A ball with a mass of $200 g$ $200 g$ 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 $\frac{3}{5} m$ $3/5 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-11-18T09:13:14

The height is $= 1.1 m$ $=1.1m$

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

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

The potential energy is

$P E = \frac{1}{2} \cdot 12 \cdot {(\frac{3}{5})}^{2} = 2.16 J$ $PE=1/2*12*(3/5)^2=2.16J$

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

Mass of the ball is $m = 0.2 k g$ $m=0.2kg$

$P E_{b a l l} = 2.16 = 0.200 \cdot 9.8 \cdot h$ $PE_(ball)=2.16=0.200*9.8*h$

$h = 2.16 \cdot \frac{1}{0.200 \cdot 9.8}$ $h=2.16*1/(0.200*9.8)$

$= 1.1 m$ $=1.1m$

The height is $= 1.1 m$ $=1.1m$

A ball with a mass of 200 g200g200 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 12 (kg)/s^212kgs212 (kg)/s^2 and was compressed by 3/5 m35m3/5 m when the ball was released. How high will the ball go?