Cambridge A-Level Physics 9702 – Gravitational Potential Energy and Kinetic Energy
Gravitational Potential Energy and Kinetic Energy
Learning Objective
Derive, using the equations of motion, the formula for kinetic energy $E_K = \frac{1}{2} m v^2$.
Key Concepts
Work–energy principle
Newton’s second law
Equations of motion for constant acceleration
Relationship between work done by gravity and change in kinetic energy
Derivation
Consider an object of mass $m$ moving vertically under the influence of gravity. Let upward be positive, so the weight is $-mg$.
Start from Newton’s second law: $F = ma$. For gravity alone, $F = -mg$, therefore
$$a = -g.$$
Use the kinematic equation that relates velocity, displacement and acceleration for constant $a$:
$$v^2 = u^2 + 2a s,$$
where $u$ is the initial velocity, $v$ the final velocity and $s$ the displacement.
Multiply both sides by $m/2$:
$$\frac{1}{2} m v^2 = \frac{1}{2} m u^2 + m a s.$$
Recognise that $m a s = (ma)s = F s$, which is the definition of work done by a constant force over a displacement $s$. For gravity $F = -mg$, so the work done by gravity is
$$W_{g}= (-mg)s = -mg s.$$
Re‑arrange the previous equation:
$$\frac{1}{2} m v^2 - \frac{1}{2} m u^2 = -mg s.$$
The left‑hand side is the change in kinetic energy $\Delta E_K$, and the right‑hand side is the work done by gravity $W_g$. This is the work–energy theorem:
$$\Delta E_K = W_{g}.$$
If the object starts from rest ($u=0$) and falls a distance $h$ (so $s = -h$), the equation becomes
$$\frac{1}{2} m v^2 = mg h,$$
giving the familiar result
$$v = \sqrt{2 g h}.$$
The expression $\frac{1}{2} m v^2$ is identified as the kinetic energy $E_K$ of the object.
Summary Table
Quantity
Symbol
Expression
Kinetic Energy
$E_K$
$\displaystyle E_K = \frac{1}{2} m v^2$
Gravitational Potential Energy
$E_P$
$\displaystyle E_P = m g h$
Work done by gravity
$W_g$
$\displaystyle W_g = -mg s$
Suggested diagram: A mass $m$ falling a distance $h$ under gravity, showing initial velocity $u=0$, final velocity $v$, and the work done by the weight $mg$.
Practice Questions
A 2.0 kg ball is dropped from rest from a height of 5.0 m. Calculate its kinetic energy just before it hits the ground. (Take $g = 9.8\ \text{m s}^{-2}$.)
A 1500 kg car accelerates from 0 to 20 m s$^{-1}$ on a level road. Determine the change in its kinetic energy.
Show that the work done by a constant horizontal force $F$ over a distance $d$ equals the increase in kinetic energy, using the same steps as in the derivation above.
Key Take‑aways
The kinetic energy of a body moving with speed $v$ is $E_K = \frac{1}{2} m v^2$.
The work–energy theorem links the work done by all forces to the change in kinetic energy.
Gravitational potential energy $E_P = m g h$ is converted into kinetic energy when an object falls, conserving the total mechanical energy (neglecting air resistance).