A 0.5 kg cart is attached to a spring of spring constant 200 N/m on a frictionless horizontal track. The cart is pulled 0.1 m from its equilibrium position and released. Calculate the speed of the cart as it passes through the equilibrium position.

Step-by-step solution

1. State conservation of mechanical energy: total energy is constant because there is no friction.
2. At maximum displacement ($x=0.1$ m), all energy is elastic potential energy: $E=\frac{1}{2}k{x}^2$.
3. At equilibrium ($x=0$), all energy is kinetic: $E=\frac{1}{2}m{v}^2$.
4. Set the two expressions for total energy equal: $\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$.
5. Solve for $v$: $v=\sqrt{\frac{k}{m}},x$. Substitute $k=200$N/m, $m=0.5$kg, $x=0.1$m. Compute $\frac{k}{m}=200/0.5=400$. Hence $v=\sqrt{400\times 0.01}=\sqrt{4}=2$m/s.
6. The cart passes the equilibrium with a speed of 2 m/s.

Concept tested: Conservation of mechanical energy

Published 13 October 2026 · How this solution was prepared and checked