Conservation of Mechanical Energy

When only conservative forces act, the total mechanical energy — kinetic plus potential — stays constant throughout the motion. This is one of the most powerful problem-solving tools in physics: bypass the trajectory and connect initial and final states directly.

Concept

The total mechanical energy of a particle is

$E = K + U$

Principle of conservation of mechanical energy: If only conservative forces do work,

$K_i + U_i = K_f + U_f$

i.e. $E = \text{constant}$.

If non-conservative forces (friction, drag) also act, the mechanical energy decreases by the work done against them, but the total energy (including thermal) is still conserved:

$K_i + U_i = K_f + U_f + |W_{\text{nc}}|$

This is the deeper, more general law: conservation of energy.

Derivation

By the work-energy theorem, $W_{\text{net}} = \Delta K$.

Split the net work into conservative and non-conservative parts:

$W_{\text{cons}} + W_{\text{nc}} = \Delta K$

For a conservative force, $W_{\text{cons}} = -\Delta U$:

$-\Delta U + W_{\text{nc}} = \Delta K$

$\Delta K + \Delta U = W_{\text{nc}}$

$\Delta(K + U) = W_{\text{nc}}$

If $W_{\text{nc}} = 0$, mechanical energy is conserved.

Worked Example

Falling body. A 2 kg stone falls from rest from a height 20 m. Find its speed just before hitting the ground. Take $g = 10$, ignore air drag.

Solution:

Take ground as PE zero.

Initial: $K_i = 0$, $U_i = mgh = 2 \cdot 10 \cdot 20 = 400 \, \text{J}$. Total $E = 400 \, \text{J}$.

Final (at ground): $U_f = 0$, $K_f = 400 \, \text{J}$:

$\tfrac{1}{2}(2)v^2 = 400 \Rightarrow v^2 = 400 \Rightarrow v = 20 \, \text{m/s}$

Simple pendulum. A bob of mass $m$ is released from rest at angle $\theta_0$. Find the speed at the lowest point.

Height descent: $h = L(1 - \cos\theta_0)$.

Energy conservation: $mgL(1-\cos\theta_0) = \tfrac{1}{2}mv^2 \Rightarrow v = \sqrt{2gL(1-\cos\theta_0)}$.

Common Confusions

• "Energy is conserved" needs to specify which energy. Mechanical energy is conserved only without dissipative non-conservative forces. Total energy (including heat, sound, deformation) is always conserved.
• PE zero level doesn't matter. Shifting the zero of PE adds the same constant to $U_i$ and $U_f$; the equation $\Delta K = -\Delta U$ is unchanged.
• Use scalars, not vectors. Energy is a scalar; you don't worry about directions, only magnitudes.
• Be careful with springs and gravity together. A mass on a vertical spring has both $\tfrac{1}{2}kx^2$ and $-mgx$ in its PE; using the natural-length zero or the equilibrium zero matters.

Key Takeaways

• $E = K + U$; mechanical energy is conserved if only conservative forces do work.
• $K_i + U_i = K_f + U_f$ links any two instants directly, bypassing the trajectory.
• Non-conservative forces add a $W_{\text{nc}}$ term — typically dissipated as heat.
• For a falling body from rest: $v = \sqrt{2gh}$.
• For a pendulum from rest at $\theta_0$: $v_{\min} = \sqrt{2gL(1-\cos\theta_0)}$ at the lowest point.