A ball with a mass of $450 g$ 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 $4/3 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-05-08T12:05:34

The height is $=2.42m$

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

The compression is $x=4/3m$

The potential energy is

$PE=1/2*12*(4/3)^2=32/3J$

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

$KE=1/2m u^2$

The initial velocity is $=u$

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

$u^2=2/0.45*32/3=47.41$

$u=sqrt47.41=6.89ms^-1$

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,

$0=47.41-2*9.8*h$

$h=47.41/(2*9.8)=2.42m$

A ball with a mass of 450 g450 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 4/3 m4/3 m when the ball was released. How high will the ball go?