Question

A particle of mass m moving along the $x-$axis has velocity $v_x=v_0\sin (\mathrm{\pi x}/2\mathrm{L})$. How much work is done on the particle as it moves (a) from $x=0$to $x=L$and

(b) from $x=0$to$x=2L$?

Short answer

a). $\frac{m{v_0}^2}{4}$

b).$\frac{-m{v_0}^2}{4}$

Step-by-step solution

Step 1. Given Information

Mass of particle$=m$

Velocity of the particle $v_x=v_o\sin (\mathrm{\pi x}/2\mathrm{L})$

Step 2. Part a). Work done on the particle

When it moves from $x=0$to $x=L$. For that the we will proceed as

$$\begin{gathered} W={\int }_a^{b}Fdx \\ W={\int }_a^{b}madx \\ {a}_x=v_x\frac{d{v}_x}{dx} \\ \frac{d{v}_x}{dx}=v_0\frac{\mathrm{\pi }}{2L}\cos \frac{\mathrm{\pi x}}{2L} \end{gathered}$$

Now,

$$\begin{gathered} W={\int }_a^{b}madx \\ ={\int }_0^{L}m{v_0}^2\frac{\mathrm{\pi }}{2L}\sin \frac{\mathrm{\pi x}}{2L}\cos \frac{\mathrm{\pi x}}{2L}dx \\ =m{v_0}^2\frac{\mathrm{\pi }}{4L}{\int }_0^{L}2\sin \frac{\mathrm{\pi x}}{2L}\cos \frac{\mathrm{\pi x}}{2L}dx \\ =m{v_0}^2\frac{\mathrm{\pi }}{4L}{\int }_0^L\sin \frac{\mathrm{\pi x}}{L}dx \end{gathered}$$

Using identity: $\sin 2\theta =2\sin \theta \cos \theta$

$=-m{v_0}^2\frac{1}{4}\left(\cos \frac{\mathrm{\pi L}}{L}-\cos 0\right)=\frac{m{v_0}^2}{4}$

Step 3. Part b). Work done on the particle

We need to determine work done on the particle when it form$x=0$to $x=2L$

Now,

$$\begin{gathered} W={\int }_a^{b}madx \\ ={\int }_0^{2L}m{v_0}^2\frac{\mathrm{\pi }}{2L}\sin \frac{\mathrm{\pi x}}{2L}\cos \frac{\mathrm{\pi x}}{2L}dx \\ =m{v_0}^2\frac{\mathrm{\pi }}{4L}{\int }_0^{2L}2\sin \frac{\mathrm{\pi x}}{2L}\cos \frac{\mathrm{\pi x}}{2L}dx \\ =m{v_0}^2\frac{\mathrm{\pi }}{4L}{\int }_0^L\sin \frac{\mathrm{\pi x}}{L}dx \\ ={\overline{)-m{v_0}^2\cos \frac{\mathrm{\pi x}}{L}}}^{2L} \\ =-\frac{m{v_0}^2}{4}\left(\cos \frac{2\mathrm{\pi L}}{L}-\cos 0\right)=-\frac{m{v_0}^2}{4} \end{gathered}$$