(a) The acceleration of the two blocks is 0.397 m/s².
(b) The tensions in the string to the left is 7.07 N.
(c) The tension to the right of the pulley is 9.04 N.
The net force due to the first and second mass can be determined using Newton's second law of motion as shown below;
[tex]T_1 = \mu m_1 g + m_1 a[/tex]
[tex]T_2 = m_2gsin \theta - \mu m_2 g cos \theta - m_2 a[/tex]
The moment of inertia of the pulley disk is given as;
[tex]I = \frac{1}{2} MR^2[/tex]
The net torque in the pulley is calculated as;
[tex]\tau _{net} = I \alpha\\\\T_2R - T_1 R = I (\frac{a}{R} )\\\\R^2 (T_2-T_1)= I a\\\\a = \frac{R^2 }{I} (T_2 - T_1)\\\\a = \frac{R^2}{1/2MR^2} (T_2-T_1)\\\\a = \frac{2}{M} (T_2 - T_1)[/tex]
The acceleration of the two blocks is determined by substituting the given parameters;
[tex]a = \frac{2}{M} (m_2 gsin\theta - \mu m_2 g cos \theta - m_2 a - \mu m_1 g - m_1 a)\\\\a= \frac{2}{M} (m_2g sin\theta - \mu m_2g cos \theta- \mu m_1 g) - \frac{2}{M} (m_2 + m_1 )a\\\\a + \frac{2}{M} (m_2 + m_1 )a = \frac{2}{M} (m_2g sin\theta - \mu m_2g cos \theta- \mu m_1 g) \\\\a(1 + \frac{2}{M} (m_2 + m_1 )) = \frac{2}{M} (m_2g sin\theta - \mu m_2g cos \theta- \mu m_1 g) \\\\a = \frac{\frac{2}{M} (m_2g sin\theta - \mu m_2g cos \theta- \mu m_1 g)}{1 + \frac{2}{M} (m_2 + m_1 )} \\\\[/tex]
[tex]a = \frac{\frac{2}{10} (6.25\times 9.8 \times sin30 \ -\ 0.36 \times 6.25 \times 9.8cos30\ -\ 0.36 \times 1.8\times 9.8) }{1 \ + \ \frac{2}{10} (6.25 + 1.8)} \\\\a = 0.397 \ m/s^2[/tex]
The tensions in the string to the left is calculated as follows;
[tex]T_1 = \mu m_1 g + m_1 a\\\\T_1 = m_1 (\mu g + a)\\\\T_1 = 1.8(0.36\times 9.8 \ + \ 0.397)\\\\T_1 = 7.07 \ N[/tex]
The tension to the right of the pulley is calculated as follows;
[tex]T_2 = m_2g sin\theta - \mu m_2 g cos \theta - m_2 a\\\\T_2 = m_2 (g sin\theta - \mu g cos \theta - a)\\\\T_2 = 6.25(9.8 \times sin30 \ - \ 0.36 \times 9.8\times cos 30 \ - 0.397)\\\\T_2 = 9.04 \ N[/tex]
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