Chapter 5, Problem 5.103CP

A block of mass m = 2.00 kg rests on the left edge of a block of mass M = 8.00 kg. The coefficient of kinetic friction between the two blocks is 0.300, and the surface on which the 8.00-kg block rests is frictionless. A constant horizontal force of magnitude F = 10.0 N is applied to the 2.00-kg block, setting it in motion as shown in Figure P5.103a. If the distance L that the leading edge of the smaller block, travels on the larger block is 3.00 m. (a) in what lime interval will the smaller block make it to the right side of the 8.00-kg block as shown in Figure P5.103b? (Note: Both blocks are set into motion when $\stackrel{\to }{F}$ is applied.) (b) How far does the 8.00-kg block move in the process?

To determine

The time interval small block make it to the right of the larger block.

Answer to Problem 5.103CP

The time interval small block make it to the right of the larger block is $2.13\text{s}$.

Explanation of Solution

Given info: The mass of the smaller block is $2.00\text{kg}$, the mass of the larger block is $8.00\text{kg}$, the coefficient of kinetic friction between the smaller and the larger block is $0.300$, the horizontal force applied on the smaller block is $10.0\text{N}$ and the distance the leading edge of the smaller block travels on the larger block is $3.00\text{m}$.

The acceleration due to gravity is $9.8{\text{m}/\text{s}}^2$.

The net force in $y$ direction is,

$\begin{array}{c}{N}_1-mg=0\\ N_1=mg\end{array}$

Here,

$N_1$ is the normal force acting on the smaller block.

$m$ is the mass of the smaller block.

$g$ is the acceleration due to gravity.

The frictional force between the smaller block and the larger block is,

$f_{\text{k}}={\mu }_{\text{k}}{N}_1$

Here,

${\mu }_{\text{k}}$ is the coefficient of kinetic friction.

Substitute $mg$ for $N_1$ in the above equation.

$f_{\text{k}}={\mu }_{\text{k}}mg$

The net force in $x$ direction is,

$F_1-f_{\text{k}}=m{a}_1$

Here,

$f_{\text{k}}$ is the frictional force between the smaller and the larger block.

$F_1$ is the applied horizontal force.

$a_1$ is the acceleration of the smaller block.

Substitute ${\mu }_{\text{k}}mg$ for $f_{\text{k}}$ in the above equation.

$F_1-{\mu }_{\text{k}}mg=m{a}_1$

Substitute $10.0\text{N}$ for $F_1$, $0.300$ for ${\mu }_{\text{k}}$, $2.00\text{kg}$ for $m$, $9.8{\text{m}/\text{s}}^2$ for $g$ in the above equation.

$\begin{array}{c}\left(2.00\text{kg}\right)a_1=10.0\text{N}-0.300\left(2.00\text{kg}\right)\left(9.8{\text{m}/\text{s}}^2\right)\\ =4.12\text{N}\\ a_1=\frac{4.12\text{N}}{2.00\text{kg}}\\ =2.06{\text{m}/\text{s}}^2\end{array}$

Thus, the acceleration of the smaller block is $2.06{\text{m}/\text{s}}^2$.

For the larger block,

The net force in $x$ direction is,

$F_2-M{a}_2=0$

The force $F_2$ acting on the larger block in the $x$ direction is the frictional force $f_{\text{k}}$ between the blocks.

Substitute $f_{\text{k}}$ for $F_2$ in the above equation.

$f_{\text{k}}-M{a}_2=0$

Substitute ${\mu }_{\text{k}}mg$ for $f_{\text{k}}$ in the above equation.

${\mu }_{\text{k}}mg-M{a}_2=0$

Substitute $0.300$ for ${\mu }_{\text{k}}$, $2.00\text{kg}$ for $m$, $9.8{\text{m}/\text{s}}^2$ for $g$ and $8.00\text{kg}$ in the above equation.

$\begin{array}{c}0.300\left(2.00kg\right)\left(9.8{\text{m}/\text{s}}^2\right)-\left(8.00\text{kg}\right)a_2=0\\ 5.88{\text{N=8a}}_2\\ a_2=0.735{\text{m}/\text{s}}^2\end{array}$

Thus, the acceleration of the larger block is $0.735{\text{m}/\text{s}}^2$.

Consider $t$ is the time interval small block make it to the right of the larger block.

From the second equation of motion the distance moved by the smaller block in time $t$ is,

$d_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$

$u_1$ is the initial speed.

The initial speed of the smaller block is $0\text{m}/\text{s}$.

Substitute $0\text{m}/\text{s}$ for $u_1$ in the above equation.

$\begin{array}{c}{d}_1=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}{a}_1{t}^2\\ d_1=\frac{1}{2}{a}_1{t}^2\end{array}$

From the second equation of motion the distance moved by the larger block in time $t$ is,

$d_2=u_{2}t+\frac{1}{2}{a}_2{t}^2$

$u_2$ is the initial speed.

The initial speed of the larger block is $0\text{m}/\text{s}$.

Substitute $0\text{m}/\text{s}$ for $u_2$ in the above equation.

$\begin{array}{c}{d}_2=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}{a}_2{t}^2\\ d_2=\frac{1}{2}{a}_2{t}^2\end{array}$

For the smaller block to reach the right edge of the larger block,

$d_1=d_2+L$

Here,

$L$ is the distance the leading edge of the smaller block travels on the larger block.

Substitute $\frac{1}{2}{a}_1{t}^2$ for $d_1$ and $\frac{1}{2}{a}_2{t}^2$ for $d_2$ in the above equation.

$\frac{1}{2}{a}_1{t}^2=\frac{1}{2}{a}_2{t}^2+L$

Substitute $0.735{\text{m}/\text{s}}^2$ for $a_2$, $2.06{\text{m}/\text{s}}^2$ for $a_1$ and $3.00\text{m}$ for $L$ in the above equation.

$\begin{array}{c}\frac{1}{2}\left(2.06{\text{m}/\text{s}}^2\right)t^2=\frac{1}{2}0.735{\text{m}/\text{s}}^2{t}^2+3.00\text{m}\\ \left(\text{1}\text{.03}{\text{m}/\text{s}}^2\right)t^2=\left(0.3675{\text{m}/\text{s}}^2\right)t^2+3.00\text{m}\end{array}$

Rearrange the above equation for $t$.

$\begin{array}{c}\left(\text{1}\text{.03}{\text{m}/\text{s}}^{2}0-0.3675{\text{m}/\text{s}}^2\right)t^2=3.00\text{m}\\ t=\sqrt{\frac{3.00\text{m}}{0.6625}}\\ =2.127\text{s}\\ \approx \text{2}\text{.13}\text{s}\end{array}$

Conclusion:

Therefore, the time interval small block make it to the right of the larger block is $2.13\text{s}$.

To determine

The distance the larger block moves in the process.

Answer to Problem 5.103CP

The larger block moves $1.67\text{m}$ in the whole process.

Explanation of Solution

Given info: The mass of the smaller block is $2.00\text{kg}$, the mass of the larger block is $8.00\text{kg}$, the coefficient of kinetic friction between the smaller and the larger block is $0.300$, the horizontal force applied on the smaller block is $10.0\text{N}$ and the distance the leading edge of the smaller block travels on the larger block is $3.00\text{m}$.

The acceleration due to gravity is $9.8{\text{m}/\text{s}}^2$.

From part (a) the distance the larger block moves in the process is,

$d_2=\frac{1}{2}{a}_2{t}^2$

Substitute $0.735{\text{m}/\text{s}}^2$ for $a_2$ and $2.13\text{s}$ for $t$ in the above equation.

$\begin{array}{c}{d}_2=\frac{1}{2}0.735{\text{m}/\text{s}}^2{\left(2.13\text{s}\right)}^2\\ =1.667\text{m}\\ \approx \text{1}\text{.67}\text{m}\end{array}$

Conclusion:

Therefore, the larger block moves $1.67\text{m}$ in the whole process.