Question

The following data was obtained for a body of mass 1 kg dropped from a height of 5 metres Distance above ground
Velocity
5 m
0 m/s
3.2 m
6 m/s
0 m
10 m/s
Show by calculations that the above data verifies the law of conservation of energy (Neglect air resistance). (g = 10 m/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

Since, the total energy is equal in all the three cases.

Hence, the above data verifies the Law of Conservation of Energy