Single Variable Calculus, Chapter 6, 6.4, Section 6.4, Problem 4

Suppose that a force of $\displaystyle \cos \left( \frac{\pi x}{3} \right)$ newtons acts on a particle located a distance $x$ meters from the origin. How much work is done in moving the particle from $x = 1$ to $x = 2$? Interpret your answer by considering the work done from $x = 1$ to $x = 1.5$ and from $x = 1.5$ to $x = 2$.


$\begin{equation} \begin{aligned} W =& \int^2_1 \cos \frac{\pi x}{3} dx \\ \\ \text{Let } u =& \frac{\pi x}{3} \\ \\ du =& \frac{\pi}{3} dx \\ \\ W =& \frac{3}{\pi} \int^2_1 \cos u du \\ \\ W =& \frac{3}{\pi} \left[ \sin \frac{\pi x}{3} \right]^2_1 \\ \\ W =& 0 \text{ Joules} \end{aligned} \end{equation}$



If we consider the work done from $x = 1$ to $x = 1.5$ we'll get a work equal in magnitude but opposite in sign to the work done from $x = 1.5$ to $x = 2$.


$\begin{equation} \begin{aligned} W_1 =& \int^{1.5}_1 \cos \frac{\pi x}{3} = \frac{3}{\pi} \left( 1 - \frac{\sqrt{3}}{2} \right) J \\ \\ W_2 =& \int^2_{1.5} \cos \frac{\pi x}{3} = \frac{3}{\pi} \left( \frac{\sqrt{3}}{2} - 1 \right) J \end{aligned} \end{equation}$


Thus, resulting to a total work of 0 Joles.

We can conclude that at $x = 1.5$ to $x = 2$, the force opposes the motion of the particle, that way take its Kinetic Energy to decrease.