The force diagram of the given system is shown below:
where,
N: normal force = Wy
Wy: component of the weight W perpendicular to the incline = Wcosθ
Wx: component of the weight W parallel to the incline = Wsinθ
W: weight of block 1 = m1*g
W2: weight of block 2 = m2*g
T: tension on the cord
Fr: friction force = μN
i) The acceleration of the two blocks is the same.
The sum of forces on the first block along the direction of the incline is (equation 1):
[tex]\begin{gathered} T-F_r-W_x=m_1\cdot a \\ T-\mu N-m_1g\sin \theta=m_1\cdot a \end{gathered}[/tex]
The sum of forces along the direction perpendicular to the incline is (equation 2):
[tex]\begin{gathered} W_y+N=0 \\ N=W_y=m_1g\cos \theta \end{gathered}[/tex]
The sum of forces on the second block is (equation 3):
[tex]m_2g-T=m_2\cdot a[/tex]
Now, add the first and third equation, solve for a, and replace the expression for N, as follow:
[tex]\begin{gathered} T-\mu N-m_1g\sin \theta+m_2g-T=m_1a+m_2a \\ a(m_1+m_2)=m_2g-\mu N-m_1g\sin \theta \\ a=\frac{m_2g-\mu m_1g\cos\theta-m_1g\sin\theta}{m_1+m_2} \end{gathered}[/tex]
The previous expression is the result for the acceleration of the two blocks.
ii) And the tension is:
[tex]\begin{gathered} T=m_2g-m_2a \\ T=m_2g-m_2(\frac{m_2g-\mu m_1g\cos\theta-m_1g\sin\theta}{m_1+m_2}) \end{gathered}[/tex]
