For an object of mass \(m\) at height \(h\) above a reference level in a uniform gravitational field \(g\), the gravitational potential energy is
\[ U = m g h \]
where \(U\) is in joules (J), \(m\) in kilograms (kg), \(g\) in metres per second squared (m s⁻²), and \(h\) in metres (m).
The kinetic energy of an object moving with speed \(v\) is given by
\[ E_{\text{K}} = \frac{1}{2} m v^{2} \]
where \(E_{\text{K}}\) is in joules (J), \(m\) in kilograms (kg), and \(v\) in metres per second (m s⁻¹).
In the absence of non-conservative forces (e.g., friction), the total mechanical energy remains constant:
\[ E_{\text{K}_i} + U_i = E_{\text{K}_f} + U_f \]
where the subscripts \(i\) and \(f\) denote initial and final states.
Starting from the work–energy theorem, the work done by a constant force \(F\) over a displacement \(d\) is
\[ W = F d \]
For a mass \(m\) accelerated from rest to speed \(v\) by a constant force \(F = m a\), the displacement is \(d = \frac{v^{2}}{2a}\). Substituting gives
\[ W = m a \left(\frac{v^{2}}{2a}\right) = \frac{1}{2} m v^{2} \]
Thus, the work done equals the kinetic energy gained.
| Quantity | Symbol | Formula | Units | Example |
|---|---|---|---|---|
| Gravitational Potential Energy | U | \(m g h\) | J | U = 2 kg × 9.81 m s⁻² × 5 m = 98.1 J |
| Kinetic Energy | EK | \(\tfrac{1}{2} m v^{2}\) | J | EK = 0.5 kg × (10 m s⁻¹)² = 50 J |
| Total Mechanical Energy | Emech | \(E_{\text{K}} + U\) | J | Emech = 50 J + 98.1 J = 148.1 J |