91Ó°ÊÓ

• The weight of blocks A and B, $w=25N$.
• The coefficient of kinetic friction, ${\mathrm{μ}}_k=0.35$.

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

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

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

Draw the free-body diagram for block A.

Here, $T_1$is the tension in the rope between blocks A and B , $f_1$is the frictional force between the block A and the surface, and $N_1$ is the normal reaction between the block A and the surface.

Draw the free-body diagram for block B.

Here, $T_1$ is the tension in the rope between blocks A and B , $T_2$ is the tension in the rope between blocks B and C , n is the normal reaction on the block B , and f is the frictional force between the block B and the surface.

The force along the y-axis for the block A is given by,

$N_1-w=0$

The force along the x-axis for the block A is given by,

$T_1-f_1=0$

The frictional force between the block A and the surface is given by,

$$\begin{gathered} f_1={\mathrm{μ}}_k{N}_1 \\ ={\mathrm{μ}}_{k}w \end{gathered}$$

Substitute the frictional force expression in the equation $T_1-f_1=0$, and we get,

$$\begin{gathered} T_1-{\mathrm{μ}}_{k}w=0 \\ {T}_1={\mathrm{μ}}_{k}w.......\left(1\right) \end{gathered}$$

Substitute 0.35 for ${\mathrm{μ}}_k$ and 25 N for w in equation (1), and we get,

$$\begin{gathered} T_1=\left(0.35\right)\left(25N\right) \\ =8.75N \end{gathered}$$

Therefore, the tension in the rope connecting blocks A and B is 8.75 N .

Draw the free-body diagram for block C.

Here, the weight of the block C is $w_c$.

The force equation for the block C is given by,

$w_c-T_2=0.......\left(2\right)$

The force along the x-axis for the block B is given by,

$T_2-T_1-f-w\sin \mathrm{θ}=0........\left(3\right)$

The force along the y-axis for the block B is given by,

$N=w\cos \mathrm{θ}......\left(4\right)$

The frictional force between the block B and the surface is given by,

$f={\mathrm{μ}}_{k}N$

From equation (4),

$f={\mathrm{μ}}_k\left(w\cos \mathrm{θ}\right).....\left(5\right)$

Noting that block C descends down at constant velocity, so its acceleration is zero.

$$\begin{gathered} w_c-T_2=0 \\ {w}_c=T_2 \end{gathered}$$

Substitute equation (5) in (3), and we get,

$$\begin{gathered} T_2-T_1-{\mathrm{μ}}_k\left(w\cos \mathrm{θ}\right)-w\sin \mathrm{θ}=0 \\ {w}_c=T_1+{\mathrm{μ}}_k\left(w\cos \mathrm{θ}\right)+w\sin \mathrm{θ}.....\left(6\right) \end{gathered}$$

Substitute 8.75 N for $T_1$, $36.9°$for $\mathrm{θ}$, 0.35 for ${\mathrm{μ}}_k$, and 25 N for w in equation (6), and we get,

$$\begin{gathered} w_c=8.75N+\left(0.35\right)\left(25N\right)\cos \left(36.9°\right)+\left(25N\right)\sin \left(36.9°\right) \\ =30.75N \end{gathered}$$

Therefore, the weight of the block C is 30.75N .

Consider being the acceleration of both the masses B and C after the rope is cut, and T is the new tension between blocks B and C .

Draw the free-body diagram of the block B after the rope is cut between A and B .

The force along the x-axis for the block B is given by,

$T-f-w\sin \mathrm{θ}=m_{B}a..........\left(7\right)$

The force along the y-axis for the block B is given by,

$N-w\sin \mathrm{θ}=0.......\left(8\right)$

The frictional force between the block B and the surface is given by,

$f={\mathrm{μ}}_{k}N........\left(9\right)$

Substitute equations (8) and (9) in the equation (7), and we get,

$T-{\mathrm{μ}}_{k}w\sin \mathrm{θ}-w\sin \mathrm{θ}=m_{B}a........\left(10\right)$

Draw the free-body diagram of the block C after the rope is cut between A and B .

The new force for the block C is given by,

$w_c-T=m_{c}a........\left(11\right)$

From equations (10) and (11),

$a=\frac{w_c-{\mathrm{μ}}_k\left(w\cos \mathrm{θ}\right)-w\sin \mathrm{θ}}{m_c+m_B}........\left(12\right)$

Calculate the mass of lock C as:

$$\begin{gathered} m_c=\frac{w_c}{g} \\ =\frac{30.75N}{9.8m/s^2} \\ =3.14kg \end{gathered}$$

Calculate the mass of lock B as:

$$\begin{gathered} m_B=\frac{w_B}{g} \\ =\frac{25N}{9.8m/s^2} \\ =2.55kg \end{gathered}$$

Substitute 30.75N for $w_c$, 0.35 for ${\mathrm{μ}}_k$, 25 N for w , 3.14 kg for $m_c$, 2.55kg for $m_B$ , and $36.9°$ for $\mathrm{θ}$ in equation (12), and we get,

$$\begin{gathered} a=\frac{30.75N-\left(0.35\right)\left(25N\right)\cos \left(36.9°\right)-\left(25N\right)\sin \left(36.9°\right)}{3.14kg+2.55kg} \\ =1.54m/s^2 \end{gathered}$$

Therefore, the acceleration of the block C is $1.54m/s^2$.