A ball with a mass of $70 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $5 (kg)/s^2$ and was compressed by $7/6 m$ when the ball was released. How high will the ball go?

Question

1447 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-06T08:38:37

The height is $=4.96m$

![http://www.mwit.ac.th](https://useruploads.socratic.org/BDdx3oLBRcinM95cLQLj_spring%20pot%20energy.gif)

The spring constant is $k=5kgs^-2$

The compression is $x=7/6m$

The potential energy in the spring is

$PE=1/2*5*(7/6)^2=3.40J$

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

$KE_(ball)=1/2m u^2$

Let the height of the ball be $=h$

The acceleration due to gravity is $g=9.8ms^-2$

Then ,

The potential energy of the ball is $PE_(ball)=mgh$

Mass of the ball is $m=0.070kg$

$PE_(ball)=3.40=0.070*9.8*h$

$h=3.40*1/(0.07*9.8)$

$=4.96m$

The height is $=4.96m$

A ball with a mass of 70 g70 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 5 (kg)/s^25 (kg)/s^2 and was compressed by 7/6 m7/6 m when the ball was released. How high will the ball go?