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What is the impulse of friction?

  1. May 16, 2008 #1

    Jason03

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    I was working on a basic momentum/impulse problem where a block is pushed by a force P over a time T that is plotted on a graph....Kinetic Friction is given but how would I incorporate that into the equation to find the velocity?

    I know the impulse is the area under the curve....

    [tex]mv_{1} + F_{avg}\Delta{t} = mv_{} [/tex]
     
    Jason03, May 16, 2008
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  3. May 16, 2008 #2

    lzkelley

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    What is the force of friction?
    What is the impulse of friction?
    How does that change your equation?
     
    lzkelley, May 16, 2008
  4. May 16, 2008 #3

    Jason03

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    Actually I figured it out, pretty easy...I just over thought it as usual....
    This equation allowed me to come up with the correct answer

    [tex] mv_{1} + P-T(\Delta{t}) = mv_{2}[/tex]
     
    Jason03, May 16, 2008
  5. May 16, 2008 #4

    Jason03

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    Im having trouble coming up with velocity for Time at 8 seconds.....I came up with the Velocity at 5 seconds easily using the equation above.....

    heres the graph for the problem....http://img91.imageshack.us/img91/6935/graphmp3.jpg [Broken]


    im trying to find the average Force at T = 8 seconds.....but its not working
     
    Last edited by a moderator: May 3, 2017
    Jason03, May 16, 2008
  6. May 16, 2008 #5

    Doc Al

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    Staff: Mentor

    Show what you did. To find the impulse of the applied force, find the area under the curve. What about the impulse of the friction force?
     
    Doc Al, May 16, 2008
  7. May 16, 2008 #6

    Jason03

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    I used this equation to solve for time at 5 seconds....and it worked...

    [tex] mv_{1} + P-T(\Delta{t}) = mv_{2}[/tex]


    but I tried solving for time at 8 seconds... finding the area under the curve which is just a rectange and triangle...but the answer comes up incorrect

    as far as area it should just be [tex](L * W) + (.5*b*h)[/tex]
     
    Last edited: May 16, 2008
    Jason03, May 16, 2008
  8. May 16, 2008 #7

    Doc Al

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    Staff: Mentor

    That equation is not quite right. That second term (P) is a force, not an impulse. (It should be the area under that curve.)


    Looks fine to me. Show me the details. What did you get for the impulse due to the force P? The impulse due to the friction force T?
     
    Doc Al, May 16, 2008
  9. Jul 9, 2010 #8

    Couthbaron

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    Re: impulse/momentum

    Don't forget that Impulse is NET force * time. So if you have the graph of the force of the engine vs. time, you need to subtract out kinetic friction, which is considered constant over the time.

    If the MR2 that kept up well with the viper was a Spyder, that is I think 137 hp and not 115 hp.
     
    Couthbaron, Jul 9, 2010
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