91影视

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.

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,
$鈭�F_1=F鈭�f_{\mathrm{s}}$
As the upper block remains on the lower block,the web force thereon is capable zero. So, the above equationis also rearranged as follows:
$0=F鈭�f_{\mathrm{s}}$

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

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

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

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

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.

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,
$\mathrm{危}{F}_1=f_{\mathrm{S}}鈭�f_{\mathrm{K}}$
The kinetic friction is expressed in terms of coefficient of kinetic friction ${渭}_{\mathrm{K}}$ and so the conventional force. So, the kinetic friction force between the lower block and floor is,
$f_K={渭}_{\mathrm{K}}{n}_2$
Replace $n_2$with$\left(m_{1+}{m}_2\right)+g$ .
$f_{\mathrm{K}}={渭}_{\mathrm{K}}\left(m_1+m_2\right)g$
The net force of the lower block may additionally be expressed in terms of its mass and acceleration as follows:
$\mathrm{危}{F}_1=m_{2}a$
Here, $a$ is that the acceleration of the system.
Replace $鈭�F_1$with $m_{2}a,$$f_{\mathrm{K}}$with ${渭}_{\mathrm{K}}\left(m_1+m_2\right)g\text{,}$and $f_{\mathrm{s}}$with ${渭}_{\mathrm{S}}{m}_{1}g$in $鈭�F_1=f_{\mathrm{S}}鈭�f_{\mathrm{K}},$and rearrange the resulting equation for$a$.
$\begin{array}{c}{m}_{2}a=\left({渭}_{\mathrm{S}}{m}_{1}g\right)鈭�\left({渭}_{\mathrm{K}}\left(m_1+m_2\right)g\right)\\ \end{array}$

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

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

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

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

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

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