Question

The following data was obtained for a body of mass 1 kg dropped from a height of 5m :

Distance above the ground Velocity

5m - 0m/s

3.2m - 6m/s

0m - 10m/s

Show by calculations that the above data verifies the law of conservation of energy.

(Neglect air resistance) g = 10m/s2

Answer 1

Answer:

Given: g = 10m/s2

Mass, m = 1 kg

h = 5m, v = 0 m/s

Case 1: Kinetic Energy, K.E. = mv2 = * 1 * 0 = 0

Potential Energy, P.E. = m * g * h = 1 * 10 * 5 = 50J

Total Energy = P.E. + K.E. = 50 + 0 = 50J

Case 2: h = 3.2

v = 6 m/s

K.E. = mv2 = * 1 * (6)2 = 18J

P.E. = m * g * h = 1 * 10 * 3.2 = 32J

Total Energy = P.E. + K.E. = 32 + 18 = 50J

Case 3: h = 0

v = 10 m/s

K.E. = mv2 = * 1 * (10)2 = 50J

P.E. = m * g * h = 1 * 10 * 0 = 0J

Total Energy = P.E. + K. E. = 50 + 0 = 50J