A ball with a mass of $125 g$ $125 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $8 \frac{k g}{s^{2}}$ $8 (kg)/s^2$ 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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A ball with a mass of $125 g$ $125 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $8 \frac{k g}{s^{2}}$ $8 (kg)/s^2$ 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 · 2017-03-26T23:02:10

1.17m high

Explanation:

$E_{p} = \frac{1}{2} k x^{2}$ $E_p = 1/2kx^2$ :
$E_{p} = \frac{1}{2} (8) (\frac{3}{5}) = 1.44 J$ $E_p=1/2(8)(3/5)=1.44J$
$E_{p} = E_{g}$ $E_p=E_g$ , and $E_g=mgh; :. h=E_p/(mg)=(1.44)/(0.125*9.81)=1.17431192661=1.17m$

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A ball with a mass of $125 g$ $125 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $8 \frac{k g}{s^{2}}$ $8 (kg)/s^2$ 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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Explanation:

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A ball with a mass of $125 g$ $125 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $8 \frac{k g}{s^{2}}$ $8 (kg)/s^2$ and was compressed by $\frac{3}{5} m$ $3/5 m$ when the ball was released. How high will the ball go?