A ball with a mass of 256 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32 (kg)/s^2 and was compressed by 12/4 m when the ball was released. How high will the ball go?

Apr 17, 2016

$57.4 m$

Explanation:

Potential energy is stored in a spring as

$P . E . = \frac{1}{2} k {x}^{2}$,

where $k$ is the spring constant and $x$ is how far it has been compressed. Using the values given in the question, $x = \frac{12}{4} m = 3 m$ and $k = 32 \frac{k g}{s} ^ 2$,

$P . E . = \frac{1}{2} \cdot 32 \frac{k g}{s} ^ 2 \cdot 9 {m}^{2}$
$= 144 \frac{k g {m}^{2}}{s} ^ 2 = 144 N m$

Due to conservation of energy, we know that the energy before a reaction must be equal to the energy after it, or

$\frac{1}{2} k {x}^{2} = \frac{1}{2} m {v}^{2}$,

because $\frac{1}{2} m {v}^{2}$ is the equation for kinetic energy afterwards.

Substituting in values we know already and rearranging to make ${v}^{2}$ the subject,

$144 = \frac{1}{2} \cdot 0.256 \cdot {v}^{2}$
$1125 = {v}^{2}$

Using the suvat or uvats equation

${v}^{2} = {u}^{2} + 2 a s$,

and knowing that $u = 0 m {s}^{-} 1$, $a = 9.8 m {s}^{-} 2$ and $s$ is the height ($h$) that we want to find,

$1125 {m}^{2} / {s}^{2} = 0 \frac{m}{s} + 2 \cdot 9.8 \frac{m}{s} ^ 2 \cdot h m$
$h = \left(\frac{1125}{19.6}\right) m = 57.4 m$