A ball with a mass of $640 g$ $640 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $36 \frac{k g}{s^{2}}$ $36 (kg)/s^2$ and was compressed by $\frac{5}{8} m$ $5/8 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-04-18T14:25:40

The height is $= 1.12 m$ $=1.12m$

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The spring constant is $k = 36 k g s^{- 2}$ $k=36kgs^-2$

The compression is $x = \frac{5}{8} m$ $x=5/8m$

The potential energy is

$P E = \frac{1}{2} \cdot 36 \cdot {(\frac{5}{8})}^{2} = 7.03 J$ $PE=1/2*36*(5/8)^2=7.03J$

This potential energy will be converted to kinetic energy when the spring is released

$K E = \frac{1}{2} m u^{2}$ $KE=1/2m u^2$

The initial velocity is $= u$ $=u$

$u^{2} = \frac{2}{m} \cdot K E = \frac{2}{m} \cdot P E$ $u^2=2/m*KE=2/m*PE$

$u^{2} = \frac{2}{0.64} \cdot 7.03 = 21.97$ $u^2=2/0.64*7.03=21.97$

$u = \sqrt{21.97} = 4.69 m s^{- 1}$ $u=sqrt21.97=4.69ms^-1$

Resolving in the vertical direction ${↑ ⏐ ⏐}^{+}$ $uarr^+$

We apply the equation of motion

$v^{2} = u^{2} + 2 a h$ $v^2=u^2+2ah$

At the greatest height, $v = 0$ $v=0$

and $a = - g$ $a=-g$

So,

$0 = 21.97 - 2 \cdot 9.8 \cdot h$ $0=21.97-2*9.8*h$

$h = \frac{21.97}{2 \cdot 9.8} = 1.12 m$ $h=21.97/(2*9.8)=1.12m$

A ball with a mass of 640g640 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 36kgs236 (kg)/s^2 and was compressed by 58m5/8 m when the ball was released. How high will the ball go?