Prove that change in kinetic energy of a paticle is equal to the work done on it by a variable force.

Explanation

Let 'K' represent kinetic energy. We know that `K=1//2mv^(2)`. The time rate of change of kinetic energy is given by `(dK)/(dt)=(d)/(dt)((1)/(2)mv^(2))=(1)/(2)m2v((dv)/(dt))=mv(dv)/(dt)=v(ma)=fv`
i.e., `(dK)/(dt)=F(dx)/(dt)` where `v=(dx)/(dt)`= time rate of displacement of particle.
or dK = Fdx
Integrating from the initial to final position.
we write `underset(k_(i))overset(k_(f))intdK=underset(x_(i))overset(x_(f))intFdx`
i.e., `(K_(f)-K_(i))=underset(x_(i))overset(x_(f))intF(x)dx`
Since `'DeltaK'` is a scalar quantity, the direction contained in Newton's II law is absent.