A ball with a mass of $160 g$ $160 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $24 \frac{k g}{s^{2}}$ $24 (kg)/s^2$ and was compressed by $\frac{3}{4} m$ $3/4 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2018-06-01T05:39:58

The height reached by the ball is $= 4.30 m$ $=4.30m$

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The spring constant is $k = 24 k g s^{- 2}$ $k=24kgs^-2$

The compression of the spring is $x = \frac{3}{4} m$ $x=3/4m$

The potential energy stored in the spring is

$P E = \frac{1}{2} k x^{2} = \frac{1}{2} \cdot 24 \cdot {(\frac{3}{4})}^{2} = 6.75 J$ $PE=1/2kx^2=1/2*24*(3/4)^2=6.75J$

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 acceleration 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 $m = 0.160 k g$ $m=0.160kg$

$P E_{b a l l} = 6.75 = 0.160 \cdot 9.8 \cdot h$ $PE_(ball)=6.75=0.160*9.8*h$

$h = 6.75 \cdot \frac{1}{0.160 \cdot 9.8}$ $h=6.75*1/(0.160*9.8)$

$= 4.30 m$ $=4.30m$

The height reached by the ball is $= 4.30 m$ $=4.30m$

A ball with a mass of 160g160 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 24kgs224 (kg)/s^2 and was compressed by 34m3/4 m when the ball was released. How high will the ball go?