A ball with a mass of $240 g$ $240 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $12 \frac{k g}{s^{2}}$ $12 (kg)/s^2$ and was compressed by $\frac{3}{2} m$ $3/2 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-08-27T15:04:02

The height is $= 5.74 m$ $=5.74m$

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The spring constant is $k = 12 k g s^{- 2}$ $k=12kgs^-2$

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

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 12 \cdot {(\frac{3}{2})}^{2} = 13.5 J$ $PE=1/2*12*(3/2)^2=13.5J$

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

The mass of the ball is $= 0.240 k g$ $=0.240kg$

$P E_{b a l l} = 13.5 = 0.240 \cdot 9.8 \cdot h$ $PE_(ball)=13.5=0.240*9.8*h$

$h = 13.5 \cdot \frac{1}{0.240 \cdot 9.8}$ $h=13.5*1/(0.240*9.8)$

$= 5.74 m$ $=5.74m$

A ball with a mass of 240g240 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 12kgs212 (kg)/s^2 and was compressed by 32m3/2 m when the ball was released. How high will the ball go?