The work-energy theorem counts the work of every force. Gravity's work and a spring's work can be tracked instead as changes in gravitational and elastic potential energy. If nothing else does work, the total stays fixed.
Key idea
When only gravity and springs do work, the mechanical energy is conserved: $K_i + U_i = K_f + U_f$, with $U = U_g + U_s$.
Why it works
Gravity's work on the object is $-\Delta U_g$ and the spring's is $-\Delta U_s$. If they are the only forces doing work, the work-energy theorem gives $-\Delta U_g - \Delta U_s = \Delta K$, so $\Delta K + \Delta U = 0$. Normal forces on a frictionless track and the tension in a pendulum string are perpendicular to the motion, so they do no work and don't spoil this.
Worked examples
Worked example: Down a slide
A child starts from rest at the top of a frictionless slide $5\ \text{m}$ high. Find the speed at the bottom.
- $mgh = \tfrac12 mv^2$; the mass cancels.
- $$v = \sqrt{2gh} = \sqrt{2(9.8)(5)} \approx 9.90\ \text{m/s},$$ whatever the slide's shape.
Answer: about $9.90\ \text{m/s}$
Worked example: From hill to hill
A coaster moves at $4\ \text{m/s}$ on a $30\ \text{m}$ hill. Ignoring friction, how fast is it on the next hill, $18\ \text{m}$ high?
- $\tfrac12 v_1^2 + gh_1 = \tfrac12 v_2^2 + gh_2$ (per kilogram).
- $$v_2 = \sqrt{4^2 + 2(9.8)(12)} = \sqrt{251.2} \approx 15.8\ \text{m/s}.$$
Answer: about $15.8\ \text{m/s}$
Worked example: A spring launcher
A spring with $k = 800\ \text{N/m}$ compressed $10\ \text{cm}$ launches a $0.2\ \text{kg}$ ball horizontally. Find its launch speed.
- $U_s = \tfrac12(800)(0.1)^2 = 4\ \text{J}$.
- $4 = \tfrac12(0.2)v^2$, so $v = \sqrt{40} \approx 6.32\ \text{m/s}$.
Answer: about $6.32\ \text{m/s}$
- Choose the system and check that only gravity and springs do work on it.
- Pick a zero height for $U_g$; any level works, but keep it for the whole problem.
- Write $K_i + U_i = K_f + U_f$ and solve for the unknown.
Common mistake
Thinking a heavier object reaches the bottom of a frictionless ramp with a higher speed. It has more energy, but it needs more energy for each m/s: $mgh = \tfrac12 mv^2$, and the mass cancels. Check: $v = \sqrt{2gh}$ has no $m$ in it.
Try it in the Science Lab's Trebuchet Workshop: a falling counterweight's potential energy becomes the projectile's kinetic energy.