Chapter 5, Problem 55P

(a)

To determine

The time interval in which the smaller block make it to the right side of the $8.00\text{kg}$ block.

Answer to Problem 55P

The time interval in which the smaller block make it to the right side of the $8.00\text{kg}$ block is $\underset{\_}{2.13\text{s}}$.

Explanation of Solution

The free body diagram of the top block is shown in Figure 1.

The free diagram of the bottom block is shown in Figure 2.

From the Figure 1, write the expression for net force in the $x$ direction.

    ${\displaystyle \sum F_x}=F-f$        (I)

Here, $F$ is the applied horizontal force, and $f$ is the frictional force.

Apply Newton’s law in the $x$ direction.

    ${\displaystyle \sum F_{\text{x}}}=m{a}_{\text{T}}$        (II)

Here, $a_{\text{T}}$ is the acceleration of the top block, and $m$ is the mass of top block.

Equate the equation (I) and (II).

    $F-f=m{a}_{\text{T}}$        (III)

From the Figure 2, write the expression for net force in the $x$ direction.

    ${\displaystyle \sum F_x}=f$        (IV)

Apply Newton’s law in the $x$ direction.

    ${\displaystyle \sum F_x}=M{a}_{\text{B}}$        (V)

Here, $a_{\text{B}}$ is the acceleration of the bottom block, and $M$ is the mass of bottom block.

Equate the equation (IV) and (V).

    $f=M{a}_{\text{B}}$        (VI)

Write the expression for frictional force.

    $f={\mu }_{\text{k}}mg$        (VII)

Use equation (VII) in (III).

    $F-{\mu }_{\text{k}}mg=m{a}_{\text{T}}$        (VIII)

Use equation (VII) in (VI).

    ${\mu }_{\text{k}}mg=M{a}_{\text{B}}$        (IX)

Consider the Figure 3.

In the time $t$, the top block moves a distance $d_{\text{T}}=\frac{1}{2}{a}_{\text{T}}{t}^2$, and bottom block moves a distance $d_{\text{B}}=\frac{1}{2}{a}_{\text{B}}{t}^2$.

The necessary condition to reach the top block at the right edge of the bottom block is,

    $d_{\text{T}}=d_{\text{B}}+L$        (X)

Here, $d_{\text{T}}$ is the distance travelled by the top block, $d_{\text{B}}$ is the distance travelled by the bottom block, and $L$ is the leading distance of the top block with respect to the bottom block.

Use $d_{\text{T}}=\frac{1}{2}{a}_{\text{T}}{t}^2$ and $d_{\text{B}}=\frac{1}{2}{a}_{\text{B}}{t}^2$ in the equation (X).

    $\frac{1}{2}{a}_{\text{T}}{t}^2=\frac{1}{2}{a}_{\text{B}}{t}^2+L$        (XI)

Conclusion:

Substitute, $10.0\text{N}$ for $F$, $0.300$ for ${\mu }_{\text{k}}$, $2.00\text{kg}$ for $m$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (VIII), and solve for $a_{\text{T}}$.

    $\begin{array}{c}10.0\text{N}-0.300\left(2.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(2.00\text{kg}\right)a_{\text{T}}\\ a_{\text{T}}=\frac{10.0\text{N}-5.88\text{N}}{\left(2.00\text{kg}\right)}\\ =2.06\text{m}/{\text{s}}^2\end{array}$

Substitute, $0.300$ for ${\mu }_{\text{k}}$, $8.00\text{kg}$ for $M$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (IX), and solve for $a_{\text{B}}$.

    $\begin{array}{c}0.300\left(2.00\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(8.00\text{kg}\right)a_{\text{B}}\\ a_{\text{B}}=\frac{5.88\text{N}}{8.00\text{kg}}\\ =0.735\text{m}/{\text{s}}^2\end{array}$

Substitute, $0.735\text{m}/{\text{s}}^2$ for $a_{\text{B}}$, $2.06\text{m}/{\text{s}}^2$ for $a_{\text{T}}$, and $3.00\text{m}$ for $L$ in the equation (XI), and solve for $t$.

    $\begin{array}{c}\frac{1}{2}\left(2.06\text{m}/{\text{s}}^2\right)t^2=\frac{1}{2}\left(0.735\text{m}/{\text{s}}^2\right)t^2+3.00\text{m}\\ t^2\left[\frac{1}{2}\left(2.06\text{m}/{\text{s}}^2\right)-\frac{1}{2}\left(0.735\text{m}/{\text{s}}^2\right)\right]=3.00\text{m}\\ t=\frac{2\times 3.00\text{m}}{\left(2.06\text{m}/{\text{s}}^2\right)-\left(0.735\text{m}/{\text{s}}^2\right)}\\ =2.13\text{s}\end{array}$

Therefore, the time interval in which the smaller block make it to the right side of the $8.00\text{kg}$ block is $\underset{\_}{2.13\text{s}}$.

(b)

To determine

The distance in which the $8.00\text{kg}$ block move in the process.

Answer to Problem 55P

The distance in which the $8.00\text{kg}$ block move in the process is $\underset{\_}{1.67\text{m}}$.

Explanation of Solution

Conclusion:

Substitute, $0.735\text{m}/{\text{s}}^2$ for $a_{\text{B}}$, and $2.13\text{s}$ for $t$ in $d_{\text{B}}=\frac{1}{2}{a}_{\text{B}}{t}^2$.

    $\begin{array}{c}{d}_{\text{B}}=\frac{1}{2}\left(0.735\text{m}/{\text{s}}^2\right){\left(2.13\text{s}\right)}^2\\ =1.67\text{m}\end{array}$

Therefore, the distance in which the $8.00\text{kg}$ block move in the process is $\underset{\_}{1.67\text{m}}$.