Answer to Question #145580 in Physics for John jefferson

Question

A 30-kg rock is on top of a 50-m cliff. How much of its potential energy is lost when it falls and
reaches 20 m above the ground

Answer 1

The loss om potential energy is:

$$\Delta U = |U_2 - U_1|$$

Where $U_1$ is the initial energy and $U_2$ is the final potential energy of the rock. The initial is:

$$U_1 = mgh_1$$

where $m = 30kg$ is the mass of the rock, $g = 9.81m/s^2$ is the gravitational acceleration, and $h_1 = 50m$ is the initial height. Thus, obtain:

$$U_1 = 30\cdot 9.81\cdot 50 = 14715J$$

Similarly, the final potential energy is:

$$U_2 = mgh_2$$

where $h_2 = 20m$ is the final height of the rock. Thus, get:

$$U_2 = 30\cdot 9.81\cdot 20 = 5886J$$

Then the loss is:

$$\Delta U = |5886J-14715J| = 8829J$$

Answer. 8829 J.