Definition
Kinetic energy
Kinetic energy is energy stored by an object because it is moving.
- Every moving object has energy in its kinetic store. A stationary object has v=0v=0v=0, so its kinetic energy is zero.
- Kinetic energy depends on mass and speed. A more massive object at the same speed has more kinetic energy, and a faster object of the same mass has more kinetic energy.
Calculating kinetic energy
- Use Ek=12mv2E_{\mathrm{k}}=\dfrac{1}{2}mv^{2}Ek=21mv2. Mass mmm is measured in kg\text{kg}kg, speed vvv in m s−1\text{m s}^{-1}m s−1, and kinetic energy EkE_{\mathrm{k}}Ek in J\text{J}J.
- Kinetic energy is directly proportional to mass at constant speed. Doubling mass doubles EkE_{\mathrm{k}}Ek.
- Kinetic energy is proportional to the square of speed at constant mass. Doubling speed gives four times the energy, while tripling speed gives nine times the energy.
- This square relationship explains why high-speed collisions are severe. The brakes and safety structures must transfer a much larger amount of energy when speed increases.
Example
Calculating kinetic energy
- Situation: a 1200 kg1200\ \text{kg}1200 kg car travels at 15 m s−115\ \text{m s}^{-1}15 m s−1.
- Substitution: Ek=12×1200×152E_{\mathrm{k}}=\dfrac{1}{2}\times1200\times15^{2}Ek=21×1200×152.
- Energy: Ek=135000 J=1.35×105 JE_{\mathrm{k}}=135000\ \text{J}=1.35\times10^{5}\ \text{J}Ek=135000 J=1.35×105 J.
Example
Comparing two speeds
- Situation: a cyclist and bicycle have a combined mass of 80 kg80\ \text{kg}80 kg.
- At 4 m s−14\ \text{m s}^{-1}4 m s−1: Ek=12×80×42=640 JE_{\mathrm{k}}=\dfrac{1}{2}\times80\times4^{2}=640\ \text{J}Ek=21×80×42=640 J.
- At 8 m s−18\ \text{m s}^{-1}8 m s−1: Ek=12×80×82=2560 JE_{\mathrm{k}}=\dfrac{1}{2}\times80\times8^{2}=2560\ \text{J}Ek=21×80×82=2560 J.
- Comparison: doubling speed multiplies the kinetic energy by 444.
Rearranging for speed or mass
- For speed, rearrange and take the square root. v=2Ekmv=\sqrt{\dfrac{2E_{\mathrm{k}}}{m}}v=m2Ek.
- For mass, rearrange to m=2Ekv2m=\dfrac{2E_{\mathrm{k}}}{v^{2}}m=v22Ek.
- The calculated speed is a magnitude. Kinetic energy is a scalar, so the equation does not give a direction.
Example
Finding speed from energy
- Situation: a 0.42 kg0.42\ \text{kg}0.42 kg ball has 21 J21\ \text{J}21 J in its kinetic store.
- Rearrangement: v=2Ekmv=\sqrt{\dfrac{2E_{\mathrm{k}}}{m}}v=m2Ek.
- Speed: v=2×210.42=10 m s−1v=\sqrt{\dfrac{2\times21}{0.42}}=10\ \text{m s}^{-1}v=0.422×21=10 m s−1.
Exam technique
Avoiding calculation losses
- Convert all quantities to SI units before substitution.
- Square the speed, not the mass and not the whole expression.
- Keep the factor 12\dfrac{1}{2}21 visible in the working.
- When rearranging for speed, take the square root only after evaluating 2Ekm\dfrac{2E_{\mathrm{k}}}{m}m2Ek.
Common Mistake
- Do not substitute speed in km h−1\text{km h}^{-1}km h−1 or mph\text{mph}mph.
- Do not treat kinetic energy as a vector. It has magnitude only and is always non-negative.
- Do not assume doubling speed doubles kinetic energy. It makes the kinetic energy four times larger.
Self review
- What is kinetic energy?
- What does each symbol mean in Ek=12mv2E_{\mathrm{k}}=\dfrac{1}{2}mv^{2}Ek=21mv2?
- How does kinetic energy change when mass doubles at constant speed?
- How does kinetic energy change when speed doubles at constant mass?
- How is the equation rearranged to find speed?
