Kinetic energy
Work, Energy & Power - OCR A-Level Physics
Key Definition
Kinetic energy
The energy stored in an object because of its motion. The amount of energy in the kinetic energyThe energy an object possesses due to its motion. Measured in joules (J). store depends on the object's mass $m$ and speed $v$.
The energy stored in an object because of its motion. The amount of energy in the kinetic energyThe energy an object possesses due to its motion. Measured in joules (J). store depends on the object's mass $m$ and speed $v$.
$$E_k = \tfrac{1}{2}mv^{2}$$
Derivation from work done
Start from $W = Fx$ for a constant force accelerating an object from rest. Substitute $F = ma$ from Newton's second lawThe resultant force on an object equals the rate of change of its momentum. For constant mass, $F = ma$., then use $v^{2} = u^{2} + 2ax$ with $u = 0$ to give $a = \frac{v^{2}}{2x}$. Therefore $W = m \cdot \frac{v^{2}}{2x} \cdot x = \tfrac{1}{2}mv^{2}$, which is the kinetic energy gained.
- Kinetic energy is a scalarA quantity with magnitude only and no direction (e.g. mass, time, temperature). quantity and is always positive (or zero), even when velocity is negative.
- $E_k \propto v^{2}$. Doubling the speed quadruples the kinetic energy. Tripling the speed gives nine times the kinetic energy.
- This is why braking distance grows much faster than speed: a car at $60 \text{ mph}$ has four times the $E_k$ of one at $30 \text{ mph}$ and needs four times the braking distance for the same braking force.
- The work-energy theorem states that the net work done on an object equals its change in kinetic energy: $W_{\text{net}} = \Delta E_k$.
Common Mistake
MEDIUM
Wrong: Thinking that doubling the speed doubles the kinetic energy.
Right: $E_k \propto v^{2}$, not $v$. Doubling the speed gives four times the kinetic energy; tripling the speed gives nine times. The factor of $\tfrac{1}{2}$ is fixed, so it is the $v^{2}$ that matters.
Right: $E_k \propto v^{2}$, not $v$. Doubling the speed gives four times the kinetic energy; tripling the speed gives nine times. The factor of $\tfrac{1}{2}$ is fixed, so it is the $v^{2}$ that matters.
Worked Example [2 marks]
A $1200 \text{ kg}$ car travels at $25 \text{ m s}^{-1}$. Calculate its kinetic energy.
Show Solution
1
$E_k = \tfrac{1}{2}mv^{2} = \tfrac{1}{2} \times 1200 \times 25^{2}$.
[1]2
$E_k = 0.5 \times 1200 \times 625 = 3.75 \times 10^{5} \text{ J}$.
[1]Answer
$3.75 \times 10^{5} \text{ J}$
Examiner Tips and Tricks
- If asked to derive $E_k = \tfrac{1}{2}mv^{2}$, start from $W = Fx$, substitute $F = ma$, and use $v^{2} = u^{2} + 2ax$ with $u = 0$. Show every algebraic step.
- Check units: kg multiplied by $(\text{m s}^{-1})^{2}$ gives $\text{kg m}^{2} \text{ s}^{-2}$, which is the joule.