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Object sliding down an inclined plane

  1. Nov 30, 2014 #1
    1. The problem statement, all variables and given/known data
    First of all, here is how it goes: an object with a starting velocity of 3 m/s reaches the height of 0.3m (inclined plane) and then it "stops" and slides down. What will be the velocity when it reaches the bottom of the inclined plane ?
    (Sorry if the terms are wrong, English is not my native language :P).

    2. Relevant equations
    a=g(sin*alpha-mu*cos*alpha) ??
    v^2=v0^2-2a*h ??
    v^2=2as ??
    i seriously dont know :/

    3. The attempt at a solution
    I seriously tried a lot of things. The problem is, I don't have enough data to calculate it (or do i?, im fairly new so i dont know). No angle, no length, no time, no acceleration etc...
     
  2. jcsd
  3. Nov 30, 2014 #2

    haruspex

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    You're right, there's not enough information. Have you quoted the whole question word for word?
    Does it start at the bottom of the slope? If not, you'll need to know how high it starts.
    Is there any friction?
     
  4. Nov 30, 2014 #3
    Its pushed from the bottom, starting speed of 3m/s and reaches the height of 0,3m. No info on friction.
     
  5. Nov 30, 2014 #4

    PeroK

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    If you assume the object starts at the bottom of the slope and that there is friction on the slope, then you should have enough information to solve this.

    Think about acceleration going up the slope compared with acceleration coming down the slope. The formula you have above is correct for one of these.
     
  6. Nov 30, 2014 #5
    I still dont quite get it, can you solve it ?
     
  7. Nov 30, 2014 #6

    PeroK

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    Yes. You need to work through the equations and trust that the unknowns (angle and coeff of friction) will cancel out, which they do!

    You have the necessary equations above.
     
  8. Nov 30, 2014 #7
    All 3 ? Or just the 2nd and 3rd ?
     
  9. Nov 30, 2014 #8

    PeroK

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    You only need ##v^2 = 2as##. And, of course, the formula for acceleration up and down.
     
  10. Nov 30, 2014 #9
    But i dont have s, only h (height)
     
  11. Nov 30, 2014 #10

    haruspex

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    Which is exactly what you need. Put in unknowns for whatever you don't know (coefficient of friction, angle of slope) and see what happens.
     
  12. Dec 1, 2014 #11
    I don't get it xD
     
  13. Dec 1, 2014 #12

    PeroK

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    It's possible to solve physics problems when everything is unknown: initial speed, angle, coeff of friction, height gained. The solution, however, will be a function of all the variables. That, after all, is what the equations of motion are: general equations, where all quantities are unknown.

    So, it must be possible to solve this problem. The answer will be a function of all the variables. So, it might be something like:

    v (at bottom of slope) = velocity at start - gh*coeff of friction*tan(angle) - ... (some long complicated function)

    There's nothing stopping you working out this formula, even if it gets very complicated.

    But, it's possible that it doesn't get complicated at all! It's possible that things like angle and coeff of friction might cancel out and leave a simple formula.

    How I started was this:

    Let ##u## be the initial velocity at the bottom of the slope (u = 3m/s)

    Let ##h## be the height gained (h = 0.3m)

    Let ##\theta## be the angle of the slope

    Let ##s## be the distance travelled up the slope

    Let ##\mu## be the coefficient of friction.

    Let ##g## be the acceleration due to gravity.

    Let ##v## be the final velocity at the bottom of the slope.

    I then worked out ##v## in terms of ##u, h, \theta, s, \mu \ and \ g##

    But, ##\theta, s, \mu## all cancelled out, leaving me with a nice equation for ##v## in terms of ##u, g \ and \ h##

    Then, of course, I plugged ##u = 3## and ##h = 0.3## into the equation.

    This is a good problem, as it shows the power of maths and general formulas to solve problems, even if you only know some of the variables.
     
  14. Dec 1, 2014 #13
    So there isnt a "real" solution to this ? Like, the velocity when the object reaches the bottom of the slope is 7 m/s (random number) ?
     
  15. Dec 1, 2014 #14

    PeroK

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    One answer to that question is that you'll never find out unless you try to solve the problem yourself.

    But, yes there is an answer in terms of an actual number of m/s. If that's important to you!
     
  16. Dec 1, 2014 #15
    I seriously dont understand what formula to use. This isnt a homework or anything btw, i found this in a book and it got me interested so yeah. But seems like its un-solvable, atleast for me. Im only 15 tho, havent studied about half of the things required to solve this :/
     
  17. Dec 1, 2014 #16

    PeroK

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    You worked out ##a = gsin(\theta) - \mu gcos(\theta)##

    That is actually the acceleration on the way down. So, can you see what the deceleration is on the way up?

    Then just use ##u^2 = 2as## for the motion up and ##v^2 = 2as## on the way down.

    (I tend to use all positive quantites for a problem like this, as I find it easier. So, everything above is positive.)
     
  18. Dec 1, 2014 #17
    Yeah but, what is a then ?
    I am very confused, sorry xd
     
  19. Dec 1, 2014 #18

    PeroK

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    Maybe this problem is a bit hard.

    On the way up, both gravity and friction are slowing the block, so we have:

    ##a_1 = gsin(\theta) + \mu gcos(\theta)##

    On the way down, gravity is accelerating the block and friction is slowing it down, so we have:

    ##a_2 = gsin(\theta) - \mu gcos(\theta)##

    Do you want to try to progress it from there?
     
  20. Dec 1, 2014 #19
    I understand that, but we DONT KNOW what the sin and cos of the angles are, that's the thing.
     
  21. Dec 1, 2014 #20

    PeroK

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    Neither do I. And I don't know what ##\mu## is either. You can't let an unknown angle stop you!

    Seriously, just keep going. You'll get formulas with ##\theta## and ##\mu##for a while, but eventually you can get rid of them.
     
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