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Find the distance traveled by a block pushed up a slope.

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data
    "A block is projected up a frictionless inclined plane with initial speed v0 = 3.50 m/s. The angle of incline is θ = 32.0°. (a) How far up the plane does the block go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?"

    2. Relevant equations
    Fg = mg⋅sinθ
    Fx = -Fg
    Assume: arbitrary mass of block is 1-kg

    3. The attempt at a solution
    (b) Fg = (1kg)(9.8m/s2)(sin (-32°)) = -5.1932N
    5.1932N = (1kg)(3.5m/s)/t
    t = (3.5 kg⋅m/s)/5.1932N = 0.674 s

    I've only figured out how to find the time it takes for the block to reach that point on the slope before it slips back down. The book says its speed when it gets back to the bottom is 3.50 m/s; same as the initial velocity. I have a vague idea how to solve for how far the block goes. Basically, I figure that once v0 = 0, it has reached the maximum point it can reach at the initial velocity at 3.50 m/s before it comes down due to gravity, anyway. At that point, gravitational force and the force pushing the box should be equal. But I have no idea how I would apply this.
     
  2. jcsd
  3. Jan 31, 2016 #2

    cnh1995

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    You don't need mass here. Just use basic kinematical equations. You have initial velocity, acceleration and final velocity. Which kinematical eauation connects these three?
     
  4. Jan 31, 2016 #3
    It's not in my book, but I think it's: "d = v0t + ½at2"? Okay, I've just done it...

    d = (3.5 m/s)(0.674 s) + ½[(0 m/s - 3.5 m/s)/(0.674 s - 0 s)](0.674 s)2
    = (2.359 m) + ½(-5.1932 m/s2)(0.674 s)2
    = 2.359 m - 1.180 m = 1.18 m


    Now why is the velocity as the block slides down the same as the initial velocity, can I ask?
     
  5. Jan 31, 2016 #4

    cnh1995

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    Right. I expected v2=v02+2ad, without involving time parameter. But since time is also asked, your method is right.
    The block starts sliding down from rest i.e. v0=0. You need to use the exact same equation to get final velocity now(part c).
     
  6. Jan 31, 2016 #5
    To plug in acceleration, though, I would need the final velocity, which is the whole point of the equation, no?
     
  7. Jan 31, 2016 #6

    cnh1995

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    You are asked to find the final velocity. You do have acceleration. What is pulling the block and making it slide downward?
     
  8. Jan 31, 2016 #7
    Oh, gravity.
    a = sin(-32°)(9.8 m/s2) = -5.1932 m/s2
    v2 = v02+2ad
    = (0)2 + 2(|-5.1932| m/s2)(1.18 m)
    v2 = (12.26 m2/s2)
    √v2 = √(12.26 m2/s2)
    v = 3.5 m/s

    Additionally, how would I go about making my equations look something like: https://upload.wikimedia.org/math/f/9/1/f9160e375f1f6fc949a99985fb1e6a72.png? I've seen it done, and I want to know if there's a guide that tells me how to do that.
     
  9. Jan 31, 2016 #8

    cnh1995

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  10. Jan 31, 2016 #9
    No, I meant the font styling. How do I change it?
     
  11. Jan 31, 2016 #10

    cnh1995

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    Oh:-p! I thought you wanted to derive kinematical equations using calculus.. I am not sure but it's LaTex. You'll find the LaTex/BBcode guide under the editor space here (below "have something to add?" space).
     
  12. Jan 31, 2016 #11
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