Question 8.4: A Block Pulled on a Rough Surface A 6.0-kg block initially a......

(A) Conceptualize This example is Example 7.6 (page 178), modified so that the surface is no longer frictionless. The rough surface applies a friction force on the block opposite to the applied force. As a result, we expect the speed to be lower than that found in Example 7.6.

Categorize The block is pulled by a force and the surface is rough, so we model the block-surface system as nonisolated with a nonconservative force acting.

Analyze Active Figure 8.8a illustrates this situation. Neither the normal force nor the gravitational force does work on the system because their points of application are displaced horizontally. Find the work done on the system by the applied force just as in Example 7.6:

\sum W_{\text {other forces }}=W_{F}=F \Delta x

Apply the particle in equilibrium model to the block in the vertical direction:

\sum F_{y}=0 \rightarrow n-m g=0 \rightarrow n=m g

Find the magnitude of the friction force:

f_{k}=\mu_{k} n=\mu_{k} m g=(0.15)(6.0 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)=8.82 \mathrm{~N}

Find the final speed of the block from Equation 8.14:

K_{f}=K_{i}-f_{k}d+\sum W_{\text{other forces }}      (8.14)

\begin{aligned}& \frac{1}{2} m v_{f}^{2}=\frac{1}{2} m v_{i}^{2}-f_{k} d+W_{F} \\& v_{f}=\sqrt{v_{i}^{2}+\frac{2}{m}\left(-f_{k} d+F \Delta x\right)}\end{aligned}

Substitute numerical values:

Finalize As expected, this value is less than the 3.5 \mathrm{~m} / \mathrm{s} found in the case of the block sliding on a frictionless surface (see Example 7.6). The difference in kinetic energies between the block in Example 7.6 and the block in this example is equal to the increase in internal energy of the block-surface system in this example.

(B) Conceptualize You might guess that \theta=0 would give the largest speed because the force would have the largest component possible in the direction parallel to the surface. Think about \overrightarrow{\mathbf{F}} applied at an arbitrary nonzero angle, however. Although the horizontal component of the force would be reduced, the vertical component of the force would reduce the normal force, in turn reducing the force of friction, which suggests that the speed could be maximized by pulling at an angle other than \theta=0

Categorize As in part (A), we model the block-surface system as nonisolated with a nonconservative force acting.

Analyze Find the work done by the applied force, noting that \Delta x=d because the path followed by the block is a straight line:

\sum W_{\text {other forces }}=W_{F}=F \Delta x \cos \theta=F d \cos \theta

Apply the particle in equilibrium model to the block in the vertical direction:

\sum F_{y}=n+F \sin \theta-m g=0

Solve for n :

n=m g-F \sin \theta

Use Equation 8.14 to find the final kinetic energy for this situation:

Maximizing the speed is equivalent to maximizing the final kinetic energy. Consequently, differentiate K_{f} with respect to \theta and set the result equal to zero:

\begin{aligned} \frac{d K_{f}}{d \theta}=-\mu_{k}(0-F \cos \theta) d-F d \sin \theta & =0 \\ \mu_{k} \cos \theta-\sin \theta & =0 \\ \tan \theta & =\mu_{k} \end{aligned}

Evaluate \theta for \mu_{k}=0.15 :

\theta=\tan ^{-1}\left(\mu_{k}\right)=\tan ^{-1}(0.15)=8.5^{\circ}

Finalize Notice that the angle at which the speed of the block is a maximum is indeed not \theta=0. When the angle exceeds 8.5^{\circ}, the horizontal component of the applied force is too small to be compensated by the reduced friction force and the speed of the block begins to decrease from its maximum value.