Question

If you drop a 2.5kg ball that initially has 31J of potential energy, how much kinetic energy will it have before it hits the ground? Round your answer to the nearest whole and include the appropriate unit

Answer 1

We will have the following:

First, we determine the distance traveled:

$31J=(2.5kg)(9.8m/s^2)h\Rightarrow h=\frac{62}{49}m$

Now, we determine the time it tool to travel that distance:

$\begin{gathered} \frac{62}{49}m=(9.8m/s^2)t^2\Rightarrow t^2=\frac{310}{2401}s^2 \\ \\ \Rightarrow t=\frac{\sqrt{310}}{49}s \end{gathered}$

Now, we determine the velocity:

$v=(9.8m/s^2)(\frac{\sqrt{310}}{49}s)\Rightarrow v=\frac{\sqrt{310}}{5}m/s$

Now, we determine the kinetic energy:

$\begin{gathered} k=\frac{1}{2}(2.5kg)(\frac{\sqrt{310}}{5}m/s)^2\Rightarrow k=15.5J \\ \\ \Rightarrow k\approx16J \end{gathered}$

So, the kinetic energy will be approximately 16 Joules.