Kinetic energy and the work-energy theorem
A moving body has kinetic energy K = ½mv². The work done by the net force on a body equals the change in its kinetic energy.
Energy and kinetic energy
Energy is the capacity to do work. A moving body can do work as it is stopped. A moving hammer drives a nail into wood. Flowing water turns a water wheel. A moving vehicle can push an object out of its way. This energy of motion is kinetic energy: K = ½mv². It is a scalar. It is never negative, because m is positive and v² cannot be negative. It depends on v², so speed matters more than mass: doubling the speed makes K four times as large.
Derivation: the work-energy theorem for a constant force
- 1A constant net force F acts on a body of mass m along its motion. It gives an acceleration a = F/m.
- 2The speed changes from v₁ to v₂ while the body covers a distance s. From the equation of motion v₂² = v₁² + 2as, we get a = (v₂² − v₁²)/(2s).
- 3So F = ma = m(v₂² − v₁²)/(2s).
- 4Work done W = Fs = m(v₂² − v₁²)/2.
- 5W = ½mv₂² − ½mv₁² = K₂ − K₁.
- 6★So, the work done by the net force on a body equals the change in its kinetic energy. This is the work-energy theorem.
★ key step
What the theorem tells you
- Positive net work (force along the motion): kinetic energy increases, so the body speeds up.
- Negative net work (brakes, friction): kinetic energy decreases, so the body slows down.
- Zero net work (force at right angles, as in uniform circular motion): the speed stays the same.
- With the same braking force F, the stopping distance d comes from Fd = ½mv². So d grows as v²: double the speed, four times the stopping distance.
- From K = p²/(2m): two bodies with the same kinetic energy do not have the same momentum. The heavier one has more momentum.
Worked example: checking the work-energy theorem
A 5 kg body is moving at 2 m s⁻¹. A 10 N force acts on it along its motion for 3 s. Find the work done and the change in kinetic energy.
- a = F/m = 10 N ÷ 5 kg = 2 m s⁻².
- Final speed v₂ = v₁ + at = 2 + (2 × 3) = 8 m s⁻¹.
- Distance s = v₁t + ½at² = (2 × 3) + (½ × 2 × 3²) = 6 + 9 = 15 m.
- Work done W = Fs = 10 N × 15 m = 150 J.
- K₁ = ½ × 5 × 2² = 10 J and K₂ = ½ × 5 × 8² = 160 J, so K₂ − K₁ = 150 J.
Work done = 150 J = change in kinetic energy, as the theorem says
Worked example: braking force and power of the brakes
A 1000 kg car moving at 72 km h⁻¹ is stopped by its brakes in 40 m, in 4 s. Find the average braking force and the average power of the brakes.
- Speed: 72 km h⁻¹ = 72 × 1000 m ÷ 3600 s = 20 m s⁻¹.
- Kinetic energy K = ½ × 1000 kg × (20 m s⁻¹)² = 2.0 × 10⁵ J. The final kinetic energy is zero.
- Work-energy theorem: work by the brakes = 0 − 2.0 × 10⁵ J. So F × 40 m = 2.0 × 10⁵ J, giving F = 5000 N.
- Average power = work ÷ time = 2.0 × 10⁵ J ÷ 4 s = 5.0 × 10⁴ W.
Braking force 5000 N (5 kN), opposite to the motion; average power of the brakes 50 kW
Doubling a vehicle's speed makes its kinetic energy four times as large. With the same braking force it then needs about four times the distance to stop, so speed limits near markets and schools are kept low.
3 marks: “What does the work-energy theorem say? Derive it for a body pushed by a constant force along its line of motion.” Statement: work done by the net force = change in kinetic energy. Proof: v₂² = v₁² + 2as gives a = (v₂² − v₁²)/2s; F = ma; W = Fs = ½mv₂² − ½mv₁² = K₂ − K₁.
2 marks: “What happens to the kinetic energy of a body if its speed is doubled and its mass is halved?” K = ½mv². New K = ½ × (m/2) × (2v)² = 2 × ½mv². So the kinetic energy doubles.
3 marks: “A 1200 kg car moving at 54 km h⁻¹ is brought to rest over 25 m. Find the average braking force.” v = 54 × 1000 ÷ 3600 = 15 m s⁻¹. K = ½ × 1200 × 15² = 1.35 × 10⁵ J. Work by the brakes = −K, so F × 25 = 1.35 × 10⁵ and F = 5400 N, opposite to the motion.
Net work in = change in ½mv² out. Double the speed, four times the energy.