Question

Prove that mechanical energy of a freely falling body remains constant

Answer 1

Answer:

Suppose a body of mass m falls freely from a height . here AB is the height .

Suppose at the point A initial velocity , u =0 and the height of the body is h

Suppose at the point C after travelling a distance x t its velocity is v1m/s

Suppose at the point B its velocity is v2 m/s

At the point A

potential energy , PE = mgh

kinetic energy , KE = 1/2 m.u2

= 1/2 m.0

= 0

Total energy = PE +KE

= mgh +0

= 0

At the point C

potential energy ,PE = mg (h-x )

kinetic energy , KE = 1/2 m (v1)2

here (v1)2 = u2 + 2gx

(v1)2 = 0 + 2gx

( v1)2 =2gx

Now ,

KE = 1/2m2gx

= mgx

Total energy = PE + KE

= mg(h-x ) +mgx

= mgh - mgx + mgx

= mgh

At the point B

potential energy , PE = mg 0

= 0

kinetic energy , KE = 1/2m( v2)2

= mgh ( as like previous one )

Total energy = 0 + mgh

= mgh

thereforeechanical energy of afreely falling body remains constant .