Question

prove that energy on freely falling body remain

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Answer 1

Answer:

Let a body of mass m falls from a point A, which is at a height h from the ground as shown in fig.

At A,

Kinetic energy kE = 0

Potential energy Ep = mgh

Total energy E = Ep + Ek = mgh + 0= mgh

During the fall, the body is at a position B. The body has moved a distance x from A.

At B,

velocity v2 = u2 + 2as

applying, v2 = 0 + 2ax = 2ax

Kinetic energy Ek = 1/2 mv2 = 1/2 m x 2gx = mgx

Potential energy Ep = mg (h – x)

Total energy E = Ep + Ek = mg (h-x) + mgx = mgh – mgx + mgx= mgh

If the body reaches the position C.

At C,

Potential energy Ep = 0

Velocity of the body C is

v2 = u2 + 2as

u = 0, a = g, s = h

applying v2 = 0 + 2gh = 2gh

kinetic energy Ek =1/2 mv2=1/2 m x 2gh= mgh

Total energy at C

                     E = Ep + Ek

                    E = 0 + mgh

                    E = mgh

Thus we have seen that sum of potential and kinetic energy of freely falling body at all points remains same. Under the force of gravity, the mechanical energy of a body remains constant.