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Physics 1 Test Problem

  1. Oct 12, 2011 #1
    1. The problem statement, all variables and given/known data
    I took a test today and I simply have to know if I got this question right! Otherwise it will haunt me for the whole weekend.

    vil3tk.jpg
    A block is at the bottom of an inclined plane with angle 30 from the horizontal. The block has an initial velocity of 5 (from pushing or something, I forget), and the kinetic friction constant is 0.4. What distance does it travel before it stops?

    I have no idea why they put that height of 10m in the figure. I didn't use it at all.

    2. Relevant equations

    3. The attempt at a solution
    I did my free-body-diagram like so:
    14mgwt5.jpg
    (I forgot to add to that image, I made positive x to the right, and positive y up)

    Then I went to work on Newtons second law to see what I could get.

    [itex]ƩF_{x} = ma[/itex]

    [itex]0 - mgsin(30) - f_{k} = ma[/itex]

    [itex]-mgsin(30) - \mu_{k}N = ma[/itex]

    [itex]-mgsin(30) - \mu_{k}mgcos(30) = ma[/itex]

    [itex]-mgsin(30) - (0.4)mgcos(30) = ma[/itex]

    [itex]-gsin(30) - (0.4)gcos(30) = a[/itex] //Divided out the mass

    [itex]-(9.8)sin(30) - (0.4)(9.8)cos(30) = a[/itex]

    [itex]-8.29 = a[/itex]

    Then, using kinematics:

    [itex]\upsilon_{f} = \upsilon_{i} + 2a \Delta x[/itex]

    [itex]0 = 5 + 2(-8.29) \Delta x[/itex]

    [itex]-5 = 2(-8.29) \Delta x[/itex]

    [itex]\frac{-5}{2(-8.29)} = \Delta x[/itex]

    [itex]0.302 = \Delta x[/itex]

    So my solution was 0.302 meters. I may have made a mistake re-working this directly in LateX, but if that is the right idea for this problem I should have got that one on the test. Was I incorrect? I also made sure to specify that my delta x was across the hyp of the inclined plane.
     
  2. jcsd
  3. Oct 12, 2011 #2

    NascentOxygen

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    Staff: Mentor

    I get that answer, too.
     
  4. Oct 12, 2011 #3

    gneill

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    Staff: Mentor

    Shouldn't that kinematic formula be:

    [tex] {\upsilon_{f}}^2 = {\upsilon_{i}}^2 + 2a \Delta x [/tex]
     
  5. Oct 12, 2011 #4

    NascentOxygen

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    Staff: Mentor

    Yes, it should be. I didn't check OP's working; I calculated independently.

    Amazingly, I overlooked squaring the 5, in effect the same mistake, so my value is out by a factor of 5. My answer is now 1.5m up the slope.

    Thanks for pointing out QuarkCharmer's mistake, it made me review my working.
     
  6. Oct 12, 2011 #5
    Oh yeah, you are right haha. I confused it with the one that is:
    v_f = V_i + at

    But I am sure I used the correct one on the test (I specifically remember writing 0^2 as one line). I just did this one from what I could remember in the physicsforum.com input box.
     
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