1. The problem statement, all variables and given/known data A 9.00-kg hanging object is connected, by a light, inextensible cord over a light, frictionless pulley, to a 5.00-kg block that is sliding on a flat table. Taking the coefficient of kinetic friction as .200, find the tension in the string 2. Relevant equations F = ma 3. The attempt at a solution Fk = ma a = (9.8 N friction force)/(5.00 kg) = 1.96 which is approximately 2 ΣFx= T - Fk = ma T = Fk + ma T = (9.8N) + total mass 14N(1.96 m/s^2) T = 37.24 or if you use 2 for the acceleration you get 37.8 which is the right answer Did I do this right? Please help me to understand why.
I am not sure why you say that Fk = ma. Try Fk=uN. yes, looking at all forces acting in the x direction on the block sitting on the table, this equaton is correct, as it applies to the 5kg block. you must make the Fk correction. You are right now just looking at the block on the table, where m = 5, not 14, and 'a' is unknown. In problems of these types, or any types using Newton's laws, you MUST draw free body diagrams. Isolate each block and note all the forces acting on it (weight, friction, normal force, applied forces, etc., in the x and y directions). Try also drawing a FBD of the hanging object. What are the forces acting on it? Then apply Newton's laws to each block. Note that since the blocks are connected by an inextensible cord over a light frictionless pulley, that the magnitude of the accelerations of each block must be the same, and the rope tension acting on each block must be the same.