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A Sled, a slope and friction

  1. Jun 24, 2009 #1
    The problem statement, all variables and given/known data
    A sled is going down a slope of 40m, which has an angle of 30 degrees with the horizontal. The total mass of the sled is 40kg. There is a constant frictional force of 40 N holding back the movement of the sled. The sled has an initial velocity of 3m.s-1. What is the net force experienced by the sled parallel to the angle? What is the maximum speed reached by the sled at the bottom of the slope?


    3. The attempt at a solution
    The part about frictional force in confusing the heck out of me, so I don't know where to start. Do I simply add Ek((1/2)(40)((3)^2)) +Ep((40)(9.8)(sin30(40)) - (40x40)?
     
    Last edited: Jun 24, 2009
  2. jcsd
  3. Jun 24, 2009 #2

    Doc Al

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    You can certainly use energy conservation; that expression is the total mechanical energy that will be left at the bottom of the incline. (You have a typo in the potential energy term.)

    Why not do as asked? What's the net force parallel to the incline? Then use Newton's 2nd law to find the acceleration. Use both methods and compare!
     
  4. Jun 24, 2009 #3
    Sorry about the typo - just fixed it. I still dont completely understand the question "What is the net force experienced by the sled parallel to the angle?" I'm having a mental block on this one for some reason. How does the frictional force effect the speed at the bottom of the slope?
     
  5. Jun 24, 2009 #4

    Doc Al

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    Just do it step by step. What forces act on the sled? (There are only three.) What are their components parallel to the slope? Add them up to get the net force parallel to the slope.
    Take a guess. If there were no friction, would the sled be moving faster or slower? (Does friction slow it down or speed it up?)
     
  6. Jun 24, 2009 #5
    So there's gravity, friction and the kinetic energy from the start? How exactly would I calculate this?

    I understand that it'll slow it down, but how exactly would I do this calculation? Surely the friction will reduce the acceleration? How do I calculate the acceleration experienced by the sled? Should I do [(9.8(40)) - (40)] / 40 to get the new rate of acceleration?

    Sorry for the stupid questions, but I know there's something obvious here I'm simply not seeing, and I cant figure it out!
     
    Last edited: Jun 24, 2009
  7. Jun 24, 2009 #6

    Doc Al

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    Kinetic energy is not a force, so skip that one. Gravity and friction are the two forces that have components parallel to the incline. The friction force is given (what direction does it act?). What's the component of gravity parallel to the incline?

    The friction reduces the net force on the sled, thus reducing the acceleration. First find the net force; then use Newton's 2nd law to find the acceleration.
     
  8. Jun 24, 2009 #7

    Doc Al

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    Not exactly. The full weight is mg, but that points straight down. What's the component of mg parallel to the incline?

    Read this: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm" [Broken]
     
    Last edited by a moderator: May 4, 2017
  9. Jun 24, 2009 #8
    Sorry: *edit:

    Well, I think I kind of got it. To get acceleration: (392N)(sin30) - (40) = Fnet. When I do this, I get an net force of 156 and an acceleration of 3.9m.s^-2?? Is that right??
     
  10. Jun 24, 2009 #9

    Doc Al

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    Perfect! Now you can use the appropriate kinematic formula to figure out the final speed.
     
  11. Jun 24, 2009 #10
    Acceleration would be: 156/40= 1.4m.s^-2
    Then the height from the ground would be: sin30(40) = 20m.

    Vf^2 = (3)^2 + 2(1.4)(20)
    Vf^2 = Sqrt(65)
    = 8.062m.s^-2

    Is that correct?
     
  12. Jun 24, 2009 #11

    Doc Al

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    You already found the correct acceleration in your last post. (Redo the arithmetic.)

    The height from the ground is not needed. The acceleration is parallel to the incline, so it's the distance along the incline that matters.

    That's the correct formula to use, but the acceleration and distance are incorrect.
     
  13. Jun 24, 2009 #12
    So the distance is in fact 40m?

    Vf^2 = (3)^2 +2(3,9)(40)
    = Sqrt(321)
    =17.916m.s^-1
     
  14. Jun 24, 2009 #13

    Doc Al

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    Yep, as given in the first line of the problem statement.
    Looks good.
     
  15. Jun 24, 2009 #14
    Ok, thanks a lot for your help. I'm doing revision for my physics (got exams coming up). Thanks again for your help. Looking over the work again, I can see how much I still need to study. I haven't given my physics nearly enough time this year.
     
    Last edited: Jun 24, 2009
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