Two blocks of mass 3.50 kg and 8.00 kg are connected by a mass less string that passes over a frictionless pulley (fig. p5.68). the inclines are frictionless. find (
a. the magnitude of the acceleration of each block and (
b. the tension in the string.

The mass on the left has a downslope weight of 
W1 = 3.5kg * 9.8m/s² * sin35º = 19.7 N 
The mass on the right has a downslope weight of 
W2 = 8kg * 9.8m/s² * sin35º = 45.0 N 
The net is 25.3 N pulling downslope to the right. 

(a) Therefore we need 25.3 N of friction force. 
Ff = 25.3 N = µ(m1 + m2)gcosΘ = µ * 11.5kg * 9.8m/s² * cos35º 
25.3N = µ * 92.3 N 
µ = 0.274 

(b) total mass is 11.5 kg, and the net force is 25.3 N, so 
acceleration a = F / m = 25.3N / 11.5kg = 2.2 m/s² 

tension T = 8kg * (9.8sin35 - 2.2)m/s² = 27 N 

Check: T = 3.5kg * (9.8sin35 + 2.2)m/s² = 27 N √ 

Divide 108.2 x 10^6 into 1.5 x 10 and the result will be in au (astronomical units)