# A ball with a mass of 640 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 7/8 m when the ball was released. How high will the ball go?

May 6, 2017

$\text{height" = 1.95"m}$

#### Explanation:

Begin by converting the mass of the ball to kilograms:

$m = 640 \text{g" rarr 0.64"kg}$

The reference, potential energy of the spring , gives us the equation:

${U}_{\text{el"= 1/2kx^2" [1]}}$

we are given: $k = 32 {\text{kg"/"s}}^{2}$ and $x = \frac{7}{8} \text{m}$

Assuming that all of the potential energy in the spring is transferred to the ball, the maximum height can be computed, using the equation for potential energy due to height:

$P . E . = m g h \text{ [2]}$

Set the right side of equation [1] equal to the right side of equation [2]:

$m g h = \frac{1}{2} k {x}^{2}$

where $m = 0.64 \text{kg}$ and $g = 9.8 {\text{m"/"s}}^{2}$

h = ((32"kg"/"s"^2)(7/8"m")^2)/(2(0.64"kg")(9.8"m"/"s"^2)"

$h = 1.95 \text{m}$