let the acceleration of masses is m
frictional force = 0.35/0.15 N
Now, for the net acceleration
net acceleration = net force/total effective mass
a = (1 * 9.8 - 0.60 * 9.8 - 0.35/.15)/(1 + 0.60 + 0.50 * 0.30 * 0.15^2/0.15^2)
a = 0.91 m/s^2
the acceleration of each mass is 0.91 m/s^2
yNotes 0/1 points 1 Previous Answers W86 8.065.E. Two masses are suspended from a pulley as...
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