Question

Show that in case of a body falling freely under gravity, total mechanical energy remains conserved (neglect air resistance).

Answer 1

Let a body of mass m fall freely under gravity from height h above the ground.
Let A, B, and C be the positions of the body.
Let x be the distance fallen from A to B.

At position A :

KE = 0 .......(body is at rest)

and PE = mgh

∴ Total energy = 0 + mgh = mgh ..........(i)

At position B :

Let v₁ be velocity of body, then u = 0, S = x.

From equation v² = u² + 2aS

v₁² = 0 + 2gx = 2gx

Since KE =`1/2` mv² = `1/2` m × 2gx = mgx

and PE = mg (h − x)

= mgh − mgx

∴ Total energy = mgx + mgh − mgx = mgh ..............(ii)

At position C :

Let velocity of body be v, then u = 0, S = h.

From equation v² = u² + 2gS

v² = 0 + 2gh = 2gh

Since KE = `1/2` mv² = `1/2` m × 2gh = mgh

and PE = 0

∴ Total energy = mgh + 0 = mgh ............(iii)

From (i), (ii) and (iii) it is clear that sum of mechanical energy remains same at any point in the path of free fall of a body.