In the figure below, m1 = 10.5 kg and m2 = 4.0 kg. The coefficient of static fri
ID: 2234601 • Letter: I
Question
In the figure below, m1 = 10.5 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30. (a) If the system is released from rest, what will its acceleration be? (b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?Explanation / Answer
FOLLOW THIS orce on m1=> Ffric = (0.6)(10.5 x 9.81) = 61.8 N Force on m2=> Fgrav = (9.81)(4) = 39.24 N Ffric > Fgrav ? If this system was set up from rest, it would not accelerate, rather, it would remain static. (b) Force on m1=> Ffric = (0.3)(10.5 x 9.81) = 30.9 N Force on m2=> Fgrav = (9.81)(4) = 39.24 N Ffric > Fgrav ? This system will accelerate, with m1 accelerating toward the pulley, and m2 accelerating downward. Force = mass x acceleration. So the total force acting on the system = total mass x acceleration (39.24 N - 30.9 N) = (4 + 10.5) x a a = 8.52/14.5 = 0.588 m/s2