A ball with a mass of $360 g$ $360 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $8 \frac{k g}{s^{2}}$ $8 (kg)/s^2$ and was compressed by $\frac{4}{3} m$ $4/3 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-05-10T06:57:58

The height is $= 2.0 m$ $=2.0m$

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The spring constant is $k = 8 k g s^{- 2}$ $k=8kgs^-2$

The compression is $x = \frac{4}{3} m$ $x=4/3m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 8 \cdot {(\frac{4}{3})}^{2} = \frac{64}{9} J$ $PE=1/2*8*(4/3)^2=64/9J$

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 $P E_{b a l l} = m g h$ $PE_(ball)=mgh$

$P E_{b a l l} = \frac{64}{9} = 0.36 \cdot 9.8 \cdot h$ $PE_(ball)=64/9=0.36*9.8*h$

$h = \frac{64}{9} \cdot \frac{1}{0.36 \cdot 9.8}$ $h=64/9*1/(0.36*9.8)$

$= 2.0 m$ $=2.0m$

A ball with a mass of 360g360 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 8kgs28 (kg)/s^2 and was compressed by 43m4/3 m when the ball was released. How high will the ball go?