# A ball with a mass of 400 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 32 (kg)/s^2 and was compressed by 3/7 m when the ball was released. How high will the ball go?

Jul 2, 2016

$h = \frac{32}{49} m$

#### Explanation:

In the given situation, one notices that the spring energy is converted into kinetic energy which inturn is converted into potential energy. So in the end spring energy is converted into potential energy.
We'll use the near-earth potential energy equation.

So the object is placed on a spring of spring constant $k = 32 k \frac{g}{s} ^ 2$ and the spring is compressed by $x = \frac{3}{7} m$
So, $E = \frac{1}{2} k {x}^{2} = \frac{1}{2} \cdot 32 \cdot {\left(\frac{3}{7}\right)}^{2} = 16 \cdot \frac{9}{49}$

Given that the object is of mass $m = 0.4 k g$ and we shall take gravity $g = 10 \frac{m}{s} ^ 2$, so substituting that to the near-earth potential equation and equating it to the spring energy, we get
${\cancel{16}}^{4} \cdot \frac{9}{49} = {\cancel{0.4}}^{1} \cdot {\cancel{10}}^{1} \cdot h$

Simplify that further and you'll get the answer I gave up there.