Question

A body of mass 6 g falls under action of gravity. At initial position ‘A’ =6m its P.E. is 480 J and K.E. is 0 J. During its downward journey at point ‘B’ =3m its energies will be (g = 10 ms-2):

Answer 1

Given info : A body of mass 6 g falls under action of gravity. At initial position A = 6m , its P.E = 480 J and K.E = 0

To find : energies of the body at point B = 3m , will be ...

Solution : initially, kinetic energy is zero. So total energy of body, T.E = K.E + P.E = 0 + 480 J = 480 J

now body is released from point A = 6m to B = 3m means, body travelled half distance of given path.

So, potential energy must be half at point B. I.e., P.E = 480/2 = 240 J

From energy conservation theorem,

Initial Total energy = final total energy

⇒480 J = 240 J + K.E

⇒K.E = 240 J

Therefore Potential energy is 240 J and kinetic energy is 240 J at point B.