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 $21 \frac{k g}{s^{2}}$ $21 (kg)/s^2$ and was compressed by $\frac{4}{5} m$ $4/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-11-06T07:43:34

The height is $= 1.43 m$ $=1.43m$

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The spring constant is $k = 21 k g s^{- 2}$ $k=21kgs^-2$

The compression is $x = \frac{4}{5} m$ $x=4/5m$

The potential energy in the spring is

$P E = \frac{1}{2} k x^{2} = \frac{1}{2} \cdot 21 \cdot {(\frac{4}{5})}^{2} = 6.72 J$ $PE=1/2kx^2=1/2*21*(4/5)^2=6.72J$

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 $= 0.480 k g$ $=0.480kg$

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

$P E_{b a l l} = 6.72 = 0.480 \cdot 9.8 \cdot h$ $PE_(ball)=6.72=0.480*9.8*h$

The height is $h = 6.72 \cdot \frac{1}{0.480 \cdot 9.8}$ $h=6.72*1/(0.480*9.8)$

$= 1.43 m$ $=1.43m$

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 21kgs221 (kg)/s^2 and was compressed by 45m4/5 m when the ball was released. How high will the ball go?