A ball with a mass of $140 g$ $140 g$ is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of $49 \frac{k g}{s^{2}}$ $49 (kg)/s^2$ and was compressed by $\frac{7}{6} m$ $7/6 m$ when the ball was released. How high will the ball go?

Question

1478 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-18T11:16:20

The height is $= 24.3 m$ $=24.3m$

enter image source here

The spring constant is $k = 49 k g s^{- 2}$ $k=49kgs^-2$

The compression is $x = \frac{7}{6} m$ $x=7/6m$

The potential energy in the spring is

$P E = \frac{1}{2} \cdot 49 \cdot {(\frac{7}{6})}^{2} = 33.35 J$ $PE=1/2*49*(7/6)^2=33.35J$

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

$K E_{b a l l} = \frac{1}{2} m u^{2}$ $KE_(ball)=1/2m u^2$

Let the height of the ball be $= h$ $=h$

Then ,

The potential energy of the ball is $P E_{b a l l} = m g h$ $PE_(ball)=mgh$

$P E_{b a l l} = 33.35 = 0.140 \cdot 9.8 \cdot h$ $PE_(ball)=33.35=0.140*9.8*h$

$h = 33.35 \cdot \frac{1}{0.140 \cdot 9.8}$ $h=33.35*1/(0.140*9.8)$

$= 24.3 m$ $=24.3m$

A ball with a mass of 140 g140g140 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 49 (kg)/s^249kgs249 (kg)/s^2 and was compressed by 7/6 m76m7/6 m when the ball was released. How high will the ball go?