A ball with a mass of $200 g$ $200 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $64 \frac{k g}{s^{2}}$ $64 (kg)/s^2$ and was compressed by $\frac{7}{4} m$ $7/4 m$ when the ball was released. How high will the ball go?

Question

1724 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-22T07:03:01

The height is $= 0.5 m$ $=0.5m$

enter image source here

The spring constant is $k = 64 k g s^{- 2}$ $k=64kgs^-2$

The compression is $x = \frac{7}{4} m$ $x=7/4m$

The potential energy is

$P E = \frac{1}{2} \cdot 64 \cdot {(\frac{7}{4})}^{2} = 98 J$ $PE=1/2*64*(7/4)^2=98J$

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 $= u$ $=u$

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

$u^{2} = \frac{2}{0.2} \cdot 98 = 9.8$ $u^2=2/0.2*98=9.8$

$u = \sqrt{9.8} = 3.13 m s^{- 1}$ $u=sqrt9.8=3.13ms^-1$

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,

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

$h = \frac{9.8}{2 \cdot 9.8} = 0.5 m$ $h=9.8/(2*9.8)=0.5m$

A ball with a mass of 200g200 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 64kgs264 (kg)/s^2 and was compressed by 74m7/4 m when the ball was released. How high will the ball go?