What would friction be in this pulley system

[spectravoid]

1. You have a mass hanging off of a frictionless pulley (massless string). The other mass is on a table. The weight of the mass causes the block on the table to slide at an acceleration. There is friction between the block and the table. What would the coefficient of kinetic friction be for this set up?



2. F = ma
w = mg



3. Well Since the system is accelerating, I thought it was

T(rope) - fk = ma
Acceleration and the masses are known. T isn't.

But you can set this up for both masses

T(rope) - ukm1g = m1a
T(rope) - m2g = m2a
T(rope) = m2g + m2a
m2g + m2a - ukm1g = m1a
-ukm1g = m1a -m2a - m2g
-uk(0.483)(10) = 0.483(1.66) - 0.25(1.66) - 0.25(10)
uk = 0.438

Does this make sense?

 

[rl.bhat]

In the second equation g and a are in the same direction. Hence they must have the same sign.

 

[kerimek]

I would respond by saying that the coefficient of kinetic friction for this setup would be 0.438. This value was calculated using the equations for Newton's second law (F=ma) and the forces acting on the system (tension in the rope and friction between the block and table). The negative sign in the equation indicates that the direction of friction is opposite to the direction of motion. This value is reasonable and makes sense in this scenario. However, it is important to note that the coefficient of friction can vary depending on the surface materials and conditions, so this value may not be the same for all pulley systems. Further experimentation and analysis would be needed to determine the exact coefficient of friction for this specific system.