Question

A camera weighing 21 N falls from a small drone hovering 28 m overhead and enters free fall. What is the gravitational potential energy change (in J) of the camera from the drone to the ground if you take the following reference points?

Answer 1

Answer:

$\Delta U_g=-588J$

Explanation:

The change of gravitational potential energy is given by:

$\Delta U_g=U_{gf}-U_{go}$

$\Delta U_g=m.g.h_f-m.g.h_i\\Fc=21N=m.g\\\Delta U_g=21N*(0)-21N*(28m)\\\Delta U_g=-588J$

The preceding result is negative because the camera was on a higher point from where it ended.