Answer:
The acceleration of [tex]M_2[/tex] is [tex]a = 0.7156 m/s^2[/tex]
Explanation:
From the question we are told that
The mass of first block is [tex]M_1 = 2.25 \ kg[/tex]
The angle of inclination of first block is [tex]\theta _1 = 43.5^o[/tex]
The coefficient of kinetic friction of the first block is [tex]\mu_1 = 0.205[/tex]
The mass of the second block is [tex]M_2 = 5.45 \ kg[/tex]
The angle of inclination of the second block is [tex]\theta _2 = 32.5^o[/tex]
The coefficient of kinetic friction of the second block is [tex]\mu _2 = 0.105[/tex]
The acceleration of [tex]M_1 \ and\ M_2[/tex] are same
The force acting on the mass [tex]M_1[/tex] is mathematically represented as
[tex]F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
=> [tex]M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1[/tex]
Where T is the tension on the rope
The force acting on the mass [tex]M_2[/tex] is mathematically represented as
[tex]F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
[tex]M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2[/tex]
At equilibrium
[tex]F_1 = F_2[/tex]
So
[tex]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[/tex]
making a the subject of the formula
[tex]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}[/tex]
substituting values [tex]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}[/tex]
=> [tex]a = 0.7156 m/s^2[/tex]
