The speed of Block A is 7.18 ft/s.
From energy conservation we know that the initial kinetic energy is equal to final kinetic energy.
Block A has weight of 60 lb.
Block B has weight of 10 lb.
In the question it is said that block A moves from rest at a distance of 5ft.
Relative motion constraint-
2sA + sB =l
differentiating we get:
So, 2ΔsA + ΔsB = 0
→ ΔsB = -2ΔsA
2 vA + vB =0
→ vB = -2vA
Now applying principle of work and energy
TA0 +TBO + ΣwO-AB = TA+TB
1/2 mA v²A₀ + 1/2 mBv²B₀ + wAΔsA + wBΔsB = 1/2 mAv²A + 1/2 mBv²B
0 + 0 +6(5) - 10(10) = 1/2 (-60/32-2) v²A +1/2 (10/32.2) 2v²A
vA = 7.18 ft/s
So, the speed of block A is 7.18 ft/s.
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