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Box dragged along resistant surface, work = ?

  1. Oct 1, 2006 #1
    box dragged along resistant surface, work == ??

    1) We have a box with mass m being dragged over a rough horizontal surface by a constant force F at an angle @ above the horizontal; it travels d meters and there is a kinetic frictional coefficient of u. What is the work done on the block?

    2) What is the work done by the force of friction.

    3) Determine it's final speed assuming it's initial velocity was 0.

    Attempted Answer:

    Fx = F cos @ - uN = m a_x (1)
    Fy = F sin @ + N - m g = 0 (2)

    From (2) we have N = m g - F sin @, substituting into (1) we have

    F cos @ - u (m g - F sin @) = m a_x

    Therefore, W = [ F cos @ - u (m g - F sin @)] d

    But this is the wrong answer?!?!

    2) I don't know how to set this up; what are we solving for?

    3) If I can come up with the right answer for (1), I can determine the acceleration and determine it's final speed via one of the one-dimensional motion equations.

    please help! :-)
  2. jcsd
  3. Oct 1, 2006 #2
    Think about it this way: how much work must be done on the block for it to move d meters?

    Pretend that there isn't friction. The amount of work done on the block is now [tex]\vec{F} \cdot \Delta \vec{x}[/tex], or F*d*cos@.

    However, there is friction. So the amount of work done on the block is greater than it would have been if there were no friction.

    Look at your expression:
    W = [ F cos @ - u (m g - F sin @)] d

    This means that it takes less work to move the block if there is friction, which does not make much sense.
    You are solving for the work done by friction. Use the relation that [tex]W = \vec{F} \cdot \Delta \vec{x}[/tex] to calculate the work done by friction. Don't forget that friction is a nonconservative force.

    You don't need to answer (1) to get the answer for (3). You can already find the acceleration.
  4. Oct 1, 2006 #3
    Thank you for your reply and aid. I solved (2) and (3), but have yet to correctly answer (1). I tried W = [ F cos @ + u (m g - F sin @)] d but to no avail. Any ideas?
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