91Ó°ÊÓ

The vertical axis is set by kinetic energy as $K_{\text{s}}=4.0\text{J}$.

The block is connected to the spring; hence, we can use the concept of the work by the spring on the block. We can use the work-kinetic energy theorem.

Formula:

$$\begin{gathered} F=-kx \\ {W}_{\text{s}}=\frac{1}{2}k{x}_{\text{i}}^{\text{2}}-\frac{1}{2}k{x}_{\text{f}}^{\text{2}} \end{gathered}$$

The work by the spring on the block:

From the graph, for vertical axis of$1\text{m}$,$\mathrm{Δ}K=4.0J$

For vertical axis of$x_{\text{i}}=2 \text{m}$,$\mathrm{Δ}{K}_{\text{s}}=0\text{J}$$x_{\text{i}}=2 \text{m}$,$$" width="9">

The work by the spring is,

According to the work-kinetic energy theorem,

$W_{\text{s}}=\mathrm{Δ}{K}_{\text{s}}$

Hence,

$\begin{array}{rcl}\mathrm{Δ}{K}_{\text{s}}& =& \frac{1}{2}k\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)\\ k& =& \frac{2×\mathrm{Δ}{K}_{\text{s}}}{\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)}\\ & & \end{array}$

Substitute all the value in the above equation.

$$\begin{gathered} k=\frac{2×4.0 \text{J}}{\left[0^2-{\left(1.0 \text{m}\right)}^2\right]} \\ k=8.0 \text{N/m} \end{gathered}$$

Hence the value of k is, $8.0 \text{N/m}$.

According to Hooke’s law,

$F=-kx$

Substitute all the value in the above equation.

$$\begin{gathered} F=-8.0 \text{N/m}×1.0 \text{m} \\ F=8.0 \text{N} \end{gathered}$$

Hence the value of F is, $8.0 \text{N}$.