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Three objects with masses m1 = 4.0 kg, m2 = 11 kg, and m3 = 18 kg, respectively, are attached by strings over frictionless pulleys as indicated in the figure below. The horizontal surface exerts a force of friction of 30 N on m2. If the system is released from rest, use energy concepts to find the speed of m3 after it moves down 4.0 m. m/s   Edit Community
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m1 = 4 kg
W1 = (4)(9.81) = 39.24 N

m2 = 11 kg
F2 = friciton force = 30 N

m3 = 18 kg
W3 = (18)(9.81) = 176 N

M = system total mass = m1 + m2 + m3 = 4+11+18 = 33 kg

Work Done by each Mass in moving h = 4 m:
m1: W1(h) = (39.24)(4) = -156.96 J
m2: Friction Work = - (30)(4) = -120.J
m3: W2(h) = (176 )(4) =704 J
Total (Net) Work = 704 - (156.96 + 120) =427.04 J
Final speed of system (all masses) = V
1/2MV� = 427.04
V� = 427.04/(0.5)(M) = 854.08/M = 854.08/33 =25.8812121212121212
V = 5.0873580689010009176m/s <= all masses move together {at m3 speed} ANS  Edit Community
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