# A ball with a mass of 80 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 9 (kg)/s^2 and was compressed by 3/5 m when the ball was released. How high will the ball go?

Jul 2, 2016

$h = \frac{81}{8} m$

#### Explanation:

Given that the entire spring energy is converted to kinetic energy which later turns to potential energy.
So that means the entire spring energy is turns to potential energy
To solve, we'll use the near-earth potential equation.

Now, spring energy equation is $E = \frac{1}{2} k {x}^{2}$
We have, $k = 9 k \frac{g}{s} ^ 2$, $x = \frac{3}{5} m$. So $E = \frac{1}{2} \cdot 9 \cdot {\left(\frac{3}{5}\right)}^{2}$
So, we end up with $E = {3}^{4} / 10 J$

Now, The near-earth potential energy is $E = m g h$
$m = 80 g = 80 \cdot {10}^{- 3} k g$, $g = 10 \frac{m}{s} ^ 2$, so we need to find $h$

So, equating the two, we get ${3}^{4} / {\cancel{10}}^{1} = {\cancel{80}}^{8} \cdot {\cancel{10}}^{1} \cdot {\cancel{10}}^{- 3} \cdot h$

Re-arranging gives us the answer that I have up there. You'll need a calculator to find it in decimals.