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Finding coefficient of kinetic friction from thermal energy?

  1. Oct 15, 2016 #1
    1. The problem statement, all variables and given/known data
    "A rope is used to pull a ##\frac{357}{100}kg## block at constant speed ##\frac{203}{50}m## across a horizontal floor. The force on the block from the rope is ##\frac{768}{100}N## and directed ##15°## above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and the floor?"

    2. Relevant equations
    ##W=ΔE_mec+ΔE_th=(ΔK+ΔU)+ΔE_th##
    ##ΔU=0##
    ##K=\frac{1}{2}mv^2##
    ##f_k=(mg)(sin\theta)##

    3. The attempt at a solution
    Wwwah4q.png
    (a) ##W=ΔK=F⋅Δx=(F)(cos\theta)(Δx)=(\frac{768}{100}N)(cos15°)(\frac{203}{50}m)=30.12J##.

    (b) Then I use the relation ##W_{block-floor system}=-ΔE_{mec}+ΔE_{th}=0## to equate those two terms to each other, since the block releases mechanical energy, and gains thermal energy. Someone please correct me because I'm very sure that I am incorrect with this reasoning.

    ##ΔE_{th}=ΔE_{mec}=30.12J##

    (c) Now I move onto finding the coefficient of static friction using the relation:

    ##ΔE_{th}=(f_k)(Δx)=(mg)(sin\theta)(μ_k)(Δx)##
    ##μ_k=\frac{ΔE_{th}}{(mg)(sin\theta)(Δx)}=\frac{(F)(cos\theta)(Δx)}{(mg)(sin\theta)(Δx)}=\frac{(F)(cot\theta)}{mg}=0.82##

    I don't understand what it is that I'm doing wrong.
     
  2. jcsd
  3. Oct 15, 2016 #2

    haruspex

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    Please explain these factors.
     
  4. Oct 15, 2016 #3
    That's the normal force.
     
  5. Oct 15, 2016 #4

    haruspex

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    How do you deduce that?
     
  6. Oct 15, 2016 #5
    It acts perpendicular to the motion. So I suppose that it is just ##F_N=mg##.
    Edit: According to Wikipedia, it is "perpendicular to the surface of contact".
    I'm still not getting the right numbers:

    ##μ_k=\frac{(F)(cos\theta)}{mg}=0.212##

    Sorry for the multiple edits.
     
    Last edited: Oct 15, 2016
  7. Oct 15, 2016 #6

    haruspex

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    Better, but still wrong. What forces acting on the block have at least some vertical component?
     
  8. Oct 15, 2016 #7
    I'm guessing it's the rope tugging on the block. So is normal force: ##F_N=(mg)-(sin\theta)(F)##?

    ##μ_k=\frac{(F)(cos\theta)}{(mg)-(sin\theta)(F)}=0.225##

    Awesome. Thanks!
     
  9. Oct 16, 2016 #8

    haruspex

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    Ok!
     
  10. Oct 16, 2016 #9
    Oh, one more thing. I need to verify if this is correct or not:

     
  11. Oct 16, 2016 #10

    kuruman

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    It looks like you are missing the point of the energy conservation equation. The block is moving at constant velocity on a horizontal surface. Just on that piece of information, what is its change in mechanical energy?
     
    Last edited: Oct 16, 2016
  12. Oct 16, 2016 #11
    Let's see... if there is no change in velocity, there is no change in kinetic energy. So assuming no change in potential energy, the mechanical energy does not change?
     
  13. Oct 16, 2016 #12
    Hold on; this is a flat surface. There is no (gravitational) potential energy.
     
  14. Oct 16, 2016 #13

    kuruman

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    Correct. The mechanical energy does not change, i.e. ##\Delta E_{mec}=0##. If the relation ##W_{block−floorsystem}=−ΔE_{mec}+ΔE_{th}=0## were correct, it would imply that ##\Delta E_{th}=0## which is not happening. Look again at the very first equation you posted under "Relevant Equations". It becomes ##W=ΔE_{th}##. That's the equation to start from. It says that the work done on the block is entirely converted into thermal energy when the mechanical energy is not changing.
     
  15. Oct 17, 2016 #14
    I see. Thank you for taking the time to explain this to me. It's hard to learn this without collaboration.
     
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