Work, Energy & Power
High SchoolEnergy — the universal currency
Work is done when a force moves its point of application. For a constant force at angle θ to the displacement:
Only the component of force along the motion does work; a force ⟂ to motion (like the normal force, or gravity on a horizontal slide) does zero work.
The work–energy theorem states the net work done on a body equals its change in kinetic energy:
Gravitational potential energy near Earth is PE = mgh. For conservative forces (gravity, springs), mechanical energy is conserved:
Power is the rate of doing work: P = W/t = Fv, measured in watts (W = J·s⁻¹).
Why energy accounting beats forces
Energy methods sidestep the details of the path. To find a roller-coaster's speed at the bottom of a loop, you don't need the shape of the track — only the height dropped. Because gravity is conservative, ½mv² = mgh regardless of the route: all that matters is the endpoints. This is enormously powerful.
Conservation of energy is one of the deepest laws in physics: energy is never created or destroyed, only transformed — chemical → kinetic → thermal, and so on. When friction "removes" mechanical energy, it hasn't vanished; it becomes heat. Friction is non-conservative because the energy it converts cannot be fully recovered as motion.
Worked example — a block sliding down a rough ramp
A 2.0 kg block slides from rest down a 5.0 m ramp inclined at 30°. Friction coefficient μk = 0.20. Find its speed at the bottom using energy.
- Height dropped:
h = 5.0 sin30° = 2.5 m. PE released:mgh = 2.0×9.81×2.5 = 49.1 J. - Normal force:
N = mg cos30° = 2.0×9.81×0.866 = 17.0 N. Friction force:f = μkN = 0.20×17.0 = 3.40 N. - Work done against friction:
Wf = f × 5.0 = 17.0 J(energy lost to heat). - Energy balance:
KE = mgh − Wf = 49.1 − 17.0 = 32.1 J. - Speed:
½mv² = 32.1 ⟹ v = √(2×32.1/2.0) = √32.1 = 5.67 m·s⁻¹.
Frictionless, it would reach v = √(2gh) = √49.05 = 7.00 m·s⁻¹. Friction cost ~19% of the speed.
You carry a heavy suitcase horizontally at constant speed for 10 m. How much work do you do on it?
Zero, in the physics sense. Your supporting force is vertical (up), the motion is horizontal, and cos90° = 0. No work is done against gravity because height doesn't change. Your muscles tire due to biological inefficiency, not mechanical work on the case.
A pendulum is released from height 0.20 m. Ignoring friction, how fast is it at the lowest point?
All PE converts to KE: ½mv² = mgh, so v = √(2gh) = √(2×9.81×0.20) = √3.92 = 1.98 m·s⁻¹. Mass cancels — the answer is independent of the bob's mass.
Car A moves at 20 m·s⁻¹, car B (same mass) at 40 m·s⁻¹. Compare their kinetic energies and stopping distances.
KE ∝ v², so B has (40/20)² = 4× the kinetic energy. With the same braking force, stopping distance ∝ KE, so B needs four times the distance to stop. Doubling speed quadruples stopping distance — a vital road-safety fact.
An engine lifts 500 kg by 20 m in 10 s. What is its useful power output?
Work = mgh = 500×9.81×20 = 98 100 J. Power = W/t = 98 100/10 = 9 810 W ≈ 9.8 kW (about 13 horsepower).
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1How much work lifts a 5.0 kg box 2.0 m? (g = 9.8)
- Force to lift = weight = m*g = 5.0*9.8 = 49 N
- W = F*d = 49 * 2.0
Example 2A 2.0 kg block slides from rest down a rough ramp, dropping 3.0 m, and 12 J is lost to friction. Find its speed at the bottom. (g = 9.8)
- PE released = m*g*h = 2.0*9.8*3.0 = 58.8 J
- KE = 58.8 - 12 = 46.8 J
- 0.5*2.0*v^2 = 46.8, v^2 = 46.8
Now you try
Work each one out first, then tap to reveal the worked answer.