A ball with a mass of $144 g$ $144 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $16 \frac{k g}{s^{2}}$ $16 (kg)/s^2$ and was compressed by $\frac{6}{4} m$ $6/4 m$ when the ball was released. How high will the ball go?

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Answer 1 · 2017-04-23T07:33:41

The height is $= 12.76 m$ $=12.76m$

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The spring constant is $k = 16 k g s^{- 2}$ $k=16kgs^-2$

The compression is $x = \frac{6}{4} m$ $x=6/4m$

The potential energy is

$P E = \frac{1}{2} \cdot 16 \cdot {(\frac{6}{4})}^{2} = 18 J$ $PE=1/2*16*(6/4)^2=18J$

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 $= u$ $=u$

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

$u^{2} = \frac{2}{0.144} \cdot 18 = 250$ $u^2=2/0.144*18=250$

$u = \sqrt{250} = 15.81 m s^{- 1}$ $u=sqrt250=15.81ms^-1$

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,

$0 = 250 - 2 \cdot 9.8 \cdot h$ $0=250-2*9.8*h$

$h = \frac{250}{2 \cdot 9.8} = 12.76 m$ $h=250/(2*9.8)=12.76m$

A ball with a mass of 144g144 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 16kgs216 (kg)/s^2 and was compressed by 64m6/4 m when the ball was released. How high will the ball go?