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Homework Help: Block jump from ramp

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A block starts at rest and slides down a frictionless track except for a small rough area on a horizontal section of the track
    It leaves the track horizontally, flies through the air, and subsequently strikes the ground. The acceleration of gravity is 9.8 m/s2 .

    What is the speed v of the block when it leaves the track?

    mass of block =0.401kg
    Starting height of block = 4.1m
    Horizontal height of block when it encounters the friction = 1.9m
    Length of the rough area on the horizontal section of track = 1.3m
    Coefficient of friction = 0.3

    2. Relevant equations

    Kinitial +Uinitial = Kfinal +Ufinal



    3. The attempt at a solution

    I set Kinitial and Ufinal to zero, yielding:
    mgh=0.5mv2d - Ffrictiond
    vfinal= [ [itex]gh+μNd/0.5(1.3)[/itex] ]1/2

    Just wondering if I went about this correctly,
  2. jcsd
  3. Feb 16, 2013 #2


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    hi racecar12! :smile:
    i'm confused :confused:

    can you check that, and write it out again, please​
  4. Feb 16, 2013 #3
    Thanks for looking at this post.

    Here was my though process as I approached this problem....

    We'll just use the conservation of energy theorm for this problem.


    Since the massive object has no initial kinetic energy at the start of the system, call it when the object is at its initial height, it has potential energy equal to mgh. m is the mass of the object, g is the acceleration due to gravity, and h is equal to the distance the object travels. In this case, I have to set h equal to a change in the initial height if the object to its height when it encounters the frictional surface. Δy=h. The left side of the equation now reads: Uinitial=mgh. Or just, mgh

    mgh=some final kinetic energy - the work done by friction
    We have an equation that relates kinetic energy with velocity...it is 0.5mv2=Kfinal...v2 is also the final velocity, which for this system, is right at the point when object leaves the surface. The object will travel in the air for some unknown distance at some other velocity, but for purposes of this problem, we won't worry about them, because those things are not what this problem is asking for. Now, I'm not sure about multiplying, by length d, but I believe we have to.

    I also know that the work done by the frictional force equals μNd. That's because work equals force x distance. We have μ and d given to us. d in this case is the distance the object travels along the rough surface only. N is unknown, but we do know that in this case N =W, which happens to = mg in this case. I may not have told you that the rough surface is flat. The minus sign comes in because the frictional force opposes the motion of the object.

    Now our equation looks like this
    mgh = 0.5mv2d - 1.532622

    Now lets isolate v. My m will cancel when I move everything over to the other side.

    (gh + 1.532622)/(0.5)(d) =v2
    PlUg it all in and take the square root...v = 5.96 m/s

    And sorry that the + is underlined..I'm not sure why it does that when I type it in.
  5. Feb 16, 2013 #4


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    ah, that's what was puzzling me

    no, you don't multiply by d (i've no idea twhere you got that idea from)

    KE = 1/2 mv2, that's all!

    try again :smile:
  6. Feb 16, 2013 #5
    Yes, you are correct that kinetic energy equals 0.5mv2. However, removing d from the equation takes me much farther from the answer, which I know to be 5.95617m/s. if I proceed with the way that I have thought, I get v to equal 5.96046. Close, but no cigar.

    I am making a mistake somewhere in my thought process, but I don't know where it is. Perhaps I'm making multiple mistakes, but I don't know where or what they are. Thanks
  7. Feb 16, 2013 #6


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    Can't tell where your error is unless you post the full details of your calculation.
  8. Feb 16, 2013 #7

  9. Feb 16, 2013 #8
    You don't need the "d" term and try adding the 1.5 term instead of subtracting since there's thermal energy at the bottom of the ramp.
  10. Feb 16, 2013 #9
    Hey, thanks for your constructive contribution to my inquiry. I attempted the solution in the way you suggested. I get v to equal 6.3228, which is unfortunately not the correct answer.

    I'm wondering if I may have to consider the total length the object travels...like adding Δy to the length of the rough surface.
  11. Feb 17, 2013 #10
    Really? The equation i get is mgh = 0.5mv2 + umgd
    Solve for v and you should get about 5.956173268m/s.
  12. Feb 17, 2013 #11
    Oh man, you are correct. I made an algebra mistake while moving everything over to solve for v. Good catch. And thanks a bunch!
  13. Feb 17, 2013 #12


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    It had been pointed out to you that mv2d was wrong. You replied that you tried taking out the d and it made matters worse, but you didn't post the calculation. There's also the matter of where 1.532622 came from. I don't see an explanation of that.
  14. Feb 17, 2013 #13


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    (just got up :zzz:)
    yes, seriously
    that is why we usually insist on you showing your full calculations

    please do so next time, and save yourself some time! :smile:
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