A CD case slides along a floor in the positive direction of an$\mathit{x}$axis while an applied force role="math" localid="1657190456443" $\overrightarrow{\mathit{F}}$acts on the case. The force is directed along the $\mathit{x}$axis and has the x component${\mathit{F}}_{\mathbf{a}\mathbf{x}}\mathbf{=}\mathbf{9}\mathit{x}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}$, within meters and${\mathit{F}}_{\mathbf{a}\mathbf{x}}$in Newton. The case starts at rest at the position $\mathit{x}\mathit{=}\mathit{0}$, and it moves until it is again at rest. (a) Plot the workrole="math" localid="1657190646760" $\overrightarrow{{\mathit{F}}_{\mathit{a}}}$does on the case as a function of x. (b) At what position is the work maximum, and (c) what is that maximum value? (d) At what position has the work decreased to zero? (e) At what position is the case again at rest?

It is given that, force is given as,

$F_{ax}=9x-3x^2$

The problem deals with the work done which is the fundamental concept of physics.Work is the displacement of an object when force is applied to it.Use the concept of work related to force and displacement and kinetic energy.Plot the graph of work done vs. displacement. From the graph, determine the positions where the work done is maximum or zero.

Formulae:

$$\begin{gathered} W=Fd \\ W=\left(\frac{1}{2}\right)m{v}_f^2-\left(\frac{1}{2}\right)m{v}_i^2 \end{gathered}$$

Where,F is force, dis displacement, m is mass, $v_{i,}{v}_f$are initial and final velocities and Wis the work done.

Find the work done by integration of the given function,

$$\begin{gathered} W=\int F_{ax}dx \\ =\int \left(9x-3x^2\right)dx \\ =\frac{9x}{2}-x^3 \end{gathered}$$

From the above graph, at x = 3, the work is maximum.

Hence, work is maximum at x =3 m.

Use the equation found in part a) to find the maximum value of work, which is at x = 3m,

$$\begin{gathered} W=\frac{9x^2}{2}-x^3 \\ =\frac{9{\left(3\right)}^2}{2}-{\left(3\right)}^3=13.5\mathrm{J} \end{gathered}$$

Hence, maximum value of work is W=13.5 J.

From the graph, work is zero at x =4.5 m.

Hence, work has decreased to zero at x =4.5 m.

The case initially starts from rest and finally comes to rest,

$$\begin{gathered} W=\left(\frac{1}{2}\right)m{v}_f^2-\left(\frac{1}{2}\right)m{v}_i^2 \\ =\left(\frac{1}{2}\right)m{\left(0\right)}^2-\left(\frac{1}{2}\right)m{\left(0\right)}^2 \end{gathered}$$

So, the case comes to rest at position$x=4.5\mathrm{m}$

Therefore, the concept of work related to force and displacement and work-energy theorem can be used.