Answer :
Consider a body m lying at a position A at a height h, above the ground. As the body falls, its kinetic energy increases at the expense of its potential energy.
At point A
The body is at rest
The kinetic energy of a body at A = KA
The potential energy of body at A = UA
The total techanical energy of body at A = EA
At point B
Suppose the body falls through a distance x and reaches B with velocity v. Then
u = 0
s = x
a = g
The kinetic energy of body at B = KB
The potential energy of a body at B = UB
The total energy of the body at B = EB
At point C
Suppose the body reaches C (on the ground) with velocity v’. Then considering motion from A to C
u = 0
s = h
a = g
v = v’
Kinetic energy of body at C = KC
Potential energy of body at C = UC
Total mechanical energy of body at C = EC
As a body falls freely under gravity P.E. decreases and K.E increases by equal amounts, however the, total mechanical energy remains constant.
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