Work, Energy & Power — Science Sage

Work Done by a Constant Force

Work is done when a force causes a displacement. It is a scalar quantity.

W = FΔx cosθ

Symbol	Meaning	Unit
W	Work done	J (joules)
F	Magnitude of the applied force	N
Δx	Magnitude of displacement	m
θ	Angle between force and displacement	°

W > 0: force has a component in the direction of motion (energy transferred into the object).
W < 0: force has a component opposing motion (energy removed — e.g. friction always does negative work).
W = 0: force is perpendicular to displacement (e.g. normal force on horizontal surface, centripetal force).

θ is the angle between the force vector and the displacement vector — not the angle with the surface.

Work-Energy Theorem

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

W_net = ΔKE = ½mv_f² − ½mv_i²

This is true regardless of path — it depends only on the net force and the displacement.

Conservative vs Non-Conservative Forces

Conservative forces: the work done is independent of the path taken; only depends on start and end positions. Examples: gravity, elastic (spring) force. These forces have associated potential energy.

Non-conservative forces: the work done depends on the path. Examples: friction, air resistance, applied forces. These forces convert mechanical energy into other forms (heat, sound).

Conservation of Mechanical Energy

When only conservative forces act, mechanical energy (ME = KE + PE) is conserved:

ME = KE + PE = ½mv² + mgh = constant
KE_i + PE_i = KE_f + PE_f

When friction or other non-conservative forces also act, they do net work that changes the mechanical energy:

W_nc = ΔME = ΔKE + ΔPE
W_nc = (KE_f + PE_f) − (KE_i + PE_i)

Friction's effect: W_friction is always negative (opposes motion). It reduces the total mechanical energy — the lost ME converts to internal energy (heat). The energy is not destroyed — it changes form.

Worked Example — Block on Incline with Friction

Problem: A 5 kg block slides 8 m down a 30° incline. μk = 0.2. Find the speed at the bottom (vi = 0).

Height lost: h = 8 sin30° = 4 m
N = mg cos30° = 5×10×0.866 = 43.3 N
fk = μkN = 0.2×43.3 = 8.66 N
Wfriction = −fk×d = −8.66×8 = −69.3 J

Using Wnc = ΔME:
−69.3 = (½×5×vf² + 0) − (0 + 5×10×4)
−69.3 = 2.5vf² − 200
2.5vf² = 130.7
vf = √52.28 = 7.23 m·s⁻¹

Power

Power is the rate at which work is done (or energy is transferred).

P = W/t = Fv

Unit: W (watt) = J/s. For a vehicle moving at constant velocity, P = Fv where F is the driving force (equals friction at constant speed).

Efficiency

η = (P_output / P_input) × 100% = (W_{useful out} / W_{total in}) × 100%

Efficiency is always between 0% and 100%. The remaining energy is wasted (usually as heat).

IEB Extension — Variable Force and F-x Graphs

When a force varies with displacement, the work done equals the area under the F-x (force vs displacement) graph. For a linear spring (Hooke's Law: F = kx), the F-x graph is a straight line and the area is a triangle:

W_spring = ½kx² (= area of triangle = ½ × base × height)

Elastic potential energy stored in a spring: Ep = ½kx². The work-energy theorem still applies: Wnet = ΔKE, where Wnet includes the work done by the spring.

IEB Extension — Springs and Hooke's Law

Hooke's Law: F = kx, where k is the spring constant (N/m) and x is the extension/compression. Elastic PE: Ep = ½kx². Energy conservation with a spring: when a mass on a spring oscillates, energy converts between KE and Ep (and gravitational PE if vertical). IEB may ask you to find the speed of a mass launched by a compressed spring using ½kx² = ½mv².

Energy Visualiser — Block on Incline

Controls

Mass (kg)5 kg

Height (m)8 m

Friction & Slope

μk0.15

Angle (°)30°

The bar chart shows KE (blue), gravitational PE (green), heat from friction (red/orange), and total ME (white outline) in real time as the block slides down the slope.

Grav. PE

—J

Heat (Wfric)

Total ME

—J

Question 1 of 8

A force of 50 N is applied at 60° to the horizontal to push a box 10 m along a flat floor. The work done by this force is (cos 60° = 0.5):

Question 2 of 8

A 2 kg object accelerates from 3 m·s⁻¹ to 7 m·s⁻¹. The net work done on it is:

Question 3 of 8

A 4 kg ball is dropped from rest at a height of 5 m. Assuming no air resistance, its speed just before hitting the ground is (g = 10 m·s⁻²):

Question 4 of 8

Friction does −80 J of work on a block as it slides down a slope. The block's change in mechanical energy is:

Question 5 of 8

A car engine produces 60 000 W of power at a constant speed of 30 m·s⁻¹. The driving force is:

Question 6 of 8

A motor inputs 5000 J of energy to lift a 40 kg load by 10 m. The efficiency of the motor is (g = 10 m·s⁻²):

Question 7 of 8

A 2 kg block starts from rest at the top of a 4 m high ramp. As it slides to the bottom, friction does −16 J of work on it. Using energy conservation, what is the block's speed at the bottom? (g = 10 m·s⁻²)

Question 8 of 8

A crane lifts a 500 kg load at constant velocity through a height of 8 m in 10 s. If the crane's motor is 80% efficient, what electrical power input does it require?

IEB Extended Questions

IEB Question 1

A spring (k = 400 N/m) is compressed by 0.2 m. The elastic potential energy stored is:

IEB Question 2

On an F-x graph, a force increases linearly from 0 N to 80 N over a displacement of 4 m. The work done by this force is:

Show all formulae, substitutions, and units. Take g = 10 m·s⁻² unless stated otherwise. Energy is a scalar — no direction required, but sign matters.

Question 1 — Work with Angle (5 marks)

A worker pulls a 30 kg crate 12 m along a horizontal floor using a rope at 25° above the horizontal. The tension in the rope is 80 N and the coefficient of kinetic friction is 0.18. (a) Calculate the work done by the applied force. (b) Calculate the work done by the friction force. (c) Calculate the net work done on the crate. (d) Use the work-energy theorem to find the final speed if the crate started from rest.

Question 2 — Work-Energy Theorem (6 marks)

A 1500 kg car moving at 20 m·s⁻¹ brakes and comes to rest over 40 m. (a) Calculate the initial kinetic energy. (b) Use the work-energy theorem to find the net work done on the car. (c) Calculate the average braking force. (d) Calculate the work done by friction. (e) Where does the kinetic energy go?

Question 3 — Conservation with Friction (7 marks)

A 3 kg block starts from rest at the top of a 6 m high ramp (length along slope = 10 m, angle = 37°, sin37°≈0.60, cos37°≈0.80). μk = 0.25. (a) Calculate the initial gravitational PE (take bottom as reference). (b) Calculate the work done by friction along the slope. (c) Use Wnc = ΔME to find the speed at the bottom. (d) Verify your answer using kinematics (Fnet = ma, then v² = u² + 2as).

Question 4 — Power of a Motor (5 marks)

A pump lifts 200 kg of water per minute from a well 15 m deep. (a) Calculate the work done against gravity per minute. (b) Calculate the minimum power required. (c) If the pump is only 70% efficient, calculate the actual electrical power input needed.

Question 5 — Efficiency Calculation (4 marks)

A car engine outputs a driving force of 3000 N at a constant speed of 25 m·s⁻¹. The engine burns fuel at a rate that releases 120 000 J of energy every second. (a) Calculate the useful power output of the engine. (b) Calculate the efficiency of the engine. (c) Calculate the power wasted as heat and sound.

Question 6 — Work Done by a Varying Force, from Data (6 marks)

A varying force acts on a 2 kg trolley (initially at rest, frictionless track). The average force measured over each 1 m interval of displacement is given below:

Displacement interval (m)	0–1	1–2	2–3	3–4
Average force (N)	40	30	20	10

(a) Calculate the work done by the force in each 1 m interval, then add these to find the total work done over the full 4 m displacement.
(b) Use the work-energy theorem to calculate the trolley's final speed.
(c) Explain, referring to the table, why a single calculation of W = F × x (using only one force value) would NOT give the correct total work done here.