Derive an expression for kinetic energy of an object. (b) If the velocity of an object is doubled. What will be change in its kinetic energy?

Suppose a body of mass m is moving with velocity v. It is brought to rest by applying a retarding force F. Suppose it traverses a distance s before coming to rest.

Kinetic energy of body, KE = Work done by retarding force to stop it.

i.e.,        Kinetic energy = F. s                 … (i)

But        Retarding force, F = ma          … (ii)

Initial velocity = v, final velocity = From the equation, v² = u² + 2as, we have

O = v² – 2as (because here a is retardation)

⟹ Distance, s = {𝑣^2 \over 2𝑎}          … (iii)

Substituting values of F and s from (ii) and (iii) in (i), we get

Kinetic energy, KE = ma  x {𝑣^2 \over 2𝑎} = {1\over2} mv²