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Work done by an external force problem help

  1. Nov 24, 2013 #1
    1. The problem statement, all variables and given/known data

    a rope is used to pull a 3.57kg block at constant speed a distance 4.06m along a horizontal floor. the force on the block from the rope is 7.68 N and is directed 15 degrees above the horizontal.
    a) calculate the work done by the ropes force
    b) calculate the increase in thermal energy of the block-floor system
    c) calculate the coefficient of kinetic friction between the block and floor.

    2. Relevant equations

    W=FDcos(theta)

    3. The attempt at a solution
    I solved for A by W=7.68*4.06*cos(5)and got A=30.1 but i am not sure what to do when solving for the thermal energy. For part C i know i need to calculate the Fn and am not sure how to find total friction force to solve for the coefficient.
     
  2. jcsd
  3. Nov 24, 2013 #2

    cepheid

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    For parts b and c:

    This problem says that the block moves at a constant speed. This means that its acceleration is zero. This means that the net force on it must be zero (Newton's 2nd law). In particular, then, the horizontal forces must sum to zero. You know what the horizontal force from the rope is. It's failing to accelerate the block, because it is exactly opposed by friction. Therefore, you know what the friction force is. ;)

    When friction slows down an object (or, in this case, just prevents it from accelerating), the kinetic energy that that object had (or would have gained) is converted into heat.
     
  4. Nov 24, 2013 #3
    so what your saying is that the thermal energy is equal to the kinetic energy and that the friction force is equal to 0?
     
  5. Nov 24, 2013 #4

    cepheid

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    No, that is totally wrong. Please read my post again, carefully. In the first paragraph, I tell you what the friction force is. It is not zero.

    In the second paragraph, I am saying that the work done by friction is converted to heat. So, you have to calculate the work done by friction.
     
  6. Nov 24, 2013 #5
    so then friction force is equal to N(mg)-fdcos(theta)?
     
    Last edited: Nov 24, 2013
  7. Nov 24, 2013 #6

    cepheid

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    f*d*cos(theta) has units of work, so it can't be equal to the friction force, since force and work are two different physical quantities. However, you are on the right track.

    Instead of just posting your latest guess, it would be useful if you could communicate a little more. Show us your work. What is the reasoning for your guess? (Don't just guess wildly). If I have some idea of what your thought process is, I can help you. Otherwise this is going to be a long and drawn out process.
     
  8. Nov 24, 2013 #7
    Was this what you wanted?

    you said the the total force is equal to zero. i thought that this would mean that the friction force is equal to fdcos(theta) because then the system would have no force in the X direction. i know that the friction force is also equal to U*N. it is around here when i get confused with rearranging the equations to get the answer.
     
  9. Nov 24, 2013 #8

    cepheid

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    Not fdcosθ. Just fcosθ. The "d" in there is distance. You can't have a force equal to a force times a distance. That's what I was saying in my previous post about units. You have exactly the right idea though:

    Correct! Sum of forces in the x direction = 0, which means that:

    Frope + Ffric = 0

    Ffric = -Fcosθ

    where F is the applied tension force in the rope.

    If (in terms of magnitudes) Fcosθ = μN, then all you have to do for part c is rearrange to solve for μ (isolated it on one side of the equation):

    μ = (Fcosθ)/N

    The tricky part is figuring out what the normal force (N) on the block is. To do this, you need to repeat the same procedure that you used for the horizontal (x) forces, but instead do it for vertical (y) forces. The block isn't accelerating in the vertical direction, so the sum of all vertical forces should also be 0. The equation you get from this summation will allow you to solve for N. Hint: draw a free body diagram: a picture of the block with all the forces that act on it (and nothing else). That way you have an inventory of all the forces, and you can be sure that you aren't missing any in your sum.
     
  10. Nov 24, 2013 #9
    oh i see so it would be sin instead of cos and then you solve for normal force in the Y direction minus the mg

    Thanks a ton.
     
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