Question

The coefficient of static friction is $0.60$ between the two blocks in FIGURE $P7.35$. The coefficient of kinetic friction between the lower block and the floor is $0.20$. Force$\overrightarrow{F}$causes both blocks to cross a distance of $5.0m$, starting from rest. What is the least amount of time in which this motion can be completed without the top block sliding on the lower block?

Short answer

The smallest amount amount of your time within which the motion will be completed is . $1.8sec$.

Step-by-step solution

Upper block free diagram

The free body diagram of the upper block is provided within the below figure.

In the above figure, $F$ is that the force applied to the upper block, $m_1$ is that the mass of upper block, $m_{1}g$ is that the burden of the upper block, $g$ is that the acceleration due to gravity, $n_1$ is thatthe traditional force on the upper block, and $f_s$ is that the force of static friction between the upper and lower bocks.

Upper block

For upper block, the standardforce balances its weight $m_g$ . So, the net force on the upper block along vertical direction iscapable zero. Thus,
$n_1=m_{1}g$
Thus,the net force on the upper block along horizontal direction is,
$\sum F_1=F-f_{\mathrm{s}}$
As the upper block remains on the lower block,the web force thereon is capable zero. So, the above equationis also rearranged as follows:
$0=F-f_{\mathrm{s}}$

$F=f_{\mathrm{s}}$

The static friction are often expressed in terms of coefficient of static friction ${\mu }_{\mathrm{s}}$and so the traditional force. So, the static friction force between the blocks is,

$f_{\mathrm{s}}={\mu }_{\mathrm{s}}{n}_1$

$f_{\mathrm{s}}={\mu }_{\mathrm{s}}{m}_{1}g$

Free-body diagram

Similarly, the free body diagram of the lower block is shown within the below figure.

In the above figure $f_K$ is that the kinetic friction force between the lower block and thus the ground , $m_2$ is that the mass of the lower block, $n_2$ is that the traditional force on the lower block, and $(m_1+m_2)g$ is that the total weight of lower and upper blocks.

Kinetic friction

For the lower block, the normal force $n_2$ balances theweightlocalid="1651389241674" $\left(m_{1+}{m}_2\right)+g$. Thus,
localid="1651389364019" $n_2$The net force on the lower block along the horizontal direction is,
$\mathrm{\Sigma }{F}_1=f_{\mathrm{S}}-f_{\mathrm{K}}$
The kinetic friction is expressed in terms of coefficient of kinetic friction ${\mu }_{\mathrm{K}}$ and so the conventional force. So, the kinetic friction force between the lower block and floor is,
$f_K={\mu }_{\mathrm{K}}{n}_2$
Replace $n_2$with$\left(m_{1+}{m}_2\right)+g$ .
$f_{\mathrm{K}}={\mu }_{\mathrm{K}}\left(m_1+m_2\right)g$
The net force of the lower block may additionally be expressed in terms of its mass and acceleration as follows:
$\mathrm{\Sigma }{F}_1=m_{2}a$
Here, $a$ is that the acceleration of the system.
Replace $\sum F_1$with $m_{2}a,$$f_{\mathrm{K}}$with ${\mu }_{\mathrm{K}}\left(m_1+m_2\right)g\text{,}$and $f_{\mathrm{s}}$with ${\mu }_{\mathrm{S}}{m}_{1}g$in $\sum F_1=f_{\mathrm{S}}-f_{\mathrm{K}},$and rearrange the resulting equation for$a$.
$\begin{array}{c}{m}_{2}a=\left({\mu }_{\mathrm{S}}{m}_{1}g\right)-\left({\mu }_{\mathrm{K}}\left(m_1+m_2\right)g\right)\\ \end{array}$

$a=\frac{\left({\mu }_{\mathrm{S}}{m}_1-{\mu }_{\mathrm{K}}\left(m_1+m_2\right)\right)g}{m_2}$

Substitute $4.0\mathrm{kg}$for localid="1651390097380" $m_1,3.0\mathrm{kg}$for $m_2$, $0.60$for${\mu }_{\mathrm{s}},$$0.20$for${\mu }_{\mathrm{K}}\text{,}$and $9.81\mathrm{m}/{\mathrm{s}}^2$for $g$

$\begin{array}{r}a=\frac{\left({\mu }_{\mathrm{S}}{m}_1-{\mu }_{\mathrm{K}}\left(m_1+m_2\right)\right)g}{m_2}\end{array}$

$\begin{array}{r}a=\frac{\left(\right(0.60\left)\right(4.0\mathrm{kg})-(0.20\left)\right(\left(4.0\mathrm{kg}\right)+\left(3.0\mathrm{kg}\right)\left)\right)\left(9.81\mathrm{m}/{\mathrm{s}}^2\right)}{\left(3.0\mathrm{kg}\right)}\\ \end{array}$

$=3.27\mathrm{m}/{\mathrm{s}}^2$

Thus, the acceleration of the system is $3.27\mathrm{m}/{\mathrm{s}}^2.$

System of blocks

The time $t$ taken for the system of blocks to cover the given distance $x$ is calculated by usingthe next kinematic equation.
$x=v_{0}t+\frac{1}{2}a{t}^2$
Here, $v_0$ is that the initial sped of the system of blocks.
Substitute $5.0m$for $x$, 0 for $v_0$, and localid="1651391198831" $3.27\mathrm{m}/{\mathrm{s}}^2$for $a$, and solve for$t$.
$\begin{array}{r}\left(5.0\mathrm{m}\right)=\left(0\right)t+\frac{1}{2}\left(3.27\mathrm{m}/{\mathrm{s}}^2\right)t^2\end{array}$

$\begin{array}{r}t=\sqrt{\frac{2\left(5.0\mathrm{m}\right)}{3.27\mathrm{m}/{\mathrm{s}}^2}}\\ \end{array}$

$=1.75\mathrm{s}$

Round off to $2$ significant figures, the time taken for the system of blocks to cover the given distance is $1.8s$.