Chapter 5, Problem 74PQ

Starting from rest, a rectangular toy block with mass 300 g slides in 1.30 s all the way across a table 1.20 m in length that Zak has tilted at an angle of 42.0° to the horizontal. a. What is the magnitude of the acceleration of the toy block? b. What is the coefficient of kinetic friction between the block and the table? c. What are the magnitude and direction of the friction force acting on the block? d. What is the speed of the block when it is at the end of the table, having slid a distance of 1.20 m?

To determine

Find the magnitude of acceleration of the toy block.

Answer to Problem 74PQ

The magnitude of acceleration of the toy block is $1.42{\text{ m/s}}^2$.

Explanation of Solution

Write the equation for acceleration.

  $\begin{array}{l}x=\frac{1}{2}a{t}^2\\ a=\frac{2x}{t^2}\end{array}$                                                                                 (I)

Here, $a$ is the acceleration, $x$ is the displacement and $t$ is the time.

Conclusion:

Substitute $1.20\text{ m}$ for $x$ and $1.30\text{ s}$ for $t$ in equation I.

    $\begin{array}{c}a=\frac{2\left(1.20\text{ m}\right)}{{\left(1.30\text{ s}\right)}^2}\\ =1.42{\text{ m/s}}^2\end{array}$

Therefore, the magnitude of acceleration of the toy block is $1.42{\text{ m/s}}^2$.

To determine

Find the coefficient of kinetic friction between the block and table.

Answer to Problem 74PQ

The coefficient of kinetic friction between the block and table is $0.706$.

Explanation of Solution

From the given condition the free body diagram is given below.

  $\begin{array}{l}\Sigma F_x=F_f-mg\sin \theta =ma\\ F_f=m\left(g\sin \theta -a\right)\end{array}$                                                                                (II)

Here, $F_x$ is the force, $F_f$ is the friction force, $m$ is the mass, $a$ is the acceleration and $\theta$ is the angle.

  $\begin{array}{l}\Sigma F_y=F_N-mg\cos \theta =0\\ F_N=mg\cos \theta \end{array}$                                                                       (III)

Here, $F_N$ is the normal force.

Write the equation for friction force.

    $F_f={\mu }_k{F}_N$                                                                                             (IV)

Here, ${\mu }_k$ is the coefficient of friction.

Substitute equation II and III in equation IV.

    $\begin{array}{c}{\mu }_k=\frac{m\left(g\sin \theta -a\right)}{mg\cos \theta }\\ =\tan \theta -\frac{a}{g\cos \theta }\end{array}$                                                                                (V)

Conclusion:

Substitute $42.0°$ for $\theta$, $9.81{\text{ m/s}}^2$ for $g$ and $1.42{\text{ m/s}}^2$ for $a$ in equation V.

    $\begin{array}{c}{\mu }_k=\tan \left(42.0°\right)-\frac{1.42{\text{ m/s}}^2}{\left(9.81{\text{ m/s}}^2\right)\cos \left(42.0°\right)}\\ =0.706\end{array}$

Therefore, the coefficient of kinetic friction between the block and table is $0.706$.

To determine

Find the direction and magnitude of friction acting on the block.

Answer to Problem 74PQ

The friction acting on the block is directed upward and its magnitude is $1.54\text{ N}$.

Explanation of Solution

Substitute equation III in IV.

  $F_f={\mu }_{k}mg\cos \theta$                                                                       (VI)

Conclusion:

Substitute $42.0°$ for $\theta$, $9.81{\text{ m/s}}^2$ for $g$, $1.42{\text{ m/s}}^2$ for $a$, $0.706$ for ${\mu }_k$ and $0.300\text{ kg}$ for $m$  in equation VI.

    $\begin{array}{c}{F}_f=\left(0.706\right)\left(0.300\text{ kg}\right)\left(9.81{\text{ m/s}}^2\right)\cos \left(42.0°\right)\\ =1.54\text{ N}\end{array}$

The friction force is directed up in the incline.

Therefore, the friction acting on the block is directed upward and its magnitude is $1.54\text{ N}$.

To determine

Find the speed of the block when it is at the end of the table.

Answer to Problem 74PQ

The speed of the block when it is at the end of the table is $1.85\text{ m/s}$.

Explanation of Solution

Write the equation of motion.

  $v_f^2=v_i^2+2a\Delta x$                                                                       (VII)

Here, $v_f$ is the final speed, $v_i$ is the initial speed, $\Delta x$ is the displacement and $a$ is the acceleration.

Conclusion:

Substitute $0$ for $v_i$, $1.42{\text{ m/s}}^2$ for $a$ and $\text{1}\text{.2 m}$ for $\Delta x$ in equation VII.

    $\begin{array}{c}{v}_f=\sqrt{0+2\left(1.42{\text{ m/s}}^2\right)\left(\text{1}\text{.2 m}\right)}\\ =1.85\text{ m/s}\end{array}$

Therefore, the speed of the block when it is at the end of the table is $1.85\text{ m/s}$.