Question
A ball having a weight of 1.5 N is dropped from a height of 4 meters. (Neglect air friction.) How much mechanical energy is "lost" just before it hits the ground?
Answers
Henry
M*g = 1.5, M = 1.5/g = 1.5/9.8 = 0.153 kg.
PE = Mg*h = 1.5 * 4 = 6.0 J.
V^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4 (m/s)^2.
KE = 0.5M*V^2 = 0.5*0.153*78.4 = 6.0 Joules.
Energy Lost = PE - KE = 6-6 = 0.
PE = Mg*h = 1.5 * 4 = 6.0 J.
V^2 = Vo^2 + 2g*h = 0 + 19.6*4 = 78.4 (m/s)^2.
KE = 0.5M*V^2 = 0.5*0.153*78.4 = 6.0 Joules.
Energy Lost = PE - KE = 6-6 = 0.