Problem 2: A body of mass 2.0 kg is dropped from a rest position 5.0 m above the ground. What is its velocity at height of 3.0 m above the ground?

Concept: At the maximum height, the body has a certain amount of energy, mgh. As it falls down, the potential energy is converted to kinetic energy until it is totally converted into kinetic energy at the lowest point. However, on the way down, the energy possessed by the body is partly potential and partly kinetic, depending on the height from the lowest point, and the sum of its K.E and P.E at any instant is equal to the P.E at the top height. So, at a height of 3.0 m, it possesses P.E less than at the height of 5.0 m and the difference in potential energies is equal to the gain in K.E. Therefore, we find the K.E and the velocity will be calculated from that.

Given, state 1                     Mass of the body = 2.0 kg,           Height, h = 5.0 m

State 2                                  Mass of the body = 2.0 kg,           Height, h2 = 3.0 m

Required                              Velocity 3 m above the ground.

Now,                                     K.E = mgh – mgh2

Put the values,                  ½ mv2 =(2 × 9.8 × 5)- (2 × 9.8 × 3)    OR     ½ × 2v2 = 98 – 58.8 = 39.2

OR                                          v2 = 39.2               OR          v = √39.2              OR          v = 6.3 ms-1 (Answer)