Question

derive kinetic energy of a moving body given by one half the product of its mass and the square of its velocity (method 1 &2)​

Answer 1

Derivation of Expression of Kinetic Energy:

We will use Work-Energy theorem to prove the expression for kinetic energy. Work-Energy theorem states that the work done by all the forces on an object is equal to the change in kinetic energy.

$\therefore \: \sf{K = work \: done}$

$= > \: \sf{K }= \displaystyle \: \sf{\int \: force \times (dr)}$

$= > \: \sf{K }= \displaystyle \: \sf{\int \: (m \times a) \times (dr)}$

$= > \: \sf{K }= \displaystyle \: \sf{\int \: m \times \dfrac{dv}{dt} \times dr}$

$= > \: \sf{K = m}\displaystyle \: \sf{\int \: \dfrac{dv}{dt} \times dr}$

$= > \: \sf{K = m}\displaystyle \: \sf{\int \: \dfrac{dr}{dt} \times dv}$

$= > \: \sf{K = m}\displaystyle \: \sf{\int \: v \times dv}$

$= > \: \sf{K = m \times \dfrac{ {v}^{2} }{2} }$

$= > \: \sf{K = \dfrac{ m{v}^{2} }{2} }$

[Hence proved]