A ball with a mass of $480 g$ $480 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $24 \frac{k g}{s^{2}}$ $24 (kg)/s^2$ and was compressed by $\frac{7}{8} m$ $7/8 m$ when the ball was released. How high will the ball go?

Question

1388 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T14:44:50

The height is $= 1.95 m$ $=1.95m$

enter image source here

The spring constant is $k = 24 k g s^{- 2}$ $k=24kgs^-2$

The compression is $x = \frac{7}{8} m$ $x=7/8m$

The potentialenergy in the spring is

$P E = \frac{1}{2} \cdot 24 \cdot {(\frac{7}{8})}^{2} = 9.1875 J$ $PE=1/2*24*(7/8)^2=9.1875J$

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 $= h$ $=h$

Then ,

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

$P E_{b a l l} = 9.1875 = 0.48 \cdot 9.8 \cdot h$ $PE_(ball)=9.1875=0.48*9.8*h$

$h = 9.1875 \cdot \frac{1}{0.48 \cdot 9.8}$ $h=9.1875*1/(0.48*9.8)$

$= 1.95 m$ $=1.95m$

A ball with a mass of 480g480 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 24kgs224 (kg)/s^2 and was compressed by 78m7/8 m when the ball was released. How high will the ball go?