Chapter 4, Problem 99P

(a)

To determine

The acceleration of the each blocks.

Answer to Problem 99P

The acceleration of the each blocks is $\underset{\_}{\text{2}\text{.5}{\text{m/s}}^{\text{2}}\text{up}\text{and}\text{2}\text{.5}{\text{m/s}}^{\text{2}}\text{down}}$.

Explanation of Solution

In this case, consider the $y-$ directions for the two blocks and each in the direction of acceleration.

For the lighter block,

Write the expression for theforce act on thelighter block,

    $F_1=+T-m_{1}g$        (I)

Here, $F_1$ is the force act on the lighter block, $m_1$ is the mass of the lighter block, $T$ is the tension force and $g$ is the acceleration due to gravity.

Substitute $m_{1}a$ for $F_1$ in the above equation,

    $\begin{array}{c}{m}_{1}a=T-m_{1}g\\ T=m_{1}a+m_{1}g\end{array}$        (II)

Here, $a$ is the acceleration of the block.

For the heavier block,

Write the expression for the force act on the heavier block,

    $F_2=T-m_{2}g$        (III)

Here, $F_2$ is the force act on the heavier block, $m_2$ is the mass of the heavier block, $T$ is the tension force and $g$ is the acceleration due to gravity.

Substitute $m_{2}a$ for $F_2$ in the above equation,

    $\begin{array}{c}{m}_{2}a=m_{2}g-T\\ T=-m_{2}a+m_{2}g\end{array}$        (IV)

Here, $a$ is the acceleration of the block.

Add equations (II) and (IV),

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

Conclusion:

Substitute $\text{3}\text{.0}\text{kg}$ for $m_1$ , $\text{5}\text{.0}\text{kg}$ for $m_2$ and $\text{9}\text{.8}{\text{m/s}}^{\text{2}}$ for $g$ in the above equation,

    $\begin{array}{c}a=\frac{\left(\text{5}\text{.0}\text{kg}-\text{3}\text{.0}\text{kg}\right)\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)}{\text{3}\text{.0}\text{kg}+5.\text{0}\text{kg}}\\ =2.5{\text{m/s}}^{\text{2}}\end{array}$

The acceleration of each block, ${\overrightarrow{\text{a}}}_1=\text{2}\text{.5}{\text{m/s}}^{\text{2}}\text{up}$ and ${\overrightarrow{\text{a}}}_2=\text{2}\text{.5}{\text{m/s}}^{\text{2}}\text{down}$.

Therefore, theacceleration of the each blocks is $\underset{\_}{\text{2}\text{.5}{\text{m/s}}^{\text{2}}\text{up}\text{and}\text{2}\text{.5}{\text{m/s}}^{\text{2}}\text{down}}$.

(b)

To determine

Thetension in the cord.

Answer to Problem 99P

The tension in the cord is $\underset{\_}{37\text{N}}$.

Explanation of Solution

Write the expression for the tension in the cord,

    $T=m_1\left(a+g\right)$        (VI)

Here, $T$ is the tension in the cord.

Substitute (V) in (VI),

    $\begin{array}{c}T=m_1\left(g+\frac{\left(m_2-m_1\right)g}{m_1+m_2}\right)\\ =\frac{2m_1{m}_2}{m_1+m_2}g\end{array}$        (VII)

Conclusion:

Substitute $\text{3}\text{.0}\text{kg}$ for $m_1$ , $\text{5}\text{.0}\text{kg}$ for $m_2$ and $\text{9}\text{.8}{\text{m/s}}^{\text{2}}$ for $g$ in the above equation,

    $\begin{array}{c}T=\frac{2\left(\text{3}\text{.0}\text{kg}\right)\left(\text{5}\text{.0}\text{kg}\right)}{\text{3}\text{.0}\text{kg}+5.\text{0}\text{kg}}\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)\\ =37\text{N}\end{array}$

Therefore, thetension in the cord is $\underset{\_}{37\text{N}}$.