Q 109P - University Physics with Modern Physics

Question

In your physics lab, a block of mass m is at rest on a horizontal surface. You attach a light cord to the block and apply a horizontal force to the free end of the cord. You find that the block remains at rest until the tension T in the cord exceeds 20.0 N. For $\mathit{T}\mathbf{>}\mathbf{20}\mathit{N}$, you measure the acceleration of the block when T is maintained at a constant value, and you plot the results (Fig. P5.109). The equation for the straight line that best fits your data is $\mathit{a}\mathbf{ }\mathbf{=}\mathbf{[}\mathbf{0}\mathbf{.}\mathbf{182}\mathbf{m}\mathbf{/}\left(N-s^2\right)\mathbf{]}\mathit{T}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{842}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. For this block and surface, what are (a) the coefficient of static friction and (b) the coefficient of kinetic friction? (c) If the experiment were done on the earth’s moon, where g is much smaller than on the earth, would the graph of a versus T still be fit well by a straight line? If so, how would the slope and intercept of the line differ from the values in Fig. P5.109? Or, would each of them be the same?

Step-by-Step Solution

Answer

a) The coefficient of static friction is 0.371.

b) The coefficient of kinetic friction is 0.290.

c) The graph will be straight line, the slope will be same, and the intercept will be smaller.

Step 1: Describe the Newton’s second law and centripetal acceleration

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, and a is acceleration of object.

 

The centripetal acceleration is given by,

 $a_c=\frac{v^2}{r}$

Here, v is velocity, and r is radius of curvature.

Step 2: Determine the coefficient of static friction

(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.

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

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

 

The equation of the line from the graph is,

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

 

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

 $\begin{array}{rcl}\frac{1}{m}& =& 0.182\mathrm{m}/(\mathrm{N}-{\mathrm{s}}^2)\\ m& =& \frac{1}{0.182\mathrm{m}/(\mathrm{N}-{\mathrm{s}}^2)}\\ & =& 5.4945 \text{kg}\end{array}$

 

The friction force of the block is given by,

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

 

Simplify further,

 ${\mu }_s=\frac{T}{mg}$

 

Substitute for , for , and  for in the above equation.

 $\begin{array}{rcl}{\mu }_s& =& \frac{20 \text{N}}{\left(5.4945 \text{kg}\right)\left(9.8 m/s^2\right)}\\ & =& 0.371\end{array}$

 

Therefore, the required coefficient of static friction is 0.371.

Step 3: Determine the coefficient of kinetic friction

(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,

 $T-{\mu }_{k}mg=ma$

Compare the equation with .

 $\begin{array}{rcl}{\mu }_{k}mg& =& 15.615\\ {\mu }_k& =& \frac{15.615}{\left(5.4945 \text{kg}\right)\left(9.8m/s^2\right)}\\ & =& 0.290\end{array}$

Therefore, the required coefficient of kinetic friction is 0.290.

Step 4: Determine the whether the graph will remain straight line in moon, and explain about the changes in slope and intercept

(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 ${\mu }_{k}g$ 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.