The assembly consists of two blocks A and B, which have a mass of 20 kg and 30 kg, respectively. Determine the distance B must d
escend in order for A to achieve a speed of 3 m>s starting from rest.
2 answers:
Answer:
ΔδB = 5.7 m
Explanation:
Given
mA = 20 Kg
mB = 30 Kg
vA = 3 m/s
vA0 = vB0 = 0 m/s (the blocks released from rest)
Since the cable length (L) is constant
3*δA + δB = L
3*ΔδA = - ΔδB
If we apply
d(3*δA + δB)/dt = dL/dt
3*vA + vB = 0 ⇒ 3*vA = - vB
vB = -3*(3 m/s) ⇒ vB = - 9 m/s
We apply Conservation of Energy
TA0 + TB0 + vA0 + vB0 = TA + TB + vA + vB
0 + 0 + 0 + 0 = 0.5*mA*vA² + 0.5*mB*vB² + WA*ΔδA + WB*ΔδB
0 = 0.5*mA*vA² + 0.5*mB*(-3*vA)² + WA*(ΔδB/3) - WB*ΔδB
0 = 0.5*20 Kg*(3 m/s)² + 0.5*30 Kg*(-9 m/s)² + (20 Kg*9.81 m/s²)*(ΔδB/3) - (30 Kg*9.81 m/s²)*ΔδB
⇒ ΔδB = 5.7 m
The pic shown can help us to understand the question.
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