Chapter 8.5, Problem 42E

To determine

To Show : Net work done by the force is $W=\frac{1}{2}m{v}_2{}^2-\frac{1}{2}m{v}_1{}^2$

Explanation of Solution

Given information :

Concept used : $v=\frac{dx}{dt}\text{ and }F=m\left(\frac{dv}{dt}\right)$

The body is moving from $x_1$ to $x_2$

Calculation :

We have work done by force to move from $x_1$ to $x_2$ is $W={\displaystyle \underset{x_1}{\overset{x_2}{\int }}f\left(x\right)}dx$

Here,

  $\begin{array}{c}f\left(x\right)=mf\text{ where }f\text{ is acceleration}\\ \text{=m}\frac{dv}{dt}\\ =mv\frac{dv}{dx}\end{array}$

  $\begin{array}{c}\therefore W={\displaystyle \underset{x_1}{\overset{x_2}{\int }}\left(mv\frac{dv}{dx}\right)}dx\\ =m{\displaystyle \underset{v_1}{\overset{v_2}{\int }}vdv\left(\text{as }{v}_1\text{ and }{v}_2\text{ are velocities when the body is at }{x}_1\text{ and }{x}_2\text{ position}\right)}\\ =m{\displaystyle \underset{v_1}{\overset{v_2}{\int }}vdv}\\ =m{\left[\frac{v^2}{2}\right]}_{v_1}^{v_2}\\ =\frac{1}{2}m{v}_2{}^2-\frac{1}{2}m{v}_1{}^2\text{ (shown)}\end{array}$