A ball with a mass of $256 g$ $256 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $32 \frac{k g}{s^{2}}$ $32 (kg)/s^2$ and was compressed by $\frac{24}{5} m$ $24/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-07-03T15:34:01

The height is $= 146.9 m$ $=146.9m$

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The spring constant is $k = 32 k g s^{- 2}$ $k=32kgs^-2$

The compression is $x = \frac{24}{5} m$ $x=24/5m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 32 \cdot {(\frac{24}{5})}^{2} = 368.64 J$ $PE=1/2*32*(24/5)^2=368.64J$

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

$P E_{b a l l} = 368.64 = 0.256 \cdot 9.8 \cdot h$ $PE_(ball)=368.64=0.256*9.8*h$

$h = 368.64 \cdot \frac{1}{0.256 \cdot 9.8}$ $h=368.64*1/(0.256*9.8)$

$= 146.9 m$ $=146.9m$

A ball with a mass of 256 g256g256 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32 (kg)/s^232kgs232 (kg)/s^2 and was compressed by 24/5 m245m24/5 m when the ball was released. How high will the ball go?