**Work-Energy Theorem**

**Statement:** Total work done by a force acting on a body is the total change in its kinetic energy.

**Proof:** Suppose a body of mass m is moving on a smooth horizontal surface with a constant velocity, u. Let a constant force F acts on the body from point A to B as shown in the figure such that the velocity increases to v. The work done by the force is

W = F s

Where 's' is the displacement of the body.

From Newton’s second law of motion,

F =ma

Then, work done is given by

W = ma s

Let the initial kinetic energy be K.E

_{1}= ½mu^{2}and final kinetic energy be K.E_{2}= ½mv^{2},Then,

From the equation of motion v

^{2}= u^{2}+ 2as,v

^{2}– u^{2}= 2asor, as = ½ (v

^{2}– u^{2})since, W = mas = m. ½ (v

^{2}– u^{2})= ½mv

^{2}–½mu^{2}= K.E

_{2}– K.E_{1}So, the work done on moving the body from A to B by applying force F is equal to the increase in Kinetic energy of the body. Again, we can write the above equation as,

W = ½mv

^{2}–½mu^{2}½mv

^{2}= W +½mu^{2}That is, the final kinetic energy of the body is increased and it is the sum of the work done by the body and its initial kinetic energy.

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