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A box slides down with initial V, force needed to stop it?

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  1. Mar 29, 2017 #1
    1. The problem statement, all variables and given/known data
    20170329_204509_zpssbxpjqdm.jpg
    Vo = 5m/s
    W to the left = 6N
    Ffriction = 2N
    F to the left = W - Friction = 4 N
    m = 1kg
    g = 10 m/s^2

    Ramp length = 10m
    sin a = 3/5
    cos a = 4/5
    2. Relevant equations
    F = ma
    Vt ^2 = Vo^2 + 2as
    Vt = Vo + at

    3. The attempt at a solution
    Outside force (1) to cancel acceleration from weight
    = 4N
    So now a = 0

    Now we calculate how much decceleration is need to stop the initial 5m/s
    Vt ^2 = Vo ^2 + 2as
    0 = 25 + 20a
    a = 1,25 decceleratiom from outside force (2)
    F2 = m.a = 1.25 N
    So F1(to cancel initail acceleration) + F2(to cancel initial velocity) = 4 + 1.25 = 5.25N

    Is there something wrong? The answer is not in the options :/
     
    Last edited: Mar 29, 2017
  2. jcsd
  3. Mar 29, 2017 #2
    Wait why didn't the image shows up? It can't show images from google drive?
     
  4. Mar 29, 2017 #3

    gneill

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    Staff: Mentor

    If the image isn't public access the software can't get to it. It's better to UPLOAD images to our server.
     
  5. Mar 29, 2017 #4
    Hi. Welcome to Physics Forums.
    Where did this come from? Is it part of the problem statement?
     
  6. Mar 29, 2017 #5
    Okay i updated it
     
  7. Mar 29, 2017 #6

    gneill

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    Staff: Mentor

    Can you provide a translation of the problem statement? It's not obvious how long the force is allowed to operate on the block. Is it applied during the entire motion?
     
  8. Mar 29, 2017 #7
    So you need 4 N (in addition to the 2 N friction force) to counter the down-the-slope weight component.
    And you need 1.25 N to produce an up-the-slope acceleration.
    4 + 1.25 = 5.25 N.
    I can't find anything wrong with that. But it is early in the morning for me. :)
    Possibly the problem creator had a sign error and came up with 4 - 1.25 = 2.75 (which rounds to 2.8)

    In response go @gneill comment (which just came through), I assumed the force was applied for the entire 10 m of the slope.
     
  9. Mar 29, 2017 #8
    It is not mentioned at all. Everything i typed is basically the translation in math equations
     
  10. Mar 29, 2017 #9
    I typed the words into Google translate and the relevant part of the translation was:
    "so that the right thing stopping at the bottom end of the track"

    Per my previous post, I think your answer is correct.
     
  11. Mar 29, 2017 #10

    gneill

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    Staff: Mentor

    In future please make sure that the problem has been described in English in the problem statement section. Translations of relevant material in images is fine. Note that this is a forum rule.
     
  12. Mar 29, 2017 #11

    gneill

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    Staff: Mentor

    Your answer may end up closer to 5.0 N if you use a more accurate value for g.
     
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