91Ó°ÊÓ

Mass of the block,$\mathrm{m}=4.10\mathrm{kg}$

Force,$\mathrm{F}=40.0\mathrm{N}$

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. To initiate, the acceleration of the block can be found from the graph and then the free body diagram for the block can be drawn. By using Newton's 2nd law of motion, coefficient of friction between the table and box can be found.

Formula:

$F_{net}=∑ma$

where, F is the net force, mis mass and a is an acceleration.

Free Body Diagram of the system:

Acceleration of the block from graph,

$$\begin{gathered} \mathrm{a}=\frac{∆\mathrm{v}}{∆\mathrm{t}} \\ =\frac{(5.0)-(0.5)}{(1.0)-\left(0\right)} \\ =4.5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

By using Newton’s 2nd law of motion along the horizontal direction,

$$\begin{gathered} F-f_k=ma \\ {f}_k=F-ma \\ {f}_k=40.0-(4.10)(4.5) \\ =21.55N\left(i\right) \end{gathered}$$

By using Newton’s 2nd law of motion along the vertical direction,

$$\begin{gathered} N+mg=0 \\ N=mg \end{gathered}$$

Thus, the frictional force,

$$\begin{gathered} f_k={\mathrm{μ}}_{s}N \\ ={\mathrm{μ}}_k\left(mg\right) \end{gathered}$$ (ii)

From Eq. (1) and (2),

$21.55={\mathrm{μ}}_k\left(mg\right)$

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{k}}=\frac{21.55}{\mathrm{mg}} \\ =\frac{21.55}{(4.10)(9.81)} \\ =0.53 \end{gathered}$$

Hence, the coefficient of kinetic friction between the block and the floor is 0.53.