Calculate the coefficient of friction between the two bundles of candy

[TheLegace]

Homework Statement

Mrs. K has just received a large shipment of candy at the local Blockbuster. Wanting to get out of the store as fast as possible (so that she can return to her loving husband) she uses the shrink wrap machine to bundle the candy into two blocks, one of 8 kg and the other of 2 kg. The 8 kg bundle is 4 m in length and the 2 kg bundle is only 20 cm in length. She places the smaller bundle on the 8 kg bundle such that the front faces of both are flush with each other. She pushes on the 8 kg bundle with a constant force of 100 N and the system begins to move. The coefficient of friction between the 8 kg bundle and the floor is only 0.1. She notices that as she pushes on the candy that the 2 kg bundle is slowly drifting towards her. When she has pushed a total distance of 10 m, the 2 kg bundle has moved so that it is now flush with the rear of the 8 kg block. Calculate the coefficient of friction between the two bundles of candy.

Homework Equations

There is a constant force being applied, thereforre work is being done.
W = Fcosd

The Attempt at a Solution

I can calculate the work being done by the person on the entire system.

W= 100N(cos0)(10m) = 1000J.
As well work being done by friction on the system
W = (0.1)(100N)(cos180)(10m) = -100J
I am not sure where to go from here.
The work done by friction on the 2kg box, I could then figure out the coefficient of friction, knowing the force of friction and Fn. But other then that I don't know where to go...

Any help is appreciated thank you.

 

[Texag]

First of all check your work done by the friction on the big block.

Secondly, I'm quite sure this problem can be done without bringing work into the picture.
Hint: What would the little block's acceleration be if there had been no friction between the 2 blocks?

 

[michalll]

Hint: What would the little block's acceleration be if there had been no friction between the 2 blocks?

That would be probably zero, I don't know if it is helpful in any way.

 

[Texag]

I apologize, you do have to bring work into this. The way that comes to me to do this problem is by switching reference frames. You know the what the work done on the block as it moves from rest to (10m minus the length along top of block) related to the mu(along the top). This is equal to the work done by the fictitious force minus the friction force, along the top of the block only. The fictitious force is the mass of the small block times the acceleration of the large block. You are basically equating the work done in the stationary reference frame to one which is accelerating with the large block. Somebody tell me if I have this correct.

 

[alphysicist]

I would do this problem without calculating the work. You can draw separate force diagrams for each of the blocks, and therefore get two equations, each with the coefficient of friction that you're looking for and the acceleration of each block. Then, since you can find the distance each one travels (in the same amount of time), you can find the ratio of their accelerations.

At that point you can solve for the desired coefficient. What do you get?

 

[tiny-tim]

… the system begins to move … the 2 kg bundle is slowly drifting towards her.

As a matter of interpretation of examination questions, don't those two phrases mean that both the accelerations are to be taken as zero?

 

[alphysicist]

Hi tiny-tim,

I don't think we can assume that the acceleration of the large package is approximately zero unless we know that the coefficient of kinetic and static friction between the package and the floor are approximately equal.

I was considering about how to use the fact that the smaller box has close to the same acceleration as the large box (since it appears to be drifting slowly backwards), but the coefficient that we are looking for would be related to how close those accelerations are (in terms of their ratio), so I concluded the direct (exact) method would be best.