Q. 1 - Calculus

Question

Work as an integral of force and distance: Find the work done in moving an object along the x-axis from the origin to $x=\frac{\mathrm{\pi }}{2}$ if the force acting on the object at a given value of x is $\mathrm{F}\left(x\right)=x\sin x$.

Step-by-Step Solution

Answer

The work done in moving an object along the x-axis from the origin to$x=\frac{\mathrm{\pi }}{2}$ is $0$.

Step 1.Given Information

Work as an integral of force and distance: Find the work done in moving an object along the x-axis from the origin to $x=\frac{\mathrm{\pi }}{2}$ if the force acting on the object at a given value of x is data-custom-editor="chemistry" $\mathrm{F}\left(\mathrm{x}\right)=\mathrm{xsinx}$.

Step 2. The work done is W = ∫ 0 π / 2 F ( x ) d x

$W={\int }_0^{\mathrm{\pi }/2}x\sin xdx$

Firstly solving the integral

$$\begin{gathered} W=\int x\sin xdx \\ W=\left[x\int \sin x\mathrm{d}x-\left\{\int \frac{d}{dx}\left(x\right)\left(\int \sin xdx\right)\right\}\right] \\ W=\left[-x\cos x-\left\{-\frac{x^2}{2}\cos x\right\}\right] \\ W=\left[-x\cos x+\frac{x^2}{2}\cos x\right] \end{gathered}$$

Step 3. Now solving the integral W = - x cos x + x 2 2 cos x 0 π / 2

$$\begin{gathered} W={\left[-x\cos x+\frac{x^2}{2}\cos x\right]}_0^{\mathrm{\pi }/2} \\ W=\left[\left(-\frac{\mathrm{\pi }}{2}\cos \frac{\mathrm{\pi }}{2}+\frac{{\left(\frac{\mathrm{\pi }}{2}\right)}^2}{2}\cos \frac{\mathrm{\pi }}{2}\right)-\left(0\cos 0+\frac{{\left(0\right)}^2}{2}\cos 0\right)\right] \\ W=\left[\left(-\frac{\mathrm{\pi }}{2}\times 0+\frac{{\mathrm{\pi }}^2}{4·2}\times 0\right)-\left(0\cos 0+\frac{{\left(0\right)}^2}{2}\cos 0\right)\right] \\ W=\left[\left(0+0\right)-\left(0+0\right)\right] \\ W=0 \end{gathered}$$