A ball with a mass of $300 g$ $300 g$ 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 $\frac{5}{6} m$ $5/6 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-06-26T05:43:50

The height is $= 3.78 m$ $=3.78m$

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

The compression is $x = \frac{5}{6} m$ $x=5/6m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 32 \cdot {(\frac{5}{6})}^{2} = 11.11 J$ $PE=1/2*32*(5/6)^2=11.11J$

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

$P E_{b a l l} = 11.11 = 0.3 \cdot 9.8 \cdot h$ $PE_(ball)=11.11=0.3*9.8*h$

$h = 11.11 \cdot \frac{1}{0.3 \cdot 9.8}$ $h=11.11*1/(0.3*9.8)$

$= 3.78 m$ $=3.78m$

A ball with a mass of 300 g300g300 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32 (kg)/s^232kgs232 (kg)/s^2 and was compressed by 5/6 m56m5/6 m when the ball was released. How high will the ball go?