91Ó°ÊÓ

Consider the two blocks as A and B:

Weight of block A,${\mathrm{W}}_{\mathrm{A}}=3.6\mathrm{N}$

Mass of block A,$M_A=0.37\mathrm{kg}$

Weight of block B,${\mathrm{W}}_B=7.2\mathrm{N}$

Mass of block B,$M_B=0.74\mathrm{kg}$

Coefficient of kinetic friction between the lighter block (Block A) and the plane,${\mathrm{μ}}_{K_A}=0.10$

Coefficient of kinetic friction between the heavier block (Block B) and the plane,${\mathrm{μ}}_{K_S}=0.20$

To find the acceleration of the blocks and the tension on the string, draw a free body diagram for each and then apply Newton's 2nd law of motion. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\underset{}{∑}=ma$

where, F is the net force, mis mass and ais an acceleration.

Free Body Diagram of the Block A (lighter) and Block B (heavier):

By using the Newton’s 2nd law along the vertical direction to the block A, (positive x axis along the incline and the positive y along the vertical direction),

$$\begin{gathered} N_A-W_A\cos \left(30\right)=0 \\ {N}_A-W_A\cos \left(30\right) \end{gathered}$$

Thus, the kinetic frictional force is,

$$\begin{gathered} f_{k_A}={\mathrm{μ}}_{k_A}{N}_A \\ ={\mathrm{μ}}_{k_A}{W}_A\cos \left(30\right) \end{gathered}$$

Similarly, to the horizontal direction, consider acceleration as a,

$$\begin{gathered} T+f_{k_A}-W_A\sin \left(30\right)=m_{A}a \\ T+{\mathrm{μ}}_{k_A}{W}_A\cos \left(30\right)-W_A\sin \left(30\right)=m_{A}a \end{gathered}$$

…(¾±)

Now, by using Newton’s 2nd law motion along the vertical and horizontal direction to the block B,

$$\begin{gathered} N_B-W_B\cos \left(30\right)=0 \\ {N}_B-W_B\cos \left(30\right)=0 \end{gathered}$$

Thus, the kinetic frictional force is,

$$\begin{gathered} f_{K_B}={\mathrm{μ}}_{K_B}{N}_B \\ ={\mathrm{μ}}_{K_B}{W}_B\cos \left(30\right) \end{gathered}$$

Along horizontal direction, consider acceleration is a,

localid="1657173983482" $$\begin{gathered} f_{K_B}T-W_B\sin \left(30\right)=m_{B}a \\ {\mathrm{μ}}_{K_B}{W}_{B}cos\left(30\right)-T-W_B\sin \left(30\right)=m_{B}a \end{gathered}$$

…(¾±¾±)

Adding equation (i) and (ii),

$$\begin{gathered} {\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\cos \left(30\right)-{\mathrm{W}}_{\mathrm{A}}\sin \left(30\right)+{\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\cos \left(30\right)-{\mathrm{W}}_{\mathrm{A}}\sin \left(30\right)=\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{a} \\ \left({\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}+{\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\right)\cos \left(30\right)-\left({\mathrm{W}}_{\mathrm{A}}+{\mathrm{W}}_{\mathrm{B}}\right)\sin \left(30\right)=\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)\mathrm{a} \\ \mathrm{a}=\frac{\left({\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}+{\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\right)}{\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)}\cos \left(30\right)-\mathrm{g}\sin \left(30\right) \\ =\left[\frac{\left(0.10×3.6\mathrm{N}+0.20×7.2\mathrm{N}\right)}{\left(0.37\mathrm{kg}+0.74\mathrm{kg}\right)}\cos \left(30\right)\right]-\left(9.8\mathrm{m}/{\mathrm{s}}^2×\sin \left(30\right)\right) \\ =-3.5\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Negative sign shows that blocks are moving to the downhill.

Thus,the magnitude of the acceleration is $3.5m/s^2$.

By using the expression of acceleration in equation (i),

$$\begin{gathered} \mathrm{T}+{\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}\cos \left(30\right)-{\mathrm{W}}_{\mathrm{A}}\sin \left(30\right)={\mathrm{m}}_{\mathrm{A}}\left(\frac{\left({\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{A}}}{\mathrm{W}}_{\mathrm{A}}+{\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{B}}}{\mathrm{W}}_{\mathrm{B}}\right)}{\left({\mathrm{m}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}\right)}\cos \left(30\right)-\mathrm{g}\sin \left(30\right)\right) \\ \mathrm{T}=\frac{{\mathrm{W}}_{\mathrm{A}}{\mathrm{W}}_{\mathrm{B}}}{\left({\mathrm{W}}_{\mathrm{A}}+{\mathrm{W}}_{\mathrm{B}}\right)}\left({\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{B}}}-{\mathrm{μ}}_{{\mathrm{K}}_{\mathrm{A}}}\right)\cos \left(30\right) \\ =\frac{3.6\mathrm{N}×7.2\mathrm{N}}{\left(3.6\mathrm{N}+7.2\mathrm{N}\right)}\left(0.20-0.10\right)\cos \left(30\right) \\ =0.21\mathrm{N} \end{gathered}$$

Thus, the magnitude of tension on the string is 0.21 N.