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What is v and delta X after switching between frictions?

  1. Oct 10, 2016 #1
    1. The problem statement, all variables and given/known data
    A wooden pallet carrying a load of 600 kg rests on a wooden floor.
    (s= .28 and k= .17)
    a. A forklift driver decides to push it horizontally instead of lifting it. What force must be
    applied to just get the pallet moving from rest?
    b. After a bit of time, the driver pushes the pallet and it slides. How fast is the pallet
    moving after 0.5 seconds of sliding under the same force you calculated in part a?
    Draw a free body diagram – don’t forget about friction!
    c. If the forklift driver stops pushing at 0.5 seconds, how far does the pallet slide before
    coming to a stop?

    2. Relevant equations
    Fs= us (Fn)
    Fk= uk (Fn)
    F= ma
    V^2 = vo^2 + 2 a (change in X)
    V=vo+ at

    3. The attempt at a solution
    A) I know that to just get the pallet moving from rest means right when it moves so it's not moving, making it static friction so I did: Fs = .28 (600kg X 9.81 m/s^2) = 1648 N
    B) I did f= ma because the problem said under the same force from part A so 1648 N = 600 kg (a) and got a = 2.75 m/s^2
    Then I did v = vo + at so it becomes v = 2.75 m/s^2 (.5 s) = 1.37 m/s^2. However my answer is wrong because I didn't use Fk= uk as it is now kinetic friction. The correct answer is supposed to use Fs - Fk= ma and is a = 1.07 m/s^2 and v = .54 m/s but I don't know why. Why do you subtract the two frictions and what happened to using the Applied force?
    C) I'm using the correct velocity (.54 m/s) and plug it into V^2 = vo^2 + 2 a (change in X) but I don't get the right answer which is .0875 m. I did (.54 m/s)^2 = 0 + 2 (1.07m/s^2)(change in x) and got change in x = .136 m. I do not know how to get the correct answer if it's the right equation.
     
    Last edited: Oct 10, 2016
  2. jcsd
  3. Oct 10, 2016 #2

    gneill

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    Staff: Mentor

    The applied force is equal to the static friction force. Remember, the driver applied just enough force to get it moving, and continued to apply that same force for 0.5 seconds once it started to move.
    What value did you use for the acceleration while the pallet was sliding by itself?
     
  4. Oct 10, 2016 #3
    Once you overcome static friction, the force you are pushing with (1648 N) doesn't change (according to the problem statement), but the friction force drops to about 1200 N, so there is now a net force of about 450 N. Did you draw a free body diagram like they asked?
    Show us the details of your calculation please.
     
  5. Oct 10, 2016 #4
    I did (.54 m/s)^2 = 0 + 2 (1.07m/s^2)(change in x) and got change in x = .136 m but the correct answer is .0875m so I don't know how. Also why does the friction force drop that much?
     
  6. Oct 10, 2016 #5
    I used a = 1.07 m/s^2.
    Also
    I used 1.07 m/s^2.
     
  7. Oct 10, 2016 #6

    gneill

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    Staff: Mentor

    What forces are operating while the box is sliding? Are they the same as when it was being pushed?
     
  8. Oct 10, 2016 #7
    Actually, I made a mistake in arithmetic. The friction force drops to 1000 N, and the net force is 647 N.

    Is your question, "why is the coefficient of kinetic friction smaller than the coefficient of kinetic friction?"

    I match your solution on the distance. So, I'm guessing that the "correct answer" is "incorrect."
     
  9. Oct 10, 2016 #8
    There is kinetic friction, you stopped applying force which was equal to Fs so those cancel out. Gravity and normal force cancel out too. So the Fnet is kinetic friction, when it was being pushed it was Fs-Fk so it's different.
     
  10. Oct 10, 2016 #9
    You don't stop applying the force after you overcome kinetic friction. What do the words "under the same force you calculated in part a" mean to you?
     
  11. Oct 10, 2016 #10
    Okay thank you!
     
  12. Oct 10, 2016 #11

    gneill

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    Staff: Mentor

    Only kinetic friction is operating (horizontally) when the pallet is sliding freely, no longer being pushed by the forklift. So the net force then is 1000 N. That alone dictates the acceleration for part c. I get the answer expected: d ≅ 0.087 m .
     
  13. Oct 10, 2016 #12
    Is the acceleration negative because it is slowing down? And to solve for change in x, you do not use .5 s at all?
     
  14. Oct 10, 2016 #13

    gneill

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    Staff: Mentor

    The acceleration is negative. The basic kinematic formula still applies:

    ##v_f^2 - v_o^2 = 2 a d##

    Just assign your velocities to the correct variables :smile:
     
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