ek,f = final kinetic energy (J), Ek,i = initial kinetic energy (J)
SI unit: J
The work-energy theorem links the concept of force-times-displacement to the change in motion of an object. It is a scalar equation, so you add up all the work done by ALL forces (including friction and gravity) to find Wnet. A positive Wnet means the object speeds up; a negative Wnet means it slows down; Wnet = 0 means constant speed. This theorem is often easier to use than Newton's second law when forces are at angles or when distances are given rather than time.
Worked Example
A 1 200 kg car travelling at 20 m·s⁻¹ brakes and comes to rest. The braking force is 8 000 N. (a) Calculate the kinetic energy of the car before braking. (b) Use the work-energy theorem to find the braking distance.
Given
- m = 1 200 kg
- vi = 20 m·s⁻¹
- vf = 0 m·s⁻¹
- F_brake = 8 000 N
Find
(a) Ek,i (b) braking distance Δx
Solution
- 1(a) Ek,i = ½mv_i² = ½ × 1 200 × 20² = 240 000 J
- 2(b) Wnet = ΔEk = Ek,f − Ek,i = 0 − 240 000 = −240 000 J
- 3 Wnet = F_brake × Δx × cos 180° = −8 000 × Δx
- 4 −8 000 × Δx = −240 000
- 5 Δx = 30 m
Worked Example
A 2 kg block is pushed from rest along a frictionless surface by a net force of 10 N over a distance of 5 m. Calculate the final speed of the block.
Given
- m = 2 kg
- vi = 0 m·s⁻¹
- F_net = 10 N
- Δx = 5 m
Find
vf
Solution
- 1Wnet = Fnet × Δx = 10 × 5 = 50 J
- 2Wnet = ½mv_f² − ½mv_i² = ½mv_f² − 0
- 350 = ½ × 2 × v_f²
- 4v_f² = 50
- 5v_f = √50 ≈ 7,07 m·s⁻¹
Watch Out
Do NOT confuse net work with the work done by a single force. You must include the work done by ALL forces acting on the object when applying the work-energy theorem.
Practice Question
A 5 kg object is moving at 4 m·s⁻¹. A net force of 15 N acts on it in the direction of motion for 3 m. Calculate the final speed.
(5 marks)
Practice Question
Explain why a car travelling at 120 km·h⁻¹ requires four times the braking distance of an identical car travelling at 60 km·h⁻¹ (assume equal braking force).
(4 marks)
4.3 Potential Energy, Conservation of Energy and Power
Definition
Conservative force
A conservative force is one for which the work done in moving an object between two points is independent of the path taken — it depends only on the initial and final positions. Conservative forces store energy as potential energy. Examples: gravitational force, elastic (spring) force. Total mechanical energy (Ek + Ep) is conserved when only conservative forces act.
Definition
Non-conservative force
A non-conservative force is one for which the work done in moving an object between two points depends on the path taken. Non-conservative forces (friction, air resistance, tension in a string) convert mechanical energy to thermal or sound energy, so the total mechanical energy of the system decreases. The work done by friction is always negative (it removes energy from the system).
Definition
Power (P)
Power is defined as the rate at which work is done or energy is transferred: P = W/t
Formula
Gravitational potential energy
Ep = gravitational potential energy (J), m = mass (kg), g = 9,8 m·s⁻² (free-fall acceleration), h = height above reference level (m)
SI unit: J
Formula
Conservation of mechanical energy (no friction)
Ek₁ = kinetic energy at position 1, Ep₁ = gravitational PE at position 1, Ek₂ and Ep₂ at position 2
SI unit: J
Formula
Power
P = power (W), W = work done (J), t = time (s), F = force (N), v = velocity (m·s⁻¹)
SI unit: W
When only conservative forces act (gravity, elastic forces), the total mechanical energy Em = Ek + Ep stays constant. Mechanical energy is transferred between kinetic and potential forms but none is lost. When friction acts, the work done by friction (negative) reduces the mechanical energy of the system — it is converted to heat. In that case use Wnet = ΔEk, including W_friction = −f × Δx as part of Wnet.
Worked Example
A 3 kg ball is released from rest at a height of 5 m above the ground on a frictionless ramp. Calculate its speed at the bottom.
Given
- m = 3 kg
- h = 5 m
- vi = 0 m·s⁻¹
- g = 9,8 m·s⁻²
Find
vf at the bottom
Solution
- 1No friction → mechanical energy is conserved
- 2Ek₁ + Ep₁ = Ek₂ + Ep₂
- 30 + mgh = ½mv_f² + 0
- 43 × 9,8 × 5 = ½ × 3 × v_f²
- 5147 = 1,5 v_f²
- 6v_f² = 98
- 7v_f = √98 ≈ 9,90 m·s⁻¹
Worked Example
A 10 kg crate slides 8 m down a ramp inclined at 30°. The frictional force is 20 N. The crate starts from rest. Calculate the speed at the bottom of the ramp using energy methods.
Given
- m = 10 kg
- Δx = 8 m
- θ = 30°
- f = 20 N
- vi = 0 m·s⁻¹
- g = 9,8 m·s⁻²
Find
vf
Solution
- 1Height dropped: h = 8 sin 30° = 8 × 0,5 = 4 m
- 2W_gravity = mgh = 10 × 9,8 × 4 = 392 J
- 3W_friction = −f × Δx = −20 × 8 = −160 J
- 4Wnet = 392 − 160 = 232 J
- 5Wnet = ΔEk = ½mv_f² − 0
- 6232 = ½ × 10 × v_f²
- 7v_f² = 46,4
- 8v_f ≈ 6,81 m·s⁻¹
Worked Example
An electric motor lifts a 400 kg load through 12 m in 15 s at constant speed. Calculate (a) the work done by the motor and (b) the power output.
Given
- m = 400 kg
- h = 12 m
- t = 15 s
- g = 9,8 m·s⁻²
Find
(a) W (b) P
Solution
- 1(a) At constant speed, W = Ep gained = mgh = 400 × 9,8 × 12 = 47 040 J
- 2(b) P = W/t = 47 040 / 15 = 3 136 W ≈ 3,14 kW
Exam Tip
Exam tip: When a question says 'constant speed', the net force is zero. This means the applied force equals friction (or weight component). Use P = Fv directly if velocity is given.
Conservative vs Non-conservative Forces
| Property | Conservative Force | Non-conservative Force |
|---|---|---|
| Path dependence | Work independent of path | Work depends on path |
| Mechanical energy | Conserved (ΔEm = 0) | Not conserved (ΔEm < 0) |
| Examples | Gravity, elastic spring force | Friction, air resistance |
| Energy fate | Stored as PE, recoverable | Converted to heat (lost) |
Practice Question
A 2 kg stone is thrown vertically upward with an initial speed of 14 m·s⁻¹. Using conservation of energy (ignore air resistance), calculate the maximum height reached.
(4 marks)
Practice Question
A machine does 90 000 J of work in 3 minutes. Calculate its power output in watts.
(3 marks)
Real World
Real-world connection: Hybrid cars use regenerative braking — the electric motor runs in reverse during braking, converting kinetic energy back into electrical energy stored in the battery instead of wasting it as heat.