Chapter 6, Problem 133P

(a)

To determine

The velocity of the blocks moving after they travel $2.0\text{m}$ along the inclines.

Answer to Problem 133P

The velocity of the blocks moving after they travel $2.0\text{m}$ along the inclines is $\underset{\_}{1.7\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the distance travelled by the first block.

    $d_1=d\sin \theta$        (I)

Here, $d_1$ is the distance travelled by the first block, $d$ is the distance along the incline, $\theta$ is the angle making the mass $m_1$ between the horizontal and the inclined plane.

Write the expression for the distance travelled by the second block.

    $d_2=d\sin \varphi$        (II)

Here, $d_2$ is the distance travelled by the second block, $\varphi$ is the angle making the mass $m_2$ between the horizontal and the inclined plane.

Since the mass $m_1$ is greater than the mass $m_2$, the blocks slides along the incline to the right.

Write the expression for the conservation law of energy.

    $\begin{array}{c}\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2=m_{1}g{d}_1-m_{2}g{d}_2\\ \frac{1}{2}\left(m_1+m_2\right)v^2=m_{1}g{d}_1-m_{2}g{d}_2\end{array}$        (III)

Here, $v$ is the velocity of the blocks, $g$ is the acceleration due to gravity.

Use equation (I) and (II) in (III) to solve for $v$.

    $v=\sqrt{\frac{2\left[m_{1}gd\sin \theta -m_{2}gd\sin \varphi \right]}{m_1+m_2}}$        (IV)

Conclusion:

Substitute $6.00\text{kg}$ for $m_1$, $4.00\text{kg}$ for $m_2$, $9.8\text{m}/{\text{s}}^2$ for $g$, $36.9°$ for $\theta$, $45°$ for $\varphi$, $2.00\text{m}$ for $d$ in equation (IV) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{2\left[\left(6.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(2.00\text{m}\right)\sin 36.9°-\left(4.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(2.00\text{m}\right)\sin 45°\right]}{\left(6.00\text{kg}+4.00\text{kg}\right)}}\\ =\sqrt{\frac{5.51\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-2}}{10\text{kg}}}\\ =1.7\text{m}/\text{s}\end{array}$

Therefore, the velocity of the blocks moving after they travel $2.0\text{m}$ along the inclines is $\underset{\_}{1.7\text{m}/\text{s}}$.

(b)

To determine

The acceleration and velocity of each of the blocks.

Answer to Problem 133P

The acceleration of each of the blocks is $\underset{\_}{0.76\text{m}/{\text{s}}^2}$ and velocity of each of the blocks is $\underset{\_}{1.7\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for the Newton’s second law of motion for the mass $m_2$.

    $\begin{array}{c}{{\displaystyle \sum F}}_{\left(m_2\right)}=m_{2}a\\ T-m_{2}g\sin \varphi =m_{2}a\end{array}$        (V)

Here, ${\displaystyle \sum F_{\left(m_2\right)}}$ is the sum of forces acting on the mass $m_2$, $a$ is the acceleration of each blocks, $T$ is the tension of the rope.

Write the expression for the Newton’s second law of motion for the mass $m_1$.

    $\begin{array}{c}{{\displaystyle \sum F}}_{\left(m_1\right)}=m_{1}a\\ m_{1}g\sin \theta -T=m_{1}a\end{array}$        (VI)

Here, ${\displaystyle \sum F_{\left(m_1\right)}}$ is the sum of forces acting on mass $m_1$.

Use equation (VI) to solve for $T$.

    $T=m_{2}g\sin \theta -m_{1}a$        (VII)

Use equation (VII) in (V) to solve for $a$.

    $\begin{array}{c}{m}_{1}g\sin \theta -m_{1}a-m_{2}g\sin \varphi =m_{2}a\\ m_{1}g\sin \theta -m_{2}g\sin \varphi =m_{2}a+m_{1}a\\ a=\frac{g\left(m_1\sin \theta -m_2\sin \varphi \right)}{m_1+m_2}\end{array}$        (VIII)

Initially the blocks are kept stationary and the initial velocity is zero.

Write the equation of motion to solve for the $v$.

    $\begin{array}{c}{v}^2-u^2=2as\\ v=\sqrt{2as}\end{array}$        (IX)

Here, $v$ is the final velocity, $u$ is the initial velocity, $s$ is the distance.

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^2$ for $g$, $6.00\text{kg}$ for $m_1$, $36.9°$ for $\theta$, $4.00\text{kg}$ for $m_2$, $45°$ for $\varphi$ in equation (IX) to find $a$.

    $\begin{array}{c}a=\frac{\left(9.8\text{m}/{\text{s}}^2\right)\left[\left(6.00\text{kg}\right)\sin 36.9°-\left(4.00\text{kg}\right)\sin 45°\right]}{\left(6.00\text{kg}+4.00\text{kg}\right)}\\ =0.76\text{m}/{\text{s}}^2\end{array}$

Substitute $0.76\text{m}/{\text{s}}^2$ for $a$, $2.00\text{m}$ for $s$ in equation (IX) to find $v$.

    $\begin{array}{c}v=\sqrt{2\left(0.76\text{m}/{\text{s}}^2\right)\left(2.00\text{m}\right)}\\ =1.7\text{m}/\text{s}\end{array}$

Therefore, the acceleration of each of the blocks is $\underset{\_}{0.76\text{m}/{\text{s}}^2}$ and velocity of each of the blocks is $\underset{\_}{1.7\text{m}/\text{s}}$.