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 calculation that the above data verifies the law of conservation of energy (Neglect air resistance). (`g = 10 m//s^(2)`).

Answer

Total energy at 5 m above ground = P.E. 50 J + K.E. 0 J = 50 J ; Total energy at 3.2 m above ground = P.E. 32 J + K.E. 18 J = 50 J ; Total energy at 0 m above ground = P.E. 0 J + K.E. 50 J = 50 J ; Since the total energy (potential energy + kinetic energy) of the freely falling body at all the three position during its fall remains the same (50 J), therefore, the given data verifies the law of conservation of energy.

Explanation

NA