A stone with mass 01 kg is dropped from a height of 80 m cal

A stone with mass 0.1 kg is dropped from a height of 80 m. calculate its potential and kinetic energies (a) at the time it was dropped, (b) after it has traveled 20 m, and (c) at the instant it touches the ground. Assume that air friction is negligible and the reference position is the ground.

Solution

\\(a) @ 80m
Potential Energy = mgh = (0.1)(9.8)(80) = 78.4 J
Kinetic Energy = 1/2(mv^2) = 0 J (since v = 0 initially)
Total Energy = 78.4 + 0 = 78.4 J

(b) @ 60m
Potential Energy = mgh = (0.1)(9.8)(80-20) = 58.8 J
Kinetic Energy = total energy - potential energy = 78.4 - 58.8 = 19.6 J

(c) @ ground level
Potential Energy = mgh = (0.1)(9.8)(0) = 0 J
Kinetic Energy = 78.4 - 0 = 78.4