Figure 7-37 gives spring force F_xversus position x for the spring–block arrangement of Fig . 7-10 . The scale is set by${\mathbf{F}}_{\mathbf{x}}\mathbf{=}\mathbf{160}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$. We release the block at$\mathbf{x}\mathbf{=}\mathbf{12}\mathbf{}\mathbf{cm}$. How much work does the spring do on the block when the block moves from ${\mathbf{x}}_{\mathbf{i}}\mathbf{=}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$ to (a) x= +5.0 cm, (b) x=-5.0 cm, (c) x=-8.0 cm, and (d) x=-10.0 cm ?

We can use the equation of Hook’s law to calculate the spring constant from the graph, hence, we can calculate the work done in each case using the equation of work done bythespring.

Formulae:

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{-}\mathbf{kx} \\ \mathbf{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{k}\left(x_i^2-x_f^2\right) \end{gathered}$$

Given:

1. The graph of ${\mathrm{F}}_{\mathrm{x}}\mathrm{vs}\mathrm{x}$
2. The scale of graph is set at, ${\mathrm{F}}_{\mathrm{s}}=160.0\mathrm{N}$

The graph of F_x vs x is given, so according to Hook’s law, the slope of this graph will give us the spring constant of the spring used.

From this graph, we get

$\begin{array}{rcl}\mathrm{slope}& =& \frac{∆\mathrm{y}}{∆\mathrm{x}}\\ & =& \frac{80\mathrm{N}}{-1\mathrm{cm}}\\ & =& -80\frac{\mathrm{N}}{\mathrm{cm}}\\ & =& -8\times {10}^3\mathrm{N}/\mathrm{m}.\end{array}$

As per Hook’s law, we have

$\mathrm{F}=-\mathrm{kx}$

So,
$\mathrm{k}=-\frac{\mathrm{F}}{\mathrm{x}}$

Hence, the spring constant becomes,

$\mathrm{k}=8\times {10}^3\mathrm{N}/\mathrm{m}.$

The equation for work done by the spring is,

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$

For this part,${\mathrm{x}}_{\mathrm{i}}=0.08\mathrm{m}\mathrm{and}{\mathrm{x}}_{\mathrm{f}}=0.05\mathrm{m}$.

So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080\mathrm{m}\right)}^2-{\left(0.050\mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& 15.6\mathrm{J}\\ & \approx & 16\mathrm{J}\end{array}$

Therefore, work done is, 16 J.

The equation for work done by spring is

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08\mathrm{m}\mathrm{and}{\mathrm{x}}_{\mathrm{f}}=-0.05\mathrm{m}$. So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080\mathrm{m}\right)}^2-{\left(-0.050\mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& 15.6\mathrm{J}\\ & \approx & 16\mathrm{J}\end{array}$

Therefore, work done is, 16 J.

The equation for work done by spring is

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08\mathrm{m}\mathrm{and}{\mathrm{x}}_{\mathrm{f}}=-0.08\mathrm{m}$. So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080\mathrm{m}\right)}^2-{\left(-0.080\mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& 0\mathrm{J}\end{array}$

Therefore, work done is, 0 J.

We have, the equation for work done by spring as,

$\mathrm{W}=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}_{\mathrm{i}}^2-{\mathrm{x}}_{\mathrm{f}}^2\right)$

For this part, ${\mathrm{x}}_{\mathrm{i}}=0.08\mathrm{m}\mathrm{and}{\mathrm{x}}_{\mathrm{f}}=-0.10\mathrm{m}$. So,

$\begin{array}{rcl}\mathrm{W}& =& \frac{1}{2}\times 8\times {10}^3\mathrm{N}.\mathrm{m}\left({\left(0.080\mathrm{m}\right)}^2-{\left(-0.10\mathrm{m}\right)}^2\right)\\ \mathrm{W}& =& -14.4\mathrm{J}\\ & \approx & -14\mathrm{J}\end{array}$

Therefore, work done is, -14 J.