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 $\frac{5}{4} m$ $5/4 m$ when the ball was released. How high will the ball go?

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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 $\frac{5}{4} m$ $5/4 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2016-04-22T14:06:23

Mass of the ball $m = 160 g = 0.16 k g$ $m=160g=0.16kg$
The spring constant $k = 24 \frac{k g}{s^{2}}$ $k=24(kg)/s^2$
Compression of spring, $x = \frac{5}{4} m$ $x=5/4m$
Let the height attained by the ball be h m

Here by the law of conservation of mechanical energy, the final potential energy of the ball at its maximum heightwill be equal to the potential energy of the compressed spring.

So
$m g h = \frac{1}{2} k x^{2}$ $mgh=1/2kx^2$
$: . h = \frac{k x^{2}}{2 m g} = \frac{24 \times {(\frac{5}{4})}^{2}}{2 \times 0.16 \times 9.8} \approx 12 m$ $: . h = (kx^2)/(2mg)=(24xx(5/4)^2)/(2xx0.16xx9.8)~~12m$

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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 $\frac{5}{4} m$ $5/4 m$ when the ball was released. How high will the ball go?

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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 $\frac{5}{4} m$ $5/4 m$ when the ball was released. How high will the ball go?