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Work problem

  1. Jun 20, 2005 #1
    A 8 kg block is being pulled horizontally on a table by a 6 kg block that hangs in the air by a cord fasten to the 8 kg block over a pulley, there is only friction between the table and the 8 kg block. Both blocks are initially moving with 0,900 m/s and they come to rest after 2 meters.
    I have to use the work-energy theorem to find the coefficient of kinetic friction.

    What I've done so far is:

    total work done on the 6 kg block must be -0.5*6*0.9*0.9 = 2.43 J
    So the net force must be -2.43J/2m = -1,215N
    Fy=-mg+T=-1,215 so the tension is 57.58 N

    Total work done on the 8 kg block must be -0.5*8*0.9*0.9 = -3.24 J
    So the net force on the block must be
    coefficient of friction must be 55.96/8*9.8=0.714

    I don't get the right answer which is 0.786

    Could someone please give me a hint to what I am doing wrong.

    With thanks in advance
  2. jcsd
  3. Jun 20, 2005 #2

    Doc Al

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    I suggest starting over and thinking in terms of the total change in energy of the system. What's the change in KE? What's the change in PE? The total change in energy must equal the work done by the friction.
  4. Jun 20, 2005 #3
    I have calculated the total kinetic energy change by using 8+6=14 as the mass and the speeds 0.9 m/s and 0 m/s That energy change is not enough to explain the friction. The book I'm reading has not mentioned potential energy so I am not supposed to use that to solve this. Sorry, I am not any closer to solving this.
  5. Jun 20, 2005 #4
    book's answer 0.786 correct

    reason your approach is not working: you're only accounting for change in kinetic energy, when in fact potential energy is also changing:

    friction work = ΔKE + ΔPE

    you calculated ΔKE correctly.
    you now need ΔPE

    fyi, in this case:
    ΔPE = mgΔH(8 kg block) + mgΔH(6 kg block)
    where ΔH is height change (0 m for 8 kg block, and 2 m for 6 kg block).

    you must account for ΔPE to get correct answer
    Last edited: Jun 20, 2005
  6. Jun 20, 2005 #5
    O.K I got it right, including the PE, geosonel. But I don't get why I have to include this energy. I can see that mg*s for the 6 kilo block is the work that is done on that block, but I did already calculate the KE for it, or -0.5*m*v*v. So the friction does work to slow the 6 kilo block down and the 8 kilo. Please explain to me :redface:
  7. Jun 20, 2005 #6

    Doc Al

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    Actually, your original method was fine. You just made an error in signs, confusing the direction of the forces.
    Realize that the net force is upward. Calling up positive, your force equation should be: T - mg = +1.215. Recalculate the tension accordingly.

    Once you use the correct tension, you'll get the right answer.

    If you had covered PE, then you could apply the "Work-Energy" theorem using total mechanical energy (KE + PE). I think that way is a bit easier, since the PE automatically takes care of gravity. But no matter, you can solve the problem just like you did, explicitly including the work done by gravity.

    Another little trick is to realize that the work done by the tension must be zero. It does positive work on one block, negative work on the other: the total is zero.
    Last edited: Jun 22, 2005
  8. Jun 21, 2005 #7
    Thank you, Doc Al. I understand this now.
    Thanks guys.
  9. Jun 21, 2005 #8


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    I'm thinking your parenthetic statement is too broad. It would be true of internal forces of a rigid object, or a system with no relative motion between masses (effectively the case in this problem) but does not apply to some cases where there is relative motion, such as inelastic collisions. Sometimes the internal forces are conservative and can be related to a potential energy, but not always.
  10. Jun 22, 2005 #9

    Doc Al

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    You are absolutely correct, Dan. Very sloppy of me. :blushing: (I will remove that statement from my earlier post.) Thanks for the correction.
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