A ball with a mass of 144 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 48  kgs^-2 and was compressed by 8/3 m when the ball was released. How high will the ball go?

Jul 11, 2016

${E}_{\text{sp}} = \frac{1}{2} k {x}^{2} = \frac{1}{2} \times 48 \times {\left(\frac{8}{3}\right)}^{2} = 170.7$ $J$

${E}_{\text{gp}} = m g h$ and, rearranging to make the height the subject, $h = {E}_{\text{gp}} / \left(m g\right) = \frac{170.7}{0.144 \times 9.8} = 121$ $m$

Explanation:

The spring potential energy is given by ${E}_{s p} = \frac{1}{2} k {x}^{2} = \frac{1}{2} \times 48 \times {\left(\frac{8}{3}\right)}^{2} = 170.7$ $J$ where $k$ is the spring constant and $x$ is the distance the spring is compressed.

The gravitational potential energy is given by ${E}_{g} p = m g h$, and we know it will be equal to the spring potential energy, because in this case one form of energy is being converted into another. Spring potential energy becomes kinetic energy becomes gravitational potential energy.

Rearrranging:

$h = {E}_{\text{gp}} / \left(m g\right) = \frac{170.7}{0.144 \times 9.8} = 121$ $m$

Note that I remembered to convert the mass of $144$ $g$ to the correct SI units, $0.144$ $k g$.