91影视

The given data is as follows:

• The mass of the first block is $m_1=4kg$.
• The mass of the second block is $m_2=8kg$.
• The angle is $胃=30掳$.
• The coefficient of kinetic friction between the first block and the plane is ${渭}_{k_1}=0.25$.
• The coefficient of kinetic friction between the second block and the plane is ${渭}_{k_2}=0.35$.

The kinetic force occurs when the object is in motion. The expression of the kinetic friction force is given by,

${\mathit{f}}_{\mathbf{k}}\mathbf{=}{\mathit{渭}}_{\mathbf{k}}\mathit{N}$

Here, ${\mathit{渭}}_{\mathbf{k}}$ is the coefficient of kinetic friction, and N is a normal force.

The forces acting on mass $m_1$ are shown in the following figure.

From the above figure, the relation of forces on the horizontal direction is given by,

$m_{1}a=m_{1}g\sin 胃-T-f_{k_1}$

Here, $m_1$is the mass of the first block, a is acceleration, g is the acceleration due to gravity, $胃$ is the angle, T is the tension force, and $f_{k_1}$is the kinetic friction due to the first block.

The kinetic friction is given by,

$f_{k_1}={渭}_{k_1}{N}_1$

Here ${渭}_{k_1}$ is the coefficient of kinetic friction and $N_1$is the normal force.

Substitute $m_{1}g\cos 胃$for $N_1$ in the equation localid="1667647933442" $f_{k_1}={渭}_{k_1}{N}_1$, and we get,

$f_{k_1}={渭}_{k_1}{m}_{1}g\cos 胃$

Rewrite the equation $m_{1}a=m_{1}g\sin 胃-T-f_{k_1}$as follows, and we get,

$$\begin{gathered} m_{1}a=m_{1}g\sin 胃-T-{渭}_{k_1}{m}_{1}g\cos 胃 \\ T=m_{1}g\left(\sin 胃-{渭}_{k_1}\cos 胃\right)-m_{1}a \end{gathered}$$

The forces acting on mass $m_2$are shown in the following figure.

Similarly, from the above figure, the relation of forces on the horizontal direction is given by,

$T=m_{2}a-m_{2}g\left(\sin 胃-{渭}_{k_2}\cos 胃\right)$

Here $m_2$ is the mass of the second block and $f_{k_2}$ is the kinetic friction due to the second block.

The tension force acts on the mass and are equal.

$$\begin{gathered} m_{1}g\left(\sin 胃-{渭}_k\cos 胃\right)-m_{1}a=m_{2}a-m_{2}g\left(\sin 胃-{渭}_{k_2}\cos 胃\right) \\ \left(m_1+m_2\right)a=m_{1}g\left(\sin 胃-{渭}_{k_1}\cos 胃\right)+m_{2}g\left(\sin 胃-{渭}_{k_2}\cos 胃\right) \\ a=\frac{m_{1}g\left(\sin 胃-{渭}_{k_1}\cos 胃\right)+m_{2}g\left(\sin 胃-{渭}_{k_2}\cos 胃\right)}{\left(m_1+m_2\right)}..........\left(1\right) \end{gathered}$$

Substitute $9.8m/s^2$for g , 4 kg for $m_1$, 8 kg for $m_2$, $30掳$for $胃$, 0.25 for ${渭}_k$, and 0.35 for ${渭}_{k_2}$in equation (1).

$$\begin{gathered} a=\frac{\left(9.8m/s^2脳4kg\right)\left(\sin 30掳-0.25脳\cos 30掳\right)+\left(9.8m/s^2脳8kg\right)\left(\sin 30掳-0.35脳\cos 30掳\right)}{4kg+8kg} \\ =\frac{11.1129+15.4363}{12}m/s^2 \\ =2.212m/s^2 \end{gathered}$$

Therefore, the acceleration of each block is $2.212m/s^2$.

Substitute $9.8m/s^2$ for g , 4kg for $m_1$, $30掳$for $胃$, 0.25 for ${渭}_{k_1}$, 0.35 for ${渭}_{k_2}$, and $2.212m/s^2$ for a in the equation $T=m_{1}g\left(\sin 胃-{渭}_{k_1}\cos 胃\right)-m_{1}a$, and we get,

$$\begin{gathered} T=\left(9.8m/s^2脳4kg\right)\left(\sin 30掳-0.25脳\cos 30掳\right)-4kg脳2.212m/s^2 \\ =2.265N \end{gathered}$$

Therefore, the tension in the string is 2.265 N .

The force on is given by,

$F_1=m_{1}g\left(\sin 胃-{渭}_k\cos 胃\right).......\left(2\right)$

Substitute $9.8m/s^2$ for g , 4 kg for $m_1$, $30掳$ for $胃$, 0.25 for ${渭}_{k_1}$, and $2.212m/s^2$ for a in the equation (2).

$$\begin{gathered} F_1=\left(4kg\right)\left(9.8m/s^2\right)\left[\sin 30掳-0.25\cos 30掳\right] \\ =11.11N \end{gathered}$$

The force on $m_2$ is given by,

$F_2=m_{2}g\left(\sin 胃-{渭}_{k_2}\cos 胃\right).......\left(3\right)$

Substitute $9.8m/s^2$ for g , 8 kg for $m_2$, $30掳$ for $胃$, 0.35 for ${渭}_{k_1}$, and 2.212$m/s^2$ for a in the equation (2), and we get,

$$\begin{gathered} F_2=\left(8kg\right)\left(9.8m/s^2\right)\left[\sin 30掳-0.35\cos 30掳\right] \\ =15.44N \end{gathered}$$

If the position of the block is revered, the acceleration can be calculated as follows.

Calculate the acceleration of mass ${\mathrm{m}}_1$ as follows.

$$\begin{gathered} a_1=\frac{F_1}{m_1} \\ =\frac{11.11N}{4Kg} \\ =2.78m/s^2 \end{gathered}$$

Calculate the acceleration of mass $m_2$ as follows.

$$\begin{gathered} a_2=\frac{F_2}{m_2} \\ =\frac{15.44N}{8kg} \\ =1.93m/s^2 \end{gathered}$$

Therefore, if the positions of the block are reversed, then the acceleration of the blocks is $2.78m/s^2$, and $1.93m/s^2$.