Question

A compressed spring is held near a small toy car of mass 0.15 kg. On the release of the spring, the toy car moves forward with a velocity of 10 ms⁻¹. Find the potential energy of the spring.

Answer 1

By conservation of energy, the spring’s potential energy changes into the toy car’s kinetic energy.

\[\text{P.E. of spring} = \text{K.E. of car} = \frac{1}{2}mv^2\]

\[= \frac{1}{2} \times 0.15 \times 10^2\]

= 7.5 J

Therefore, the potential energy of the spring is 7.5 J.