A ball with a mass of $144 g$ $144 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $64 \frac{k g}{s^{2}}$ $64 (kg)/s^2$ and was compressed by $\frac{8}{3} m$ $8/3 m$ when the ball was released. How high will the ball go?

Question

1471 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T07:00:06

The height is $= 161.2 m$ $=161.2m$

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

The spring constant is $k = 64 k g s^{- 2}$ $k=64kgs^-2$

The compression is $x = \frac{8}{3} m$ $x=8/3m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 64 \cdot {(\frac{8}{3})}^{2} = 227.6 J$ $PE=1/2*64*(8/3)^2=227.6J$

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

$K E_{b a l l} = \frac{1}{2} m u^{2}$ $KE_(ball)=1/2m u^2$

Let the height of the ball be $= h$ $=h$

The sacceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

Then ,

The potential energy of the ball is $P E_{b a l l} = m g h$ $PE_(ball)=mgh$

Mass of the ball is $m = 0.144 k g$ $m=0.144kg$

$P E_{b a l l} = 227.6 = 0.144 \cdot 9.8 \cdot h$ $PE_(ball)=227.6=0.144*9.8*h$

$h = 227.6 \cdot \frac{1}{0.144 \cdot 9.8}$ $h=227.6*1/(0.144*9.8)$

$= 161.2 m$ $=161.2m$

The height is $= 161.2 m$ $=161.2m$

A ball with a mass of 144g144 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 64kgs264 (kg)/s^2 and was compressed by 83m8/3 m when the ball was released. How high will the ball go?