Question

A bullet of an air gun weighs 0.01 kg. It is propelled out from the air gun with a velocity of 40 ms⁻¹. Calculate the potential energy of the spring.

Answer 1

\[ \text{Kinetic energy of bullet} = \frac{1}{2}\, mv^{2} \]

\[= \frac{1}{2}\, (0.01\ \mathrm{kg})\, (40\ \mathrm{m\,s^{-1}})^{2} \]

= 8 J

By the law of conservation of energy, Potential energy of spring = Kinetic energy of bullet = 8 J.