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Block On Incline Projected Up With Initial Speed

  1. Nov 10, 2011 #1
    The problem statement, all variables and given/known data

    A block with mass m = 7.4 kg slides down an inclined plane of slope angle 43.8 ° with a constant velocity. It is then projected up the same plane with an initial speed 3.10 m/s. How far up the incline will the block move before coming to rest?


    The attempt at a solution

    You need to find the force parallel of the block on the incline, so

    Fpara = mgsinθ
    Fpara = (7.4)(9.8)sin(43.8)
    Fpara = 50.194 N

    Acceleration down is
    a = F/m
    a = 50.194/7.4
    a = 6.783

    If the block begins moving up at Vi = 3.10 m/s, and Vf = 0, then
    Vf^2 = Vi^2 + 2ad
    0 = 3.1^2 + (2)(-6.783)d
    d = -(3.1)^2 / -13.566
    d = 0.708

    This is not correct. Obviously, I'm doing this wrong, but I'm not sure of any other way to do it. I feel like having such a large acceleration down can't possibly be correct, and if that is indeed the case, how do I find acceleration of the block as it goes up the incline?
     
  2. jcsd
  3. Nov 10, 2011 #2
    You stated that the block slides at a constant velocity down, and since gravity is acceleration that means there must be a frictional force acting on the block which you must find. If the block is moving at a constant velocity downwards this means that Ftot has to equal zero, so this will allow you to calculate the force of frictiona and the coefficient of kinetic friction.

    F// = Fpara = mgsinθ = Ff = μkFN

    so you need to solve for the FN force normal to find μk and the normal force is perpendicular to the plane.

    [You do not really need to find μk just the force of friction.] Knowing the force of friction and the force of gravity acting against the block as it moves up you will be able to solve for the acceleration of the block and then be able to find the distance at which it stops at.



    EDIT*** what you need to do is now the force of friction and force of gravity are acting in the same direction and initially when the block is sliding down they are equal and opposite but when the block is sliding up then they are equal and the same. So you need to sum them up and then solve for the acceleration
     
  4. Nov 10, 2011 #3

    As it doesn't mention friction in the problem, aren't we supposed to assume that the incline is frictionless?

    Either way, I'm working it out like you said now.
     
  5. Nov 10, 2011 #4
    if it was frictionless then the block would be accelerating down the plane and not moving at a constant velocity.
     
  6. Nov 10, 2011 #5
    Good point.

    Since on the way down,
    Fpara = Ff
    50.194 = 50.194

    and both of them are acting in the same direction going the opposite way, Ff becomes 100.388, the acceleration back is ridiculously high. It ends up as 13.56 m/s^2. Which is almost 50 km/h. Which doesn't seem in the realm of possibility.

    What am I doing wrong?
     
  7. Nov 10, 2011 #6
    looks right to me. I wouldn't worry about it seeming real considering it's just a made up question not a real life problem
     
  8. Nov 10, 2011 #7
    When you plug that acceleration into the equation Vf^2 = Vi^2 + 2ad, you get 0.345 which is incorrect.

    Is there something to this that I'm missing?
     
  9. Nov 10, 2011 #8
    Draw a free body diagram for the block on the incline. Note that while gravity acts straight down from the block, the normal force has components. As such, if we sum the forces acting in the y direction, you will get an expression for FN (the normal force) in terms of gravity (I leave it to you to figure out which trigonometric ratio to use).

    Also, you may or may not be aware that if an object slides with constant velocity down a slope, its coefficient of friction is equal to tanθ. With this knowledge, and the normal force, we can find the frictional force.

    Using Newton's Second Law, if we return to the x direction, and sum those forces as we did for the y direction, we should be able to find the acceleration of the block, at which point you can use a basic kinematics equation that involves known variables of acceleration, final and initial speed and the required unknown distance to solve for the distance. (I believe you have already made reference to this equation.)

    Hope this helps! :)
     
  10. Nov 10, 2011 #9

    I like Serena

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    Homework Helper

    Hi Becca93! :smile:

    How did you get Fpara?
    I get a different value (52.34 N).
    Edit: sorry, your value is right.

    Btw, your acceleration is not ridiculously high.
    In particular you can not convert it to a speed of 50 km/h, since it is not a speed, it's an acceleration (that's just a little over the standard acceleration of gravity).
     
    Last edited: Nov 10, 2011
  11. Nov 10, 2011 #10
    Hey! Thanks for pointing that out. I used mgcosθ to get Fn, and mgsinθ to get Fpara.

    I did get the answer to the question though! It ended up being 0.354 m, not 0.345 m like I got earlier.

    Thanks to everyone for the help! It was seriously appreciated!
     
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