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 descend in order for A to achieve a speed of 3 m>s starting from rest.

5.70 m

Explanation:

3sₐ + sB = L

3∆sₐ = - ∆sB

3vₐ = -vB

3*3 = -vB

vB = -9 ms⁻²

T₁ + V₁ = T₂ + V₂

(0 + 0) + (0 + 0) =  1/2(20)(3)² + 1/2(30)(-9)² + 20*9.81*(sB/3) + 30*9.81* sB

0 =  90  +  1215  - 228.9sB

-1305 = - 228.9sB

sB = 1305/228.9 = 5.70 m

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.

answer:

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explanation:

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