Newton's second law for translational and rotational motion we can find the answer for the acceleration of the body is 0.04 m / s²
Newton's second law says that the sum of the external forces is equal to the product of the mass and the acceleration of a body.
Reference systems are coordinate systems with respect to which to measure and decompose the vectors, in this case we will use a regency system with the x=axis parallel to the plane and a positive direction in the direction of movement, eg the y-axis is perpendicular to the plane. .
In the Attachment we have a free-body diagram of the systems
y-axis
     N -[tex]W_y[/tex] = 0
     N = [tex]W_y[/tex]
x-axis
     Wₓ -fr - T = m a
The friction force is the macroscopic manifestation of the microscopic interactions between the two surfaces, it is represented by the relationship
     fr = μ N
Let's use trigonometry to break down the weight
     cos 37 = [tex]\frac{W_y}{W}[/tex]
     sin 37 = [tex]\frac{W_x}{W}[/tex]
      [tex]W_y[/tex] = W cos 37
      Wₓ = W sin 37
Let's  substitute
      N = mg cos 37
      mg sin 37 - μ m g cos 37 - T = m a       (1)
We use Newton's second law for rotational motion, in the pulley
      τ = I α
       T R =  I α        (2)
The weight and the normal of the pulley pass through its axis of rotation, so the distance is zero and they do not create torque
Linear and rotational variables
       a = α R
       α = a / R
Let's substitute in equation 2
      T R = I a / R
       T = I / R² a
Let's substitute in equation 1
      mg sin 37 - very m g cos 37 - [tex]\frac{I}{R^2}\ a[/tex] = m a
       a = [tex]\frac{g ( sin 37 - \mu \ cos 37 )}{ 1 + \frac{I}{m R^2} }[/tex]
Let's calculate
       a =  [tex]\frac{ 9.8 ( sin 37 - 0.2 cos 37) }{1 + \frac{0.4}{ 1.0 \ 0.20^2} }[/tex]
       a = [tex]\frac{4.33246}{11}[/tex]
       a = 0.39 m / s²
In conclusion using Newton's second law for translational and rotational motion we can find the answer for the eceleration of the body is
        0.39 m /s²
Learn more about Newton's second law here:
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