A ball with a mass of $240 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $12 (kg)/s^2$ and was compressed by $1/2 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-05-05T08:38:01

The height is $=0.64m$

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The spring constant is $k=12kgs^-2$

The compression is $x=1/2m$

The potential energy is

$PE=1/2*12*(1/2)^2=1.5J$

This potential energy will be converted to kinetic energy when the spring is released

$KE=1/2m u^2$

The initial velocity is $=u$

$u^2=2/m*KE=2/m*PE$

$u^2=2/0.24*1.5=12.5$

$u=sqrt12.5=3.54ms^-1$

Resolving in the vertical direction $uarr^+$

We apply the equation of motion

$v^2=u^2+2ah$

At the greatest height, $v=0$

and $a=-g$

So,

$0=12.5-2*9.8*h$

$h=12.5/(2*9.8)=0.64m$

A ball with a mass of 240 g240 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 12 (kg)/s^212 (kg)/s^2 and was compressed by 1/2 m1/2 m when the ball was released. How high will the ball go?