Question

The velocity of a body moving in a straight line is increased by applying a constant force $F$, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

Answer 1

Step 1: Given,

Let us consider the mass of the body is $m$, the initial velocity is $u$, the final velocity is $v$, the distance covered by the body is $S$, and the acceleration caused by the force $F$ is $a$.

Step 2: Formula used,

$$\begin{gathered} v^2=u^2+2aS.....(3rdequationofmotion) \\ Wokdoneisgivenby, \\ W=F\times S \end{gathered}$$

Step 3: Calculation,

$$\begin{gathered} As,W=F\times S \\ \Rightarrow F=\frac{W}{S} \\ \Rightarrow ma=\frac{W}{S} \\ \Rightarrow a=\frac{W}{mS} \\ Now,onsubstitutingvalueof\text{'}a\text{'}in3rdequationofmotion.Weget, \\ {v}^2=u^2+2\times \left(\frac{W}{mS}\right)\times S \\ {v}^2=u^2+2\times \left(\frac{W}{m}\right) \\ \frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2+W.............Multiplyiingbothsideby\left(\frac{1}{2}m\right) \\ \frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2=W \\ Finalkineticenergy-initialkineticenergy=Workdonebyexternalforce \end{gathered}$$

Hence, from the above expression, we can say that the increase in the kinetic energy of the body is equal to the work done by the force on the body.