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/s 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)
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.
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