A ball with a mass of $650 g$ $650 g$ 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 $\frac{7}{8} m$ $7/8 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-05-23T13:05:18

The height is $= 1.92 m$ $=1.92m$

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

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

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 32 \cdot {(\frac{7}{8})}^{2} = 12.25 J$ $PE=1/2*32*(7/8)^2=12.25J$

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

$P E_{b a l l} = 12.25 = 0.65 \cdot 9.8 \cdot h$ $PE_(ball)=12.25=0.65*9.8*h$

$h = 12.25 \cdot \frac{1}{0.65 \cdot 9.8}$ $h=12.25*1/(0.65*9.8)$

$= 1.92 m$ $=1.92m$

A ball with a mass of 650g650 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32kgs232 (kg)/s^2 and was compressed by 78m7/8 m when the ball was released. How high will the ball go?