Figure 7-40 gives the acceleration of a 2.00 kg particle as an applied force $\overrightarrow{{\mathbf{F}}_{\mathbf{a}}}$ moves it from rest along an x axis from$\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{}\mathbf{to}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$. The scale of the figure’s vertical axis is set by${\mathbf{a}}_{\mathbf{s}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. How much work has the force done on the particle when the particle reaches (a) x=4.0 m, (b) x=7.0 m, and (c) x=9.0 m? What is the particle’s speed and direction of travel when it reaches (d) x=4.0 m, (e) x=7.0 m, and (f) x=9.0 m ?

1. The mass of particle is m=2.00 kg
2. The scale of acceleration is set as ${\mathrm{a}}_{\mathrm{s}}=6.0\mathrm{m}/{\mathrm{s}}^2$
3. At x=3.0 m the kinetic energy of the particle is${\mathrm{K}}_{\mathrm{f}}=11.0\mathrm{J}$

We can calculate the word done by usingthegraph and finding the area of a particular region. As the given graph is of acceleration vs position, we multiply the area by the value of mass of a particle to get the work done. We can use the equation ofthework-energy theorem and kinetic energy to find the velocity at different positions.

Formula:

The equation of work-energy theorem

$$\begin{gathered} \mathbf{W}\mathbf{=}{\mathbf{K}}_{\mathbf{f}}\mathbf{-}{\mathbf{K}}_{\mathbf{i}} \\ \mathbf{W}\mathbf{=}{\mathbf{\int }}_{{\mathbf{x}}_{\mathbf{i}}}^{{\mathbf{x}}_{\mathbf{f}}}\mathbf{F}\mathbf{d}\mathbf{x} \\ \mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}} \end{gathered}$$

The equation for work done is

$\mathrm{W}={\int }_{{\mathrm{x}}_{\mathrm{i}}}^{{\mathrm{x}}_{\mathrm{f}}}\mathrm{F}d\mathrm{x}$

Also, the given graph is, a vs x, so we can find the work done by finding the area under the curve and multiplying it by mass.

To calculate the area under the curve for x=0 to x=4.0, we can divide the region in two parts as shown:

So, the equation for work will become

$\mathrm{W}=\mathrm{m}\times \left[\mathrm{Area}\mathrm{of}\mathrm{region}1+\mathrm{Area}\mathrm{of}\mathrm{region}2\right]$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{W}=2.0\mathrm{kg}\times \left[\frac{1}{2}\times 1.0\mathrm{m}\times 6.0\mathrm{m}/{\mathrm{s}}^2+3.0\mathrm{m}\times 6.0\mathrm{m}/{\mathrm{s}}^2\right] \\ \mathrm{W}=42\mathrm{J} \end{gathered}$$

Therefore, work done on the particle when the particle x=4.0m reaches is 42 J

To calculate the area under the curve from x=0 m to x=7.0 m, we can divide the graph in 5 parts as,

From this graph, we can see that the area of region 3 and 4 is equal and opposite, so both will cancel out.

$\mathrm{W}=\mathrm{m}\times \left[\mathrm{Area}\mathrm{of}\mathrm{region}1+\mathrm{Area}\mathrm{of}\mathrm{region}2+\mathrm{Area}\mathrm{of}\mathrm{region}5\right]$

Substitute all the value in the above equation.

$\begin{array}{rcl}\mathrm{W}& =& 2.0\mathrm{kg}\times \left[\frac{1}{2}\times 1\mathrm{m}\times 6.0\mathrm{m}/{\mathrm{s}}^2+3.0\mathrm{m}\times 6.0\mathrm{m}/{\mathrm{s}}^2+\left(-6.0\mathrm{m}/{\mathrm{s}}^2\times 1.0\mathrm{m}\right)\right]\\ & =& 30\mathrm{J}\end{array}$

Therefore, work done on the particle when the particle x=7.0 m reaches is 30 J

To calculate the area under the curve for x=0 m to x=9.0 m. We can divide the region in six parts as,

Here, we can see that the area of regions 1 and 3 will cancel out with the area of region 4 and 6.

So, the work done will become,

$\mathrm{W}=\mathrm{m}\times \left[\mathrm{Area}\mathrm{of}\mathrm{region}2+\mathrm{Area}\mathrm{of}\mathrm{region}5\right]$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{W}=2.0\mathrm{kg}\times \left[3.0\mathrm{m}\times 6.0\mathrm{m}/{\mathrm{s}}^2+\left(-2.0\mathrm{m}\times 6.0\mathrm{m}/{\mathrm{s}}^2\right)\right] \\ \mathrm{W}=12\mathrm{J} \end{gathered}$$

Therefore, work done on the particle when the particle x=9.0 m reaches is 12 J

According to the work-energy theorem,

$\mathrm{W}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}$

As the particle is at rest initially,${\mathrm{K}}_{\mathrm{i}}=0$, the above equation becomes,

$\mathrm{W}={\mathrm{K}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}^2$

Rearranging it for velocity, we get

$\mathrm{v}=\sqrt{\frac{2\mathrm{W}}{\mathrm{m}}}$

Form part a, we have $\mathrm{W}=42\mathrm{J}$. So,

$$\begin{gathered} \mathrm{v}=\sqrt{2\times \frac{42\mathrm{J}}{2.0\mathrm{kg}}} \\ \mathrm{v}=6.5\mathrm{m}/\mathrm{s} \end{gathered}$$

As the answer is positive, its direction is along positive x axis.

From the work-energy theorem and using initial kinetic energy as zero, we write

$\mathrm{v}=\sqrt{\frac{2\mathrm{W}}{\mathrm{m}}}$

Form part b, we have $\mathrm{W}=30\mathrm{J}$. So,

$$\begin{gathered} \mathrm{v}=\sqrt{2\times \frac{30\mathrm{J}}{2\mathrm{kg}}} \\ \mathrm{v}=5.5\mathrm{m}/\mathrm{s} \end{gathered}$$.

As the answer is positive, its direction is along positive x axis.

Following the similar procedure as in part d and e, with value of work from part c, we get

$\mathrm{v}=3.5\mathrm{m}/\mathrm{s}$

As the answer is positive, its direction is also along positive x axis.