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Q78P

Expert-verifiedFound in: Page 176

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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${{\mathit{F}}}_{\mathbf{a}\mathbf{x}}{\mathbf{=}}{\mathbf{9}}{\mathit{x}}{\mathbf{-}}{\mathbf{3}}{{\mathit{x}}}^{{\mathbf{2}}}$**

- Graph is plotted below.
- Work is maximum at x=3 m
- Maximum value of work is
*W*=13.5J. - Work has decreased to zero at
*x*=4.5m. - The case is again at rest at x =4.5m.

It is given that, force is given as,

${F}_{ax}=9x-3{x}^{2}$

**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:**

$W=Fd\phantom{\rule{0ex}{0ex}}W=\left(\frac{1}{2}\right)m{v}_{f}^{2}-\left(\frac{1}{2}\right)m{v}_{i}^{2}$

Where,* F* is force, *d *is displacement, m is mass, ${v}_{i,}{v}_{f}$are initial and final velocities and *W *is the work done.

Find the work done by integration of the given function,

$W=\int {F}_{ax}dx\phantom{\rule{0ex}{0ex}}=\int \left(9x-3{x}^{2}\right)dx\phantom{\rule{0ex}{0ex}}=\frac{9x}{2}-{x}^{3}$

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,

$W=\frac{9{x}^{2}}{2}-{x}^{3}\phantom{\rule{0ex}{0ex}}=\frac{9{\left(3\right)}^{2}}{2}-{\left(3\right)}^{3}=13.5\mathrm{J}$

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,

$W=\left(\frac{1}{2}\right)m{v}_{f}^{2}-\left(\frac{1}{2}\right)m{v}_{i}^{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{1}{2}\right)m{\left(0\right)}^{2}-\left(\frac{1}{2}\right)m{\left(0\right)}^{2}$

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.

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