Block A is on top of Block B. They are connected by a light, flexible cord passing through a fixed, frictionless pulley. Block A weighs 3.60 N and block B weighs 5.40 N. The coefficient of kinetic friction between all surfaces is 0.25. Find the magnitude of the force(adsbygoogle = window.adsbygoogle || []).push({}); Fnecessary to drag Block B to the left at a constant speed.

Ok so the key here isconstantspeed. So that means [tex] F = 0 [/tex]. So all forces must be equal and opposite one another in the same direction. Since [tex] w_{total} = 9.0 N [/tex] does that mean that the force required is [tex] f_{k} = \mu_{k}N = 0.25(9.0 N) = 2.5 N [/tex]?

Thanks

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# Friction problem

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