A spring and block are in the arrangement of Fig. $\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{10}$.When the block is pulled out to $\mathit{x}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$, we must apply a force of magnitude $\mathbf{360}\mathbf{}\mathit{N}\mathbf{}$ to hold it there.We pull the block to $\mathit{x}\mathbf{=}\mathbf{11}\mathbf{}\mathit{c}\mathit{m}\mathbf{}$and then release it. How much work does the spring do on the block as the block moves from ${\mathit{x}}_{\mathbf{i}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}\mathbf{}$to (a)$\mathit{x}\mathbf{=}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$, (b) $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$, (c)$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$ ,and (d)$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$ ?

The magnitude of the force is,$F=360\text{N}$.

The block is pulled out to $x=+4.0\text{cm}=+0.040\text{m}$.

The initial position of the block is $x_i=+5.0\text{cm}=0.050\text{m}$.

The final positions of the block are,

$$\begin{gathered} x_f=+3.0\text{cm}=+0.030\text{m} \\ {x}_f=-3.0\text{cm}=-0.030m \\ {x}_f=-5.0\text{cm}=-0.050\text{m} \\ {x}_f=-9.0\text{cm}=-0.090\text{m} \end{gathered}$$

The problem deals with the concept of Hooke’s law. It states that for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load. The block is connected to the spring; hence, we can use the concept of the work by the spring on the block.

According to Hooke’s law, the force constant can be calculated as,

$$\begin{gathered} F=-kx \\ k=\frac{F}{x} \end{gathered}$$

Here F is the applied force on the spring,and x is the displacement caused by it.

Substitute the values in the above expression, and we get,

$$\begin{gathered} k=-\frac{360\text{N}}{0.040\text{m}} \\ k=\text{9.0}\times {\text{10}}^{\text{3}}\text{N/m} \end{gathered}$$

The work done on the spring can be calculated as,

$W_s=\frac{1}{2}k\left(x_i^2-x_f^2\right)$ (1)

When the block moves from $x_i=0.050\text{m}$to $x_{\text{f}}=+0.030\text{m}$, the work by the spring can be calculated as,

$\begin{array}{rcl}{W}_{\text{s}}& =& \frac{1}{2}\times 9.0\times {10}^3\text{N/m}\times \left[{\left(0.050\text{m}\right)}^2-{\left(0.030\text{m}\right)}^2\right]\\ & =& \frac{1}{2}\times 9.0\times {10}^3\times \text{1.6}\times {\text{10}}^{-3}·\left(1\text{N/m}\times 1{\text{m}}^2\times \frac{1\text{J}}{1\text{N}·\text{m}}\right)\\ W_{\text{s}}& =& 7.2\text{J}\\ & & \end{array}$

Thus, work done is, $W_s=7.2\text{J}$

When the block moves from $x_i=0.050\text{m}$ to $x_f=-0.030\text{m}$, from equation 1, the work by the spring can be calculated as,

$\begin{array}{rcl}{W}_{\text{s}}& =& \frac{1}{2}\times 9.0\times {10}^3\text{N}\times \left[{\left(0.050\text{m}\right)}^2-{\left(-0.030\text{m}\right)}^2\right]\\ & =& \frac{1}{2}\times 9.0\times {10}^3\times \text{1.6}\times {\text{10}}^{-3}·\left(1\text{N/m}\times 1{\text{m}}^2\times \frac{1\text{J}}{1\text{N}·\text{m}}\right)\\ W_{\text{s}}& =& 7.2\text{J}\\ & & \end{array}$

Thus, work done is,$W_{\text{s}}=7.2\text{J}$

When the block moves from $x_i=0.050\text{m}$ to$x_f=-0.050\text{m}$ , from equation 1, the work by the spring can be calculated as,

$\begin{array}{rcl}{W}_{\text{s}}& =& \frac{1}{2}\times 9.0\times {10}^3\text{N}\times \left[{\left(0.050\text{m}\right)}^2-{\left(-0.050\text{m}\right)}^2\right]\\ & =& \frac{1}{2}\times 9.0\times {10}^3\times 0·\left(1\text{N/m}\times 1{\text{m}}^2\times \frac{1\text{J}}{1\text{N}·\text{m}}\right)\\ & =& 0\\ & & \end{array}$

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

When the block moves from $x_i=0.050\text{m}$to$x_f=-0.090\text{m}$ , from equation 1, the work by the spring can be calculated as,

$\begin{array}{rcl}{W}_{\text{s}}& =& \frac{1}{2}\times 9.0\times {10}^3\text{N}\times \left[{\left(0.050\text{m}\right)}^2-{\left(-0.090\text{m}\right)}^2\right]\\ & =& -\frac{1}{2}\times 9.0\times {10}^3\times \text{5.6}\times {\text{10}}^{-3}·\left(1\text{N/m}\times 1{\text{m}}^2\times \frac{1\text{J}}{1\text{N}·\text{m}}\right)\\ W_{\text{s}}& =& -25.2\text{J}\\ & & \end{array}$

Thus, work done is, $W_{\text{s}}=-25.2\text{J}$.