A ball with a mass of $125 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $8 (kg)/s^2$ and was compressed by $2/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-04-08T05:30:08

The height is $=0.52m$

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The spring constant is $k=8kgs^-2$

The compression is $x=2/5m$

The potential energy is

$PE=1/2*8*(2/5)^2=16/25=0.64J$

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

$KE=1/2m u^2$

The initial velocity is $=u$

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

$u^2=2/0.125*0.64=10.24$

$u=sqrt10.24=3.2ms^-1$

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,

$0=10.24-2*9.8*h$

$h=10.24/(2*9.8)=0.52m$

A ball with a mass of 125 g125 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 8 (kg)/s^28 (kg)/s^2 and was compressed by 2/5 m2/5 m when the ball was released. How high will the ball go?