Question

What is the change in gravitational potential energy of a 67.0 kg student as they climb from ground level to a height of 10.7 meters. Use g= 9.81 m/s²

Answer 1

First , we define the variation of the potential energy as the following expression :

$\Delta U _{grav} = U_{f} - U_{i}= mg(h_{f}-h_{i})$

Where "h" means the height. As the student climb from the ground level, the initial height is equals 0. Then:

$\Delta U _{grav} = mg(h_{f}-h_{i})= (67.0kg)(9.81\frac{m}{s^{2}})(10.7m - 0m)= \mathbf{7032.789J}$