A ball with a mass of $480 g$ 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 $7/8 m$ when the ball was released. How high will the ball go?

Question

1407 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T06:21:27

The height is $=2.6m$

enter image source here

The spring constant is $k=32kgs^-2$

The compression is $x=7/8m$

The potential energy is

$PE=1/2*32*(7/8)^2=49/4=12.25J$

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

$KE=1/2m u^2$

The initial velocity is $=u$

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

$u^2=2/0.48*12.25=51.04$

$u=sqrt51.04=7.14ms^-1$

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,

$0=51.04-2*9.8*h$

$h=51.04/(2*9.8)=2.6m$

A ball with a mass of 480 g480 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 7/8 m7/8 m when the ball was released. How high will the ball go?