Question

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2.25 kg and is on an incline of 1=43.5∘ with coefficient of kinetic friction 1=0.205 . 2 has a mass of 5.45 kg and is on an incline of 2=32.5∘ with coefficient of kinetic friction 2=0.105 . The two‑block system is in

Answer 1

Answer:

The acceleration of $M_2$ is  $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

    The mass of first block is  $M_1 = 2.25 \ kg$

    The angle of inclination of first block is  $\theta _1 = 43.5^o$

    The coefficient of kinetic friction of the first block is  $\mu_1 = 0.205$

      The mass of the second block is  $M_2 = 5.45 \ kg$

     The angle of inclination of the second block is  $\theta _2 = 32.5^o$

      The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

     $F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as    

  $F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

   $M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

  $F_1 = F_2$

So

 $T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

    $a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

    => $a = 0.7156 m/s^2$

     

Answer 2

The acceleration of the second block to the right is 2.21 m/s².

The normal force on block1 is calculated as follows;

$F_n_1 = m_1g cos(\theta_1)$

The parallel force on block 1 is calculated as;

$F_x_1 = m_1gsin(\theta)$

The frictional force on block 1 is calculated as;

$F_k_1 = \mu_k F_n = \mu_k m_1gcos\theta_1$

The net force on block 1 is calculated as;

$\Sigma F_x_1 = m_1gsin(\theta_1) - \mu_k_1m_1gcos(\theta_1)$

The normal force on block 2 is calculated as follows;

$F_n_2 = m_2gcos\theta _2$

The frictional force on block 2 is calculated as;

$F_k_2 = \mu k_2 m_2g cos\theta _2$

The net force on block 2 is calculated as follows;

$\Sigma F_x_2 = m_2a_2\\\\ m_2g_2 sin(\theta _2) - F_k_2 - \Sigma F_x_1 = m_2a_2 \\\\m_2gsin(\theta) - F_k_2 - (m_1gsin(\theta) - \mu_k _1 m_1g cos(\theta)) = m_2a_2\\\\m_2gsin(\theta) -\mu_k_2 m_2gcos(\theta) + \mu_k _1 m_1g cos(\theta) - m_1gsin(\theta) = m_2a_2\\\\5.45( 9.8) sin(32.5) -(0.105)(5.45)(9.8)cos(32.5) + \\\\0.205( 2.25) ( 9.8)cos(43.5) - 2.25( 9.8) sin (43.5) = 5.45a_2\\\\ 12.07 = 5.45a_2\\\\a_2= \frac{12.07}{5.45} \\\\a_2 = 2.21 \ m/s^2$

Thus, the acceleration of the second block to the right is 2.21 m/s².

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