Analyzing Tension Forces with friction

[Mindstein]

Homework Statement

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

Homework Equations

F = ma

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.

 

[PhanthomJay]

Homework Statement

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

Homework Equations

F = ma

The Attempt at a Solution

Fk = ma
a = (9.8 N friction force)/(5.00 kg) = 1.96 which is approximately 2

I am not sure why you say that Fk = ma. Try Fk=uN.

ΣFx= T - Fk = ma

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.

T = Fk + ma
T = (9.8N) + total mass 14N(1.96 m/s^2)

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.

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.

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.

 

[thomate1]

Your solution is correct. You used the correct formula F = ma to calculate the acceleration of the 5.00-kg block. Then, you used the equation ΣFx = T - Fk = ma to find the tension in the string, where ΣFx represents the sum of all the forces acting on the block in the horizontal direction. The tension in the string is equal to the force of friction (Fk) plus the product of the mass (m) and acceleration (a). Your answer of 37.24 N is also consistent with the solution using a = 2, which is the correct acceleration for the given problem. Great job!