State and prove work-energy relationship.

Solution

The Work- Energy Theorem states that the work done on or by the body is equal to the change in its kinetic energy.
Consider a force F applied on the body of mass m moving on the horizontal frictionless surface.
Let after travelling a distances s, the velocity of body change from u to v.
The small amount of work, dW done by force F in displacing the body by ds is,
dW = $straight F with rightwards harpoon with barb upwards on top space. space stack d s with rightwards harpoon with barb upwards on top$
$equals space straight m space fraction numerator straight d straight v with rightwards harpoon with barb upwards on top over denominator dt end fraction. space ds with rightwards harpoon with barb upwards on top space equals space straight m space. space dv with rightwards harpoon with barb upwards on top space. space fraction numerator ds with rightwards harpoon with barb upwards on top over denominator dt end fraction equals space straight m space straight v with rightwards harpoon with barb upwards on top. space dv with rightwards harpoon with barb upwards on top equals space mv space dv space$
dW = mv dv
Now, integrating both sides, we have
$integral dW space equals space integral subscript straight u superscript straight v mv space dv space equals space straight m space integral subscript straight u superscript straight v straight v. space dv straight W space equals space straight m right enclose space straight v squared over 2 end enclose subscript straight u superscript straight v space equals space straight m over 2 left parenthesis space straight v squared space minus space straight u squared right parenthesis space straight W space equals space 1 half mv squared space minus space 1 half mu squared space$
That is,
Work = Change in Kinetic energy.