91Ó°ÊÓ

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

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

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,

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

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,

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

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,

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

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