Chapter 5, Problem 79PQ

Two boxes with masses m₁ = 4.00 kg and m₂ = 10.0 kg are attached by a massless cord passing over a frictionless pulley as shown in Figure P5.79. The incline is frictionless, and θ = 30.0°.

1. a. Draw a free-body diagram for each of the boxes.
2. b. What is the magnitude of the acceleration of the boxes?
3. c. What is the tension in the cord connecting the boxes?
4. d. What is the speed of each of the boxes 3.00 s after the system is released from rest?

FIGURE P5.79

To determine

Draw the free body diagram for each of the boxes.

Answer to Problem 79PQ

The free body diagram for each of the boxes is given below.

Explanation of Solution

The free body diagram each of the boxes is given below.

Conclusion:

              

To determine

Find the acceleration of the boxes.

Answer to Problem 79PQ

The acceleration of the boxes is $\text{0}{\text{.701 m/s}}^2$.

Explanation of Solution

Applying Newton’s laws,

    $\Sigma F_x=m_{2}g\sin \theta -F_T=m_{2}a$                                                                (I)

Here, $F_x$ is the force, $F_T$ is the tension force, $m_2$ is the mass of block 2, $a$ is the acceleration, $\theta$ is the angle between horizontal and incline and $g$ is the acceleration due to gravity.

    $\begin{array}{l}\Sigma F_y=F_T-m_{1}g=m_{1}a\\ F_T=m_1\left(g+a\right)\end{array}$                                                        (II)

Here, $m_1$ is the mass of block 1.

Substitute equation II in equation I.

    $m_{2}g\sin \theta -m_1\left(g+a\right)=m_{2}a$

  $\begin{array}{c}a\left(m_1+m_2\right)=g\left(m_2\sin \theta -m_1\right)\\ a=\frac{g\left(m_2\sin \theta -m_1\right)}{m_1+m_2}\end{array}$                                                            (III)

Conclusion:

Substitute $9.81{\text{ m/s}}^2$ for $g$, $\text{4}\text{.00 kg}$ for $m_1$, $\text{10}\text{.0 kg}$ for $m_2$ and $30.0°$ for $\theta$ in equation III.

    $\begin{array}{c}a=\frac{\left(9.81{\text{ m/s}}^2\right)\left[\left(\text{10}\text{.0 kg}\right)\sin \left(30.0°\right)-4.\text{00 kg}\right]}{\left(\text{4}\text{.00 kg}+\text{10}\text{.0 kg}\right)}\\ =0.701{\text{ m/s}}^2\end{array}$

Therefore, the acceleration of the boxes is $\text{0}{\text{.701 m/s}}^2$.

To determine

Find the tension in the cord connecting the boxes.

Answer to Problem 79PQ

The tension in the cord connecting the boxes is $\text{42}\text{.0 N}$. 

Explanation of Solution

Use the equation II to find the tension in the rope connecting the two sleds.

Conclusion:

Substitute $9.81{\text{ m/s}}^2$ for $g$, $\text{4}\text{.00 kg}$ for $m_1$ and $\text{0}{\text{.701 m/s}}^2$ for $a$ in equation I.

    $\begin{array}{c}{F}_T=\left(\text{4}\text{.00 kg}\right)\left[9.81{\text{ m/s}}^2+\text{0}{\text{.701 m/s}}^2\right]\\ =42.0\text{ N}\end{array}$

Therefore, the tension in the rope connecting the two sleds is $\text{50}\text{.6 N}$.

To determine

Find the speed of the boxes $3.00\text{ s}$ after the system is released from rest.

Answer to Problem 79PQ

The speed of the boxes $3.00\text{ s}$ after the system is released from rest is $\text{2}\text{.10 m/s}$. 

Explanation of Solution

As the boxes are at rest initially, write the equation of final speed.

  $v_f=at$                                                                       (IV)

Here, $v_f$ is the final speed, $t$ is the time and $a$ is the acceleration.

Conclusion:

Substitute $\text{0}{\text{.701 m/s}}^2$ for $a$ and $3.00\text{ s}$ for $t$  in equation IV.

    $\begin{array}{c}{v}_f=\left(\text{0}{\text{.701 m/s}}^2\right)\left(3.00\text{ s}\right)\\ =2.10\text{ m/s}\end{array}$

Therefore, the speed of the boxes $3.00\text{ s}$ after the system is released from rest is $\text{2}\text{.10 m/s}$.