A ball with a mass of $144 g$ $144 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $36 \frac{k g}{s^{2}}$ $36 (kg)/s^2$ and was compressed by $\frac{6}{4} m$ $6/4 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-12-01T06:16:30

The height is $= 57.4 m$ $=57.4m$

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The spring constant is $k = 36 k g s^{- 2}$ $k=36kgs^-2$

The compression is $x = \frac{6}{4} = \frac{3}{2} m$ $x=6/4=3/2m$

The potential energy is

$P E = \frac{1}{2} \cdot 36 \cdot {(\frac{3}{2})}^{2} = 81 J$ $PE=1/2*36*(3/2)^2=81J$

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 $= h$ $=h$

The acceleration due to gravity is $g = 9.8 m s^{- 2}$ $g=9.8ms^-2$

Then ,

The potential energy of the ball is $P E_{b a l l} = m g h$ $PE_(ball)=mgh$

Mass of the ball is $m = 0.144 k g$ $m=0.144kg$

$P E_{b a l l} = 81 = 0.144 \cdot 9.8 \cdot h$ $PE_(ball)=81=0.144*9.8*h$

$h = 81 \cdot \frac{1}{0.144 \cdot 9.8}$ $h=81*1/(0.144*9.8)$

$= 57.4 m$ $=57.4m$

The height is $= 57.4 m$ $=57.4m$

A ball with a mass of 144 g144g144 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 36 (kg)/s^236kgs236 (kg)/s^2 and was compressed by 6/4 m64m6/4 m when the ball was released. How high will the ball go?