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 EK\$, and the right‑hand side is the work done by gravity \$Wg\$. This is the work–energy theorem:
\$\Delta EK = 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).