Chapter 4, Problem 92P

(a)

To determine

The acceleration of each blocks and tension in the cord.

Answer to Problem 92P

The acceleration of each blocks is $\underset{\_}{\frac{\left(m_{2}g-{\mu }_k{m}_{1}g\right)}{\left(m_1+m_2\right)}}$ and tension in the cord is $\underset{\_}{m_{2}g-m_{2}a}$.

Explanation of Solution

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

    $T-f_k=m_{1}a$        (I)

Here, $T$ is the tension in the cord, $f_k$ is the frictional force, $m_1$ is the mass of block $1$, $a$ is the acceleration of each block.

Write the expression for the $f_k$.

    $\begin{array}{c}{f}_k={\mu }_{k}N\\ ={\mu }_k{m}_{1}g\end{array}$        (II)

Here, ${\mu }_k$ is the coefficient of kinetic friction, $g$ is the acceleration due to gravity.

Use equation (II) in (I) to solve for $T$.

    $T=m_{1}a+{\mu }_k{m}_{1}g$        (III)

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

    $m_{2}g-T=m_{2}a$        (IV)

Here, $m_2$ is the mass of block $2$.

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

    $T=m_{2}g-m_{2}a$        (V)

Equate equations (III) and (IV) to solve $a$.

    $a=\frac{\left(m_{2}g-{\mu }_k{m}_{1}g\right)}{\left(m_1+m_2\right)}$        (VI)

Conclusion:

Therefore, the acceleration of each blocks is $\underset{\_}{\frac{\left(m_{2}g-{\mu }_k{m}_{1}g\right)}{\left(m_1+m_2\right)}}$ and tension in the cord is $\underset{\_}{m_{2}g-m_{2}a}$.

(b)

To determine

The acceleration and tension in the cord for $m_1\ll m_2$, $m_1\gg m_2$ and $m_1=m_2$.

Answer to Problem 92P

The acceleration for $m_1\ll m_2$ is $\underset{\_}{g}$, $m_1\gg m_2$ is $\underset{\_}{-{\mu }_{k}g}$ and $m_1=m_2$ is $\underset{\_}{\frac{g}{2}\left(1-{\mu }_k\right)}$ and the tension in the cord for $m_1\ll m_2$ is $\underset{\_}{0}$, $m_1\gg m_2$ is $\underset{\_}{m_{2}g\left(1+{\mu }_k\right)}$ and $m_1=m_2$ is $\underset{\_}{\frac{mg}{2}\left(1+{\mu }_k\right)}$.

Explanation of Solution

Use equation (VI) to solve for acceleration for $m_1\ll m_2$.

    $\begin{array}{c}a=\frac{\left(m_{2}g-{\mu }_k{m}_{1}g\right)}{\left(m_1+m_2\right)}\\ \approx \frac{m_{2}g}{m_2}\\ =g\end{array}$        (VII)

Use equation (VI) to solve for acceleration for $m_1\gg m_2$.

    $\begin{array}{c}a=\frac{\left(m_{2}g-{\mu }_k{m}_{1}g\right)}{\left(m_1+m_2\right)}\\ \approx \frac{-{\mu }_k{m}_{1}g}{m_1}\\ =-{\mu }_{k}g\end{array}$        (VIII)

Use equation (VI) to solve for acceleration for $m_1=m_2$.

    $\begin{array}{c}a=\frac{\left(m_{2}g-{\mu }_k{m}_{1}g\right)}{\left(m_1+m_2\right)}\\ =\frac{mg\left(1-{\mu }_k\right)}{2m}\\ =\frac{g}{2}\left(1-{\mu }_k\right)\end{array}$        (IX)

Use equation (V) to solve for the tension in the cord for $m_1\ll m_2$.

    $\begin{array}{c}T=m_{2}g-m_{2}a\\ =m_{2}g-m_{2}g\\ =0\end{array}$        (X)

Use equation (V) to solve for the tension in the cord for $m_1\gg m_2$.

    $\begin{array}{c}T=m_{2}g-m_{2}a\\ =m_{2}g-m_2\left(-{\mu }_{k}g\right)\\ =m_{2}g\left(1+{\mu }_k\right)\end{array}$        (XI)

Use equation (V) to solve for the tension in the cord for $m_1=m_2$.

    $\begin{array}{c}T=m_{2}g-m_{2}a\\ =m_{2}g-m_2\left[\frac{g}{2}\left(1-{\mu }_k\right)\right]\\ =\frac{m_{2}g}{2}+\frac{{\mu }_k{m}_{2}g}{2}\\ =\frac{mg}{2}\left(1+{\mu }_k\right)\end{array}$        (XII)

Conclusion:

Therefore, the acceleration for $m_1\ll m_2$ is $\underset{\_}{g}$, $m_1\gg m_2$ is $\underset{\_}{-{\mu }_{k}g}$ and $m_1=m_2$ is $\underset{\_}{\frac{g}{2}\left(1-{\mu }_k\right)}$ and the tension in the cord for $m_1\ll m_2$ is $\underset{\_}{0}$, $m_1\gg m_2$ is $\underset{\_}{m_{2}g\left(1+{\mu }_k\right)}$ and $m_1=m_2$ is $\underset{\_}{\frac{mg}{2}\left(1+{\mu }_k\right)}$.

(c)

To determine

The tension in the two blocks if the blocks slides at a constant velocity.

Answer to Problem 92P

The tension in the two blocks if the blocks slides at a constant velocity is $\underset{\_}{m_{2}g}$.

Explanation of Solution

Use equation (V) to solve for the tension in the cord for the block sliding at a constant velocity.

    $\begin{array}{c}T=m_{2}g-m_2\left(0\right)\\ =m_{2}g\end{array}$        (XIII)

Conclusion:

Therefore, the tension in the two blocks if the blocks slides at a constant velocity is $\underset{\_}{m_{2}g}$.