Work-Energy Theorem problem

[jackthehat]

Homework Statement

Hi everyone,
I have a problem that has me stumped and would appreciate some pointers as to where I am going wrong and maybe point me in the right direction for solving the problem.
The problem is in essence to use the "Work-Energy Theorem" to find the co-efficient of kinetic friction in a pulley system.

Problem - We have an 8.00 kg-block on flat horizontal tabletop attached via a rope and pulley to a hanging 6.00 kg-block. The rope and pulley have negligible mass and the pulley is friction-less. Initially the 6.00 kg-block is moving downward and the 8.00 kg-block is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving 2.00 meters.

Use the Work-Energy Theorem to calculate the coefficient of Kinetic friction between the 8.00 kg-block and the tabletop.

Homework Equations

Main equations I used to attempt to solve this problem were ..
Work = Change in Kinetic energy
Work = Force x Distance (for a constant force)
I used these separately for each of the masses (blocks) connected by the pulley system.

The Attempt at a Solution

Basically I used the two main equations of the Work-Energy Theorem to try to solve this. I first calculated the Work used in moving each block using the difference in kinetic energy over the distance traveled that is ...
W= K(2) - K(1) = 1/2 mv(2) sqrd - 1/2 mv(1) sqrd for each block, and since both blocks come to rest, each of the equations above reduce to just 1/2mv(1) sqrd for each block.
I then took the difference in the values for the work each block expended to be the work expended by friction force.
Now since (for a constant force) WORK also equals Force x distance, I equated the Work difference above to be equal to the work expended by the Kinetic Friction Force.
And so Work difference = Work(Friction force)=Kinetic friction x distance moved.
From my calculations I got W(8kg-block)=3.24 J, W(6kg-block)=2.43 J giving difference of 0.81 J as the Work of Friction force.
Now since W=f x distance then f=w/distance =0.81/2.0 = 0.405 J
I now have a value for Friction force (f) and I then used the relationship Friction = Coefficient of Friction x Normal force .. in this case 0.405=coefficient x mg (Normal force for 8kg-block = weight of block ie. 'mg')
So we have coefficient = w/mg = 0.405/(8x9.8) = 0.405/78.8 = 0.02 .
However the answer at the back of the book gives coefficient = 0.786.
I have tried doing this in slightly different ways and the nearest I get to the correct answer is .. 0.75 (which if you notice is just the mass-ratio between the 2 blocks) ?
So where have I gone wrong ?
can anyone help ?
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[TSny]

Hello. Welcome to PF.

The work-energy theorem states that the change in KE of the system is equal to the total work done on the system. Friction is not the only force that does work on the system.

(There is a formatting tool bar that you can use to type subscripts and superscripts: for example, (1/2)m₁v₁².)

 

[faradayscat]

Friction is not the only force doing work here.

The Work-Energy Theorem states that the total work done on a system is equal to the change in kinetic energy:

W_net = ½mv²_f - ½mv²_i

In this case, friction and tension act on the first block, say m₁, therefore:
W_friction + W_tension = ½m₁v²_f - ½m₁v²_i

See where I'm going with this?

 

[NascentOxygen]

Does your ∆E allow for the loss of P.E. as well as of K.E.?

 

[Doc Al]

Does your ∆E allow for the loss of P.E. as well as of K.E.?

Generally, the "work-energy" theorem relates the work done by all forces (including gravity) to the change in KE.

 

[HelpMePlease27]

This problem has given me a headache like nothing else, my homework is due tomorrow so I won't receive the answer in time however more people will need help this problem in the future. How do you solve for the work of the tension? It seems like to me that it is impossible without the work done by friction