A ball with a mass of $480 g$ $480 g$ 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 $\frac{9}{5} m$ $9/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2018-02-06T17:39:46

The height is $= 4.13 m$ $=4.13m$

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

The compression is $x = \frac{9}{5} m$ $x=9/5m$

The potential energy is

$P E = \frac{1}{2} \cdot 12 \cdot {(\frac{9}{5})}^{2} = 19.44 J$ $PE=1/2*12*(9/5)^2=19.44J$

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 sacceleration 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 $m = 0.48 k g$ $m=0.48kg$

$P E_{b a l l} = 5.76 = 0.48 \cdot 9.8 \cdot h$ $PE_(ball)=5.76=0.48*9.8*h$

$h = 19.44 \cdot \frac{1}{0.48 \cdot 9.8}$ $h=19.44*1/(0.48*9.8)$

$= 4.13 m$ $=4.13m$

The height is $= 4.13 m$ $=4.13m$

A ball with a mass of 480g480 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 95m9/5 m when the ball was released. How high will the ball go?