Question

a spring is compressed with a force of 80N and a distance of 1.5cm. if a ball (0.050kg) rests against the spring when it is released, what is the speed that it leaves the spring?

Answer 1

the potential energy stored in the spring will be transformed completely into kinetic energy of the ball when the ball leaves the spring.

Potential energy of the spring \(V = \frac{1}{2} k x^2\) with \(k = \dfrac{F_0}{x_0} = \dfrac{80\,N}{1.5 \times 10^{-2}\,m} = 5.3 \times 10^3 \dfrac{N}{m}\), so \(V = \dfrac{1}{2}5.3 \times 10^3 \dfrac{N}{m}1.5^2 \times 10^{-4} m^2 = 6 \times 10^{-1} J\)
So the kinetic energy of the ball when it leaves the spring is \(V = 6 \times 10^{-1} J = \dfrac{1}{2} m v^2\). it follows that \( v = \sqrt{\frac{2 V}{m}}\)