A ball with a mass of $650 g$ $650 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $16 \frac{k g}{s^{2}}$ $16 (kg)/s^2$ and was compressed by $\frac{3}{5} m$ $3/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-08-19T17:23:06

The height is $= 0.45 m$ $=0.45m$

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The spring constant is $k = 16 k g s^{- 2}$ $k=16kgs^-2$

The compression is $x = \frac{3}{5} m$ $x=3/5m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 16 \cdot {(\frac{3}{5})}^{2} = 2.88 J$ $PE=1/2*16*(3/5)^2=2.88J$

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 $= 0.650 k g$ $=0.650kg$

$P E_{b a l l} = 2.88 = 0.650 \cdot 9.8 \cdot h$ $PE_(ball)=2.88=0.650*9.8*h$

$h = 2.88 \cdot \frac{1}{0.650 \cdot 9.8}$ $h=2.88*1/(0.650*9.8)$

$= 0.45 m$ $=0.45m$

A ball with a mass of 650g650 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16kgs216 (kg)/s^2 and was compressed by 35m3/5 m when the ball was released. How high will the ball go?