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?

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Answer 1 · 2016-04-17T11:12:51

$57.4m$

Potential energy is stored in a spring as

$P.E.=1/2kx^2$ ,

where $k$ is the spring constant and $x$ is how far it has been compressed. Using the values given in the question, $x=12/4m=3m$ and $k=32(kg)/s^2$ ,

$P.E.=1/2*32(kg)/s^2*9m^2$
$=144(kgm^2)/s^2=144Nm$

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

$1/2kx^2=1/2mv^2$ ,

because $1/2mv^2$ is the equation for kinetic energy afterwards.

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

$144=1/2*0.256*v^2$
$1125=v^2$

Using the suvat or uvats equation

$v^2=u^2+2as$ ,

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

$1125m^2/s^2=0m/s+2*9.8m/s^2*hm$
$h=(1125/19.6)m=57.4m$

A ball with a mass of 256 g256 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32 (kg)/s^232 (kg)/s^2 and was compressed by 12/4 m12/4 m when the ball was released. How high will the ball go?