Forces problem - Please Help ASAP

Question

In the figure below, m1 = 3.2 kg, m2 = 5.8 kg, and the coefficient of kinetic friction between the inclined plane and the 3.2-kg block is μk = 0.25. Find the magnitude of the acceleration of the masses and the tension in the cord.

Answer 1

As I understand the problem m1 is sliding up a 30 degree incline.  m2 is connected to m1 by string-pulley arrangement and is moving vertically.

In this arrangement both masses have same acceleration.   The forces on m1 are gravity (the component along the slope) and the rope tension and friction.   Fn = m1gcos30   Ffrict =   μFn

a = (T- m1g*sin 30 - μm1gcos30)/m1   =  (T - 38.8N)/m1

For the block m2      a = (m2g - T)/m2.    Now set the two accelerations equal and we can find the cord tension.

(T - 38.8N)/m1  = (56.84N - T)m2 =>  m2T - 38.8N*m2 = 56.84N * m1 - m1T

(m2 + m1)T = (38.8m2 + 56.84m1)   T =  45.2 N   

a = (45.2 - 38.8)/m1 = 2 m/s^2.  

As a check  use the equation for m2 to recalc acceleration: 

a = (5.8*9.8 - 45.2)/5.8 = 2.0 m/s^2.