A ball with a mass of $280 g$ $280 g$ 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 $\frac{7}{3} m$ $7/3 m$ when the ball was released. How high will the ball go?

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A ball with a mass of $280 g$ $280 g$ 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 $\frac{7}{3} m$ $7/3 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2016-06-20T05:55:13

$spring constant (k) = 12 \frac{kg}{s^{2}}$ $"spring constant "(k)=12"kg"/s^2$
$compressed length (x) = \frac{7}{3} m$ $"compressed length "(x)=7/3m$
Potential energy of spring= $\frac{1}{2} k x^{2}$ $1/2kx^2$

$Mass of the ball (m) = 280 g = 0.28 k g$ $"Mass of the ball "(m)=280g=0.28kg$

By principle of conservation of energy

Energy of spring=energy of the ball

$\frac{1}{2} \cdot k \cdot x^{2} = m g h$ $1/2*k*x^2=mgh$
$\Rightarrow h = \frac{1}{2} \frac{k x^{2}}{m g}$ $=>h=1/2(kx^2)/(mg)$
$\Rightarrow h = \frac{1}{2} \cdot 12 \cdot (\frac{7}{3} \cdot \frac{7}{3}) \cdot \frac{1}{0.28 \cdot 9.8}$ $=>h=1/2*12*(7/3*7/3)*1/(0.28*9.8)$
$\Rightarrow h = \frac{1}{3 \cdot 0.028} = 11.9 m$ $=>h=1/(3*0.028)=11.9m$

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A ball with a mass of $280 g$ $280 g$ 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 $\frac{7}{3} m$ $7/3 m$ when the ball was released. How high will the ball go?

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A ball with a mass of $280 g$ $280 g$ 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 $\frac{7}{3} m$ $7/3 m$ when the ball was released. How high will the ball go?