A 2.48-kg rock is released from rest at a height of 28.1 m. Ignore air resistance and determine (a) the kinetic energy at 28.1 m, (b) the gravitational potential energy at 28.1 m, (c) the total mechanical energy at 28.1 m, (d) the kinetic energy at 0 m, (e) the gravitational potential energy at 0 m, and (f) the total mechanical energy at 0 m.

Answer 1

PE = m x g x h
PE = 2.48 x 9.8 x 28.1
PE = 682.94J at 28.1m
KE at 28.1m = 0, because the object is still at rest.
The kinetic energy at 0m transfers from the previously stated gravitational potential energy.
KE = 682.94J at 0m
Therefore PE = 0J at 0m
Mechanical energy is the sum of both KE and PE.

Answer 2

At 28.1 meters, the PE = m *g * h
PE=2.48 kg * 9.8 m/s^2 * 28.1 m = 683 J
KE = 0 because the object has not moved.
ME = KE + PE = 683 J

At 0 meters, the KE = PE
KE = 683 J
PE = 0 because PE = m*g*h and h = 0 meters
ME = KE + PE = 683 J