We will have the following:
3.
a. The gravitational potential energy will be:
[tex]E_g=(60kg)(9.8m/s^2)(10m)\Rightarrow E_g=5880J[/tex]
So, the gravitational potential energy at the beginning will be 5880 J.
b. The total kinetic energy just before hitting the ground will be 5880 J.
The velocity will be 14 m/s.
4.
a. First, we determine the gravitational potential energy and the kinetic energy at the bottom:
[tex]\begin{gathered} E_g=(60kg)(9.8m/s^2)(15m)\Rightarrow E_g=8820J \\ \\ and \\ \\ K=\frac{1}{2}(60kg)(4m/s)^2\Rightarrow K=480J \end{gathered}[/tex]
So, of Rhoda hit a dry spot a total of 8340 J of energy was dissipated due to friction.
b. The percentage of energy dissipated was approximately 94.56%.
