In the arrangement of Fig. $\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{10}$, we gradually pull the block from $\mathit{x}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathbf{}\mathit{x}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$, where it is stationary. Figure $7-35$gives the work that our force does on the block. The scale of the figure’s vertical axis is set by ${\mathit{W}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathit{J}$. We then pull the block out to $\mathit{x}\mathbf{=}\mathbf{+}\mathbf{}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathit{c}\mathit{m}\mathbf{}$and release it from rest. How much work does the spring do on the block when the block moves from${\mathit{x}}_{\mathbf{i}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$to, (a) $\mathit{x}\mathbf{=}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$(b) $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathit{c}\mathit{m}$, and (c) $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathit{c}\mathit{m}$?

1. The initial position of the block is $x_{\text{i}}=0\text{m}$.
2. The vertical axis of the graph shows work done by the spring on the block as, $W_{\text{s}}=1.0\text{J}$.

The block is connected to the spring; hence, we can use the concept of the work by the spring on the block, which is associated with the change in energy.

Formulae:

$W_{\text{s}}=\frac{1}{2}k{x}_{\text{i}}^{\text{2}}-\frac{1}{2}k{x}_{\text{f}}^{\text{2}}$ (1)

Here $x_{\text{f}}$ is the final position of the block and is the spring constant.

From the graph,

For the vertical axis of${\text{x}}_{\text{i}}=2\text{cm}=0.020\text{m}$, work done is .

$x_{\text{f}}$ will be 0.

The work done by the spring can be calculated as,

$$\begin{gathered} W_{\text{s}}=\frac{1}{2}k\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right) \\ k=\frac{2W_s}{\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)} \end{gathered}$$

Substitute the values and find the value of spring constant as

$$\begin{gathered} k=\frac{2\times 0.40\text{J}}{\left[{\left(0.020\text{m}\right)}^2-0^2\right]} \\ k=2\times {10}^3\text{N/m} \end{gathered}$$

When the block moves from$x_{\text{i}}=+5.0\text{cm}=0.050\text{m}$ to $x_{\text{f}}=4.0\text{cm}=0.040\text{m}$, the work by the spring can be calculated using equation 1 as,

$$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3\text{N/m}\times \left[{\left(0.050\text{m}\right)}^2-{\left(-0.040\text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3\text{N/m}\times \text{9}\times {\text{10}}^{-4}{\text{m}}^2 \\ {W}_{\text{s}}=0.90\text{J} \end{gathered}$$

Thus, the work done is, $W_{\text{s}}=0.90\text{J}$.

When the block moves from $x_{\text{i}}=+5.0\text{cm}=0.050\text{m}$to $x_{\text{f}}=-2.0\text{cm}=-0.020\text{m}$, the work done by the spring can be calculated using equation 1 as,

$$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3\text{N/m}\times \left[{\left(0.050\text{m}\right)}^2-{\left(-0.020\text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3\text{N/m}\times \text{2.1}\times {\text{10}}^{-3}{\text{m}}^2 \\ {W}_{\text{s}}=2.1\text{J} \end{gathered}$$

Thus, the work doneis,$W_{\text{s}}=2.1\text{J}$.

When the block moves from $x_{\text{i}}=+5.0\text{cm}=0.050\text{m}$to$x_{\text{f}}=-5.0\text{cm}=-0.050\text{m}$ , the work by the spring can be calculated using equation 1 as,

$$\begin{gathered} W_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3\text{N/m}\times \left[{\left(0.050\text{m}\right)}^2-{\left(-0.050\text{m}\right)}^2\right] \\ {W}_{\text{s}}=\frac{1}{2}\times 2.0\times {10}^3\text{N/m}\times 0 \\ {W}_{\text{s}}=0\text{J} \end{gathered}$$

Thus, the work done is, $W_{\text{s}}=0\text{J}$.