A ball with a mass of $140 g$ $140 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $16 \frac{k g}{s^{2}}$ $16 (kg)/s^2$ and was compressed by $\frac{7}{5} m$ $7/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-06-02T10:29:31

The height is $= 11.43 m$ $=11.43m$

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The spring constant is $k = 16 k g s^{- 2}$ $k=16kgs^-2$

The compression is $x = \frac{7}{5} m$ $x=7/5m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 16 \cdot {(\frac{7}{5})}^{2} = 15.68 J$ $PE=1/2*16*(7/5)^2=15.68J$

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

$P E_{b a l l} = 15.68 = 0.14 \cdot 9.8 \cdot h$ $PE_(ball)=15.68=0.14*9.8*h$

$h = 15.68 \cdot \frac{1}{0.14 \cdot 9.8}$ $h=15.68*1/(0.14*9.8)$

$= 11.43 m$ $=11.43m$

A ball with a mass of 140g140 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16kgs216 (kg)/s^2 and was compressed by 75m7/5 m when the ball was released. How high will the ball go?