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 $18 (kg)/s^2$ and was compressed by $4/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-05-04T12:42:05

The height is $=1.31m$

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The spring constant is $k=18kgs^-2$

The compression is $x=4/5m$

The potential energy is

$PE=1/2*18*(4/5)^2=5.76J$

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

$KE=1/2m u^2$

The initial velocity is $=u$

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

$u^2=2/0.45*5.76=25.6$

$u=sqrt25.6=5.06ms^-1$

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,

$0=25.6-2*9.8*h$

$h=25.6/(2*9.8)=1.31m$

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 18 (kg)/s^218 (kg)/s^2 and was compressed by 4/5 m4/5 m when the ball was released. How high will the ball go?