In Fig.$\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{10}$ , we must apply a force of magnitude to hold the block stationary at $\mathit{x}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$. From that position, we then slowly move the block so that our force does $\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathit{J}\mathbf{}$ of work on the spring–block system; the block is then again stationary. What is the block’s position? (Hint:There are two answers.)

1. The magnitude of the applied force is, $F_{\text{x}}=80\text{N}$.
2. The initial position of the block is,$x_{\text{i}}=-2.0\text{cm}=-2.0\times {10}^{-2}\text{m}$.
3. The work done by the applied force is $W_{\text{a}}=+4.0\text{J}$.

Use the concept of Hooke’s law. If a block attached to a spring is stationary before and after a displacement, then the work done on it by the applied force displacing it is negative of the work done on it by the spring force.

Formulae:

$F_{\text{x}}=-kx$

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

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

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

$$\begin{gathered} k=-\frac{80\text{N}}{-2.0\times {10}^{-2}\text{m}} \\ k=4.0\times {10}^3\text{N/m} \end{gathered}$$

The block is stationary before and after the displacement; therefore, the work done by the applied force is,

$W_{\text{a}}=-W_{\text{s}}$ (1)

The expression for the work by spring force is,

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

Substitute the value from equation 1, in the above expression, and we get,

$$\begin{gathered} W_{\text{a}}=\frac{1}{2}k\left(x_{\text{f}}^{\text{2}}-x_{\text{i}}^{\text{2}}\right) \\ {x}_{\text{f}}=\pm \sqrt{x_{\text{i}}^{\text{2}}+\frac{\text{2}{W}_{\text{a}}}{k}} \end{gathered}$$

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

$$\begin{gathered} x_{\text{f}}=\pm \sqrt{{\left(-2.0\times {10}^{-2}\text{m}\right)}^2+\frac{2\times \left(4.0\text{J}\right)}{4.0\times {10}^3\text{N/m}}} \\ {x}_{\text{f}}=\pm \sqrt{{\left(-2.0\times {10}^{-2}\text{m}\right)}^2+\frac{2\times \left(4.0\text{N}·\text{m}\right)}{4.0\times {10}^3\text{N/m}}} \\ {x}_{\text{f}}=\pm \sqrt{{\left(-2.0\times {10}^{-2}\text{m}\right)}^2+\frac{2\times \left(4.0{\text{m}}^2\right)}{4.0\times {10}^3}} \\ {x}_{\text{f}}=\pm 0.049\text{m}\backslash \end{gathered}$$

We can have two values of position, one is $+0.049\text{m}$, and the other is $-0.049\text{m}$.

Thus, the final position of the block is $x_{\text{f}}=\pm 0.049\text{m}$.