A ball with a mass of $240 g$ $240 g$ 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 $\frac{1}{2} m$ $1/2 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-05-05T08:38:01

The height is $= 0.64 m$ $=0.64m$

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The spring constant is $k = 12 k g s^{- 2}$ $k=12kgs^-2$

The compression is $x = \frac{1}{2} m$ $x=1/2m$

The potential energy is

$P E = \frac{1}{2} \cdot 12 \cdot {(\frac{1}{2})}^{2} = 1.5 J$ $PE=1/2*12*(1/2)^2=1.5J$

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 $= u$ $=u$

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

$u^{2} = \frac{2}{0.24} \cdot 1.5 = 12.5$ $u^2=2/0.24*1.5=12.5$

$u = \sqrt{12.5} = 3.54 m s^{- 1}$ $u=sqrt12.5=3.54ms^-1$

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,

$0 = 12.5 - 2 \cdot 9.8 \cdot h$ $0=12.5-2*9.8*h$

$h = \frac{12.5}{2 \cdot 9.8} = 0.64 m$ $h=12.5/(2*9.8)=0.64m$

A ball with a mass of 240g240 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 12kgs212 (kg)/s^2 and was compressed by 12m1/2 m when the ball was released. How high will the ball go?