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Finding distance using forces

  1. Nov 18, 2007 #1
    [SOLVED] Finding distance using forces

    1. The problem statement, all variables and given/known data

    A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.18 and the push imparts an initial speed of 4.2 m/s?

    2. Relevant equations

    x=Vit+1/2at squared

    3. The attempt at a solution

    I've been trying to figure out Fa since i know the coeffiecient. But since i cannot find time i'm trying to use another equation to find distance. I'm realy bad at finding these things without a mass given becaue i'm so used to that. Thanks for your help!
     
  2. jcsd
  3. Nov 18, 2007 #2

    hage567

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    What equation are you using to find the frictional force? That is the only force acting horizontally on the box. So it is the the net force in [tex]F_{net} = ma [/tex]. Use this approach to find the deceleration of the box. Once you have that, you can apply kinematics to figure out how far the box moved.

    You don't need to know the mass, it will cancel out in your equations.
     
  4. Nov 18, 2007 #3
    i dont understand how mass can cancel out, howe can you use Fnet=ma with two unknowns?
     
  5. Nov 18, 2007 #4

    hage567

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    What do you think is the other unknown?
     
  6. Nov 18, 2007 #5
    keep in mind that after the moment the box is given the push the only force acting on the object in the x- direction is friction from there you can apply

    SumF = ma
    -f = ma
    -(mu)mg = m*a
    therefore a = - (mu) g

    Now as hage567 said use kinematics to find the displacement of the box.
     
  7. Nov 18, 2007 #6
    i dont see how you use F=ma to solve this question, you dont have any of the variables. although i dont see how to solve it myself..
     
  8. Nov 18, 2007 #7
    well my teacher really hasn't talked much about kinematics or maybe she has..we dont ever use the book, just her notes. But ya i used the acceleration which i got -1.76 m/s2 and used it to find time t=-2Vi/a and got 4.77 s and plugged the numbers into x=Vit+1/2at2 and got 40.0 m is that correct?
     
  9. Nov 18, 2007 #8

    hage567

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    After the initial push the only force acting on the block is the frictional force, [tex]f = \mu_k N [/tex], where N is the normal force (in this case it is equal to the weight of the block, mg).

    So put that together with F = ma, the mass cancels and there is only one unknown left, which is a.
     
  10. Nov 18, 2007 #9
    is 40.0 m correct> ?
     
  11. Nov 18, 2007 #10
    oops, im tired.. thanks lol so so ma = uN

    N = mg

    ma/mg mass cancles out, a/g = u

    so a = g(u) ?
     
    Last edited: Nov 18, 2007
  12. Nov 18, 2007 #11

    hage567

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    Where is that 2 coming from? Find the right time, and your second equation should give the correct result.

    But:

    You can do this without finding the time. Look at your kinematic equations and see which one has what you know, and the one thing you need. You know the initial and final velocities, right?
     
  13. Nov 18, 2007 #12
    sorry for another interruption, so after a = gu

    solve for delta t using a = delta v/delta t
    then solve for delta x using v = delta x/delta t ?
     
  14. Nov 18, 2007 #13

    hage567

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    Yes, you can find t using your first equation. But to find x, you must use an equation that involves acceleration, since the block is not moving at a constant velocity.

    But, like I said already, you don't actually need to solve for the time directly. There is a kinematic equation you can use that will let you solve for x without knowing what t is. Perhaps it is good practice for you to try it both ways!
     
  15. Nov 18, 2007 #14
    Ok ..i don't have Vf though..if i use (Vf2-Vi2)/2a and to find Vf i need to have x don't i? hm..let me look some more or think about what i can use to find Vf ..so far i could only think of squareroot Vi+2ax ..
     
  16. Nov 18, 2007 #15
    Oh! i see! you must use the formula: [tex]\Delta X = V_{1}\Delta t + \frac{1}{2}a(\Delta t)^2[/tex] ??
     
    Last edited: Nov 18, 2007
  17. Nov 18, 2007 #16

    hage567

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    The block comes to rest, so Vf = 0.
     
  18. Nov 18, 2007 #17

    hage567

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    Yes, you can use this equation if you are going to find the time first.
     
  19. Nov 18, 2007 #18
    Awesome, Thanks a bunch!
     
  20. Nov 18, 2007 #19
    how do i find the right time?
     
  21. Nov 18, 2007 #20
    oh ..sorry i didn't see the second page okies..so it's 5.01 m?
     
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