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Work and Energy with friction. Due tomorrow!

  1. Sep 29, 2008 #1
    A box slides across a frictionless floor with an initial speed v = 3.5 m/s. It encounters a rough region where the coefficient of friction is µk = 0.6.
    a) What is the shortest length of rough floor which will stop the box?


    b) If instead the strip is only 0.41 m long, with what speed does the box leave the strip?


    I'm not sure how to start this problem. I'm kind of lost with work and energy. This is due tomorrow morning so any help is greatly appreciated.
    Thanks!
     
  2. jcsd
  3. Sep 29, 2008 #2
    Draw a free body diagram. What is the retarding force acting on the box?
     
  4. Sep 29, 2008 #3
    [tex]\mu[/tex]mg is acting in the negative x direction. V is in the positive and mg and N are opposing each other in the y direction. Is that correct?
     
  5. Sep 29, 2008 #4
    V is not a force so technically you should not put it on a free body diagram. However, in this case, just remember that the only horizontal force is the frictional force. Because you know this force, tell me what is the acceleration of the box?
     
  6. Sep 29, 2008 #5
    Hmm...I'm not sure. Can you walk me through this please?
     
  7. Sep 29, 2008 #6
    Fnetx = ma = the net force acting in the horizontal direction. What is the only force acting in the horizontal direction?
     
  8. Sep 29, 2008 #7
    The only force acting in the horizontal direction is the coefficient of friction.
     
  9. Sep 29, 2008 #8
    I think you mean the only force is the frictional force* But yes, that's the only force acting in that direction. So then what is ma in the horizontal direction?
     
  10. Sep 29, 2008 #9
    mu=ma. Fnetx is mu because its the only horizontal force right?
     
  11. Sep 29, 2008 #10
    mu*mg. Mu is not a force, it's only the coefficient of friction. Frictional force is mu times the normal force, which in this case is mg.

    So yes, mumg = ma. Then the m's cancel and you're left with acceleration.
     
  12. Sep 29, 2008 #11
    ok so i get a=5.886 m/s^2. so now i have a in the positive x direction and mumg in the negative x direction. How do i implement the Work-Energy Theorem? I know the equation is Wnet=K2 - K1. And K=1/2mv^2. i know i don't have the mass so i think i have to implement mu and change in x somehow. is that correct?
     
  13. Sep 30, 2008 #12
    If you build an equation with the work of the friction force on one side and the change in kinetic energy on the other side, you'll see that the mass of the object doesn't really matter. You can actually conclude from that, the the stopping distance of a moving object (in an ideal environment) has nothing to do with its mass.
     
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