Blocks `A, B and C` are placed as shown in Fig and connected by the rops of negligible mass . Both `A` and `B` weigh `25.0 N ` each , and the coefficient of kinetic friction between each and the surface is `0.35` blocks `C` descends with constant velocity .

a. Draw two separate free- body diagrams showing the forces acting on `A and B`
b. Find the tension in the rope connecting blocks `A and B`
c. What is the weight of blocks C?
d. If the rope connecting `A and B` were cut, what would be the acceleration of C?

Explanation

a.
b.The blocks move with constant speed, so there is no net force on blocks `A : ` the tension in the connecting `A` and `B` must be equal to the friction force on block `A` .
`T_(1) = mu_(k)n_(A) = (0.35)(25.0) = 8.75 N`
b. The weight of block`C` will be the tension in the rope connecting `B` and `C`
`w_(c) = T_(2) `
`= 9 N + w_(B) (sin 37^(@) + mu_(k) cos 37^(@))`
`= 9 N + (25.0 N) [sin 37^(@) + (0.35) cos 37^(@)]`
`= 31.0 N`
Appling Newton's second law to the remaining massese `(B` and `C)` gives
`a = (g(w_(c) - mu_(k) w_(B) cos theta - w_(B) sin theta))/((w_(B) + w_(c))) = 1.6ms^(-2)`