A ball with a mass of $160 g$ $160 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $15 \frac{k g}{s^{2}}$ $15 (kg)/s^2$ and was compressed by $\frac{1}{4} m$ $1/4 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-06-14T06:12:43

The height is $= 0.3 m$ $=0.3m$

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The spring constant is $k = 15 k g s^{- 2}$ $k=15kgs^-2$

The compression is $x = \frac{1}{4} m$ $x=1/4m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 15 \cdot {(\frac{1}{4})}^{2} = 0.46875 J$ $PE=1/2*15*(1/4)^2=0.46875J$

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

$P E_{b a l l} = 0.46875 = 0.16 \cdot 9.8 \cdot h$ $PE_(ball)=0.46875=0.16*9.8*h$

$h = 0.46875 \cdot \frac{1}{0.16 \cdot 9.8}$ $h=0.46875*1/(0.16*9.8)$

$= 0.3 m$ $=0.3m$

A ball with a mass of 160g160 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 15kgs215 (kg)/s^2 and was compressed by 14m1/4 m when the ball was released. How high will the ball go?