91Ó°ÊÓ

According to Newton’s second law, the linear force is given by,

F =ma

Here,F is linear force,m is the mass of object, anda is acceleration of object.

The centripetal acceleration is given by,

${\mathbf{a}}_{\mathbf{c}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$

Here,v is velocity, andr is radius of curvature.

(a)

Draw the free-body diagram of the block.

Initially when the block is at rest, it remains under the static friction. When the kinetic friction will act on the block then the block starts accelerating.

The equation for the block is as follows.

$$\begin{gathered} T-{\mathrm{μ}}_{k}mg=ma \\ a=\left(\frac{1}{m}\right)T-{\mathrm{μ}}_{k}g \end{gathered}$$

From the above equation the slope is $\frac{1}{m}$.

The equation of the line from the graph is,

$\mathrm{a}=[0.182 \mathrm{m}/(\mathrm{N}â‹\dots {\mathrm{s}}^2\left)\right]\mathrm{T}-2.842 \mathrm{m}/{\mathrm{s}}^2$

Compare the equation$\mathrm{a}=[0.182 \mathrm{m}/(\mathrm{N}â‹\dots {\mathrm{s}}^2\left)\right]\mathrm{T}-2.842 \mathrm{m}/{\mathrm{s}}^2$with $a=\left(\frac{1}{m}\right)T-{\mathrm{μ}}_{k}g$.

$$\begin{gathered} \frac{1}{m}=0.182m/\left(N.s^2\right) \\ m=\frac{1}{0.182m/\left(N.s^2\right)} \\ =5.4945kg \end{gathered}$$

The friction force of the block is given by,

$$\begin{gathered} T-f_s=0 \\ {f}_s=T \\ {\mathrm{μ}}_{s}N=T \\ {\mathrm{μ}}_{s}mg=T \end{gathered}$$

Simplify further,

${\mathrm{μ}}_s=\frac{T}{mg}$

Substitute 20 N for T, 5.4945 kg for ,m and $9.8\mathrm{m}/{\mathrm{s}}^2$forg in the above equation.

$$\begin{gathered} {\mathrm{μ}}_s=\frac{20N}{\left(5.4945kg\right)\left(9.8m/s^2\right)} \\ =0.371 \end{gathered}$$

Therefore, the required coefficient of static friction is 0.371.

(b)

The equation of acceleration versus tension graph is given as,

$$\begin{gathered} 0.182T-2.842=a \\ T-15.615=5.43a \end{gathered}$$

The condition of dynamic equilibrium is,

$T-{\mathrm{μ}}_{k}mg=ma$

Compare the equation$T-{\mathrm{μ}}_{k}mg=ma$ with $T-15.615=5.43a$.

$$\begin{gathered} {\mathrm{μ}}_{k}mg=15.615 \\ {\mathrm{μ}}_k=\frac{15.615N}{\left(5.4945kg\right)\left(9.8m/s^2\right)} \\ =0.290 \end{gathered}$$

Therefore, the required coefficient of kinetic friction is 0.290 .

(c)

Yes, it will remain a straight line because there is no graph dependency. If the experiment were conducted on the moon's surface, the friction opposing the motion would be less since the weight would become smaller.

But the slope of the graph drawn will be again reciprocal to the mass of the block. Even though weight varies, the mass remains constant. Thus, there would be no variation in the slope of the graph drawn.

However, the intercept${\mathrm{μ}}_{k}g$ depends on the static frictional force. Since it becomes smaller, the force required for the initial move would be lesser; hence, the graph would meet the x-axis at a lower value of tension.