Question

law of conservative energy at free falling body?
sol;

Answer 1

Consider a body of mass m placed at A.

h = AB is the height of the body above the ground

u = 0 is initial velocity at A

v1 = velocity of body at C

v = velocity of body at B, i.e., just above the ground

(i) At point A

PEA = mgh

KEA = 0

Total mechanical energy at A, EA = PEA + KEA = mgh + 0 = mgh

(ii) At point C

v12-0 = 2gs

v12 = 2gs

KEC =

PEC = mg(h-s)

Total mechanical energy at C, EC = PEC + KEC = mg(h-s) + mgs = mgh

(iii) At point B

v2-02 = 2gh

v2 = 2gh

KEB =

PEB = 0

Total mechanical energy at B, EB = PEB + KEB = 0 + mgh = mgh

The total mechanical energy of the body at A, B and C (also at any other point in the path AB) is the same. So, the total mechanical energy of the body throughout the free fall is conserved.