91Ó°ÊÓ

• Mass of block A:$m_A=4.0\mathrm{kg}$ .
• Mass of block B:$m_B=2.0\mathrm{kg}$.
• The coefficient of kinetic friction between block B and the horizontal plane is:${\mathrm{μ}}_k=0.50$ .

The problem deals with Newton’s second law of motion, which states that the acceleration, a of an object is dependent upon the net force, F acting upon the object, and the mass, m of the object.

Formula:

F = ma

Applying Newton’s second law:

For block A:

$$\begin{gathered} m_{A}g\sin \left({30}^a\right)-T=m_{A}a \\ -T=m_{A}g\sin \left(30°\right)-m_{A}a \end{gathered}$$

(i)

For block B:

$$\begin{gathered} T-f=m_{B}a \\ T=f+m_{B}a \end{gathered}$$

(ii)

(Where a is the acceleration of both blocks and f is the force due to friction)

Equating T from both equations:

$m_{A}g\sin (30°)-m_{A}a=f+m_{B}a$

The magnitude of the acceleration of the blocks can be calculated as:

Solving the above equation for acceleration a as:

$$\begin{gathered} m_{A}g\sin (30°)-f=\left(m_A+m_B\right)a \\ {m}_{A}g\sin (30°)-{\mathrm{Ï\dots }}_k×F_{n2}=\left(m_A+m_B\right)a \\ a=\frac{m_{A}g\sin (30°)-{\mathrm{Ï\dots }}_k×F_{n2}}{\left(m_A+m_B\right)} \\ =\frac{m_{A}g\sin (30°)-{\mathrm{Ï\dots }}_k×\left(m_{B}g\right)}{\left(m_A+m_B\right)} \end{gathered}$$

Substituting the values in the above expression, and we get,

$$\begin{gathered} \mathrm{a}=\frac{\left(4.0\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2×\sin (30°)\right)-\left(0.50×2.0\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2\right)}{4.0\mathrm{kg}+2.0\mathrm{g}} \\ \mathrm{a}=\frac{9.8\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{6.0\mathrm{kg}} \\ \mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the magnitude of the acceleration is$\mathrm{a}=1.6\mathrm{m}/{\mathrm{s}}^2$ .

Using equation (ii), we get,

$$\begin{gathered} T=f+m_{B}a \\ =\left({\mathrm{μ}}_k×F_{n2}\right)+m_{B}a \end{gathered}$$

Substituting the values in the above expression, and we get,

$$\begin{gathered} \mathrm{T}=\left(0.50×2.0\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2\right)+\left(2.0\mathrm{kg}×1.6\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{T}=13\mathrm{N} \end{gathered}$$

Thus, the tension in the rope is 13 N.