Question

In Fig. $\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{10}\mathit{a}$, a block of mass lies on a horizontal frictionless surface and is attached to one end of a horizontal spring (spring constant ) whose other end is fixed. The block is initially at rest at the position where the spring is unstretched $(x=0)$ when a constant horizontal force $\overrightarrow{F}$ in the positive direction of the x axis is applied to it. A plot of the resulting kinetic energy of the block versus its position x is shown in Fig.$\mathbf{7}\mathbf{}\mathbf{-}\mathbf{}\mathbf{36}$. The scale of the figure’s vertical axis is set by ${\mathit{K}}_{\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{\text{J}}$ . (a) What is the magnitude of $\overrightarrow{\mathbf{F}}\mathbf{}$? (b) What is the value of k?

Short answer

1. The magnitude of $\overrightarrow{F}$ is, $8.0 \text{N}$.
2. The value of k is, $8.0 \text{N/m}$.

Step-by-step solution

Given

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

Understanding the concept

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:

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

Calculate the magnitude of  k

The work by the spring on the block:

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

For vertical axis of$x_{\text{i}}=2 \text{m}$,$\Delta 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,

$W_{\text{s}}=\Delta K_{\text{s}}$

Hence,

$\begin{array}{rcl}\Delta K_{\text{s}}& =& \frac{1}{2}k\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)\\ k& =& \frac{2\times \Delta K_{\text{s}}}{\left(x_{\text{i}}^{\text{2}}-x_{\text{f}}^{\text{2}}\right)}\\ & & \end{array}$

Substitute all the value in the above equation.

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

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

Calculate the magnitude of  F→

According to Hooke’s law,

$F=-kx$

Substitute all the value in the above equation.

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

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