A ball with a mass of $256 g$ $256 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $27 \frac{k g}{s^{2}}$ $27 (kg)/s^2$ and was compressed by $\frac{12}{4} m$ $12/4 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-10-02T10:06:08

The height is $= 48.43 m$ $=48.43m$

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The spring constant is $k = 27 k g s^{- 2}$ $k=27kgs^-2$

The compression is $x = \frac{12}{4} m$ $x=12/4m$

The potential energy in the spring is

$P E = \frac{1}{2} k x^{2} = \frac{1}{2} \cdot 27 \cdot {(\frac{12}{4})}^{2} = 121.5 J$ $PE=1/2kx^2=1/2*27*(12/4)^2=121.5J$

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 $= 0.256 k g$ $=0.256kg$

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

$P E_{b a l l} = 121.5 = 0.256 \cdot 9.8 \cdot h$ $PE_(ball)=121.5=0.256*9.8*h$

The height is $h = 121.5 \cdot \frac{1}{0.256 \cdot 9.8}$ $h=121.5*1/(0.256*9.8)$

$= 48.43 m$ $=48.43m$

A ball with a mass of 256g256 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 27kgs227 (kg)/s^2 and was compressed by 124m12/4 m when the ball was released. How high will the ball go?