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Friction problem: finding the Fnet first

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data

    A 25 kg box is being pushed across the floor by a constant force ‹ 100, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.18. At t = 7.0 s the box is at location ‹ 11, 4, −3 › m, traveling with velocity ‹ 6, 0, 0 › m/s. What is its velocity and position at t = 8.5 s?


    2. Relevant equations
    deltaP= Fnet(deltat)


    3. The attempt at a solution
    My professor gave us a hint that once we find the the fnet, then we can use the momentum principle and the position update formula to find the velocity and the position. So my problem is how to find the fnet.

    The method that I tried to find the fnet is: 100-(25*0.18*9.8)= 55.9

    Well, there is another method (25*9.8) + (.18*100) = 263

    Is one of them correct? (I can see that the difference between these two numbers are pretty large)

    Once I find the fnet, then I can use the momentum principle, which is Pf=Pi + fnet(delta t) to find the final momentum. Then from there, find the final velocity.
    Pi = 150
    delta t = 8.5-7= 1.5

    Thanks in advance for your assistance!
     
    Last edited: Sep 25, 2011
  2. jcsd
  3. Sep 25, 2011 #2

    PeterO

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    Not sure what your second method is all about, but the first one looks appropriate.
     
  4. Sep 25, 2011 #3
    Well, I tried the answer I got from the first method and it was wrong.

    So I used the number 55.9 as fnet.

    Momentum principle: Pf= Pi + Fnet*delta t
    Pi = m*vi
    pi = 25 * 6 = 150
    Pf = 150 + 55.9(1.5)
    Pf= 233.85

    Now to find the final velocity
    Pf=Vf*m
    Vf= 233.85/25 = 9.354

    So for the final velocity vector, it would be ( 9.354, 0,0)m/s

    However, that answer is wrong. Any help would be much appreciated!!!
     
  5. Sep 25, 2011 #4

    PeterO

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    I would be happy with that answer for V. You didn't enter that as the position by any chance did you?

    Do you interpret the 3 digits as an i, j, k arrangement. In which case I found it intriguing that for the displacement at the 7 second mark, k was -3 ; so presumably the floor was below the origin???

    I assume you wrote out the question correctly?
     
  6. Sep 25, 2011 #5
    Well, I double checked everything. I copied and pasted the question, so that's definitely right. And I did enter (9.354, 0, 0) as the new velocity. And then I used that new velocity to find the new position which I got was (25.03, 4,-3). And that's wrong too.

    The way I did to find the final position was ( the change is only is the x component)
    Rf = Ri + Vf(delta t)
    Rf= 11 + 9.354(1.5)
    Rf = 25.03

    What I am not sure is if I got the Fnet right. According to my professor, I am giving one of the two forces that made up the Fnet, and I need to calculate the other force. This is where I am lost.

    The first method that i used to find the fnet, a friend told to do that. I am not sure if that's right or not.

    Any pointer would be much appreciated?
     
  7. Sep 25, 2011 #6

    PeterO

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    A 25 kg box is being pushed across the floor by a constant force ‹ 100, 0, 0 › N. The coefficient of kinetic friction for the table and box is 0.18.

    I copied the first line from your original post.
    When did the floor disappear and a table come in to it??
     
  8. Sep 25, 2011 #7
    Hahah, I didn't notice that. I think it was either a silly typo by my professor or a ninja made the switch. I will ask my professor about that tomorrow. I think I did everything correctly. Not sure what went wrong. Well, anyways, thank you very much for your assistance.
     
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