The block in Fig. 7-10a lies on a horizontal frictionless surface, and the spring constant is 50N/m. Initially, the spring is at its relaxed length and the block is stationary at position x=0. Then an applied force with a constant magnitude of 3.0 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. When that stopping point is reached, what are (a) the position of the block, (b) the work that has been done on the block by the applied force, and (c) the work that has been done on the block by the spring force? During the block’s displacement, what are (d) the block’s position when its kinetic energy is maximum and (e) the value of that maximum kinetic energy?

We can use the equation of work done bythespring and the work done by the applied force to calculate the work done by each force respectively. To calculate the position of the block at maximum kinetic energy, we can take the derivative of equation of work and set it to zero that is we can use the extreme condition.

Formula:

Given:

1. The spring constant is,$\mathrm{k}=50\mathrm{N}/\mathrm{m}$
2. The applied force is, ${\mathrm{F}}_{\mathrm{a}}=3.0\mathrm{N}$
3. The initial position of the block is, ${\mathrm{x}}_{\mathrm{i}}=0$

The equation relating the work done by the spring force and the work done by applied force is

$$\begin{gathered} {\mathrm{W}}_{\mathrm{a}}=-{\mathrm{W}}_{\mathrm{s}} \\ {\mathrm{F}}_{\mathrm{a}}.\mathrm{x}=\frac{1}{2}{\mathrm{kx}}^2 \\ \mathrm{x}=\frac{2{\mathrm{F}}_{\mathrm{a}}}{\mathrm{k}} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} \mathrm{x}=\frac{2\times 3.0\mathrm{N}}{50\mathrm{N}/\mathrm{m}} \\ \mathrm{x}=0.12\mathrm{m} \end{gathered}$$$$\begin{gathered} \mathrm{x}=\frac{2\times 3.0\mathrm{N}}{50\mathrm{N}/\mathrm{m}} \\ \mathrm{x}=0.12\mathrm{m} \end{gathered}$$

Therefore, the block is at $\mathrm{x}=0.12\mathrm{m}$.

The equation for the work done by the applied force is

${\mathrm{W}}_{\mathrm{a}}={\mathrm{F}}_{\mathrm{a}}.\mathrm{x}$

Substitute all the value in the above equation.

$$\begin{gathered} {\mathrm{W}}_{\mathrm{a}}=3.0\mathrm{N}\times 0.12\mathrm{m} \\ {\mathrm{W}}_{\mathrm{a}}=0.36\mathrm{J} \end{gathered}$$

Therefore, work done by applied force is $0.36\mathrm{J}$.

We have,

${\mathrm{W}}_{\mathrm{a}}=-{\mathrm{W}}_{\mathrm{s}}$

So,

${\mathrm{W}}_{\mathrm{s}}=-0.36\mathrm{J}$

Therefore, work done by the spring force is, $-0.36\mathrm{J}$.

Let us assume that ${\mathrm{x}}_{\mathrm{c}}$is the position at which kinetic energy is maximum

So,

$\frac{d\mathrm{K}}{d\mathrm{x}}\left(\mathrm{at}\mathrm{x}={\mathrm{x}}_{\mathrm{c}}\right)=0\left(\mathrm{Condition}\mathrm{of}\mathrm{maxima}\right)$

The work done is the change in kinetic energy.

So,

$\begin{array}{rcl}\mathrm{W}& =& {\mathrm{W}}_{\mathrm{a}}-{\mathrm{W}}_{\mathrm{s}}\\ & =& {\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}\end{array}$

As, the block is at rest initially,${\mathrm{K}}_{\mathrm{i}}=0$Hence, we get

$\begin{array}{rcl}\mathrm{K}& =& \mathrm{F}.\mathrm{x}=\frac{1}{2}{\mathrm{kx}}^2\\ \frac{d\mathrm{K}}{d\mathrm{x}}& =& \mathrm{F}-\frac{1}{2}\left(2\mathrm{kx}\right)\\ 0& =& \mathrm{F}-\frac{1}{2}\left(2\mathrm{kx}\right)\\ {\mathrm{x}}_{\mathrm{c}}& =& \frac{\mathrm{F}}{\mathrm{k}}\end{array}$

Substituting this equation in the above condition of maxima, we get

$\begin{array}{rcl}{\mathrm{x}}_{\mathrm{c}}& =& \frac{\mathrm{F}}{\mathrm{k}}\\ & =& \frac{3.0\mathrm{N}}{50\mathrm{N}/\mathrm{m}}\\ {\mathrm{x}}_{\mathrm{c}}& =& 0.06\mathrm{m}\end{array}$

Therefore, kinetic energy is the maximum at ${\mathrm{x}}_{\mathrm{c}}=0.06\mathrm{m}$.

Now, using ${\mathrm{x}}_{\mathrm{c}}=0.06\mathrm{m}$in the equation of kinetic energy, we get

$$\begin{gathered} {\mathrm{K}}_{\max }=3.0\mathrm{N}\times 0.06\mathrm{m}-\frac{1}{2}\times 50\mathrm{N}/\mathrm{m}\times {\left(0.06\mathrm{m}\right)}^2 \\ {\mathrm{K}}_{\max }=0.09\mathrm{J} \end{gathered}$$

Therefore, the value of maximum kinetic energy is $0.09\mathrm{J}$