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Another seemingly easy forces question

  1. Jan 25, 2013 #1
    Posting this from my phone, apologies.
    "A block is pushed across a horizontal surface with a coefficient of kinetic friction of 0.15 by applying a 150N horizontal force. If the block accelerates at 2.53m/s^2 find the mass of the block."

    I don't even know where to start. I know the force applied, acceleration and kinetic friction coefficient. I can't use Fnet=ma or Fnet=sum of all forces because I don't have the mass or Fnet. I can't calculate friction without mass (since normal force must be equal to gravity in this scenario, and is there equal to mg). I've drawn a fBD but that didn't really help. Can I get a hint on where to start? Perhaps a system of equations?
     
  2. jcsd
  3. Jan 25, 2013 #2

    Dick

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    Yes. A system of equations. Write the equations down in terms of the unknown mass 'm' and try to solve for it.
     
  4. Jan 25, 2013 #3
    Yeah I can't quite seem to get it even knowing that. I tried using:
    Fnet=ma and Fnet= Ff+Fa
    Since they both equal Fnet, they must be equal:
    Ma=Ff+Fa
    Ma=(mu)mg+Fa
    Then I get stuck. I tried rearranging and factoring out mass and I get 58kg (the answer is 38kg)
     
  5. Jan 25, 2013 #4

    Dick

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    You need to keep track of the direction of the forces. Do the frictional force and the applied force both point in the same direction as the acceleration?
     
  6. Jan 25, 2013 #5

    tms

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    Maybe you're confusing yourself by using two symbols for the same mass. All you have to do is solve your equation for [itex]m[/itex]. Also be careful of the signs of the forces.
     
  7. Jan 25, 2013 #6
    Well acceleration and the force applied are in the same direction. Friction is opposite to that. I don't understand how that changes my calculations though :/
     
  8. Jan 25, 2013 #7

    Dick

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    It would change ma=mu*m*g+Fa to ma=Fa-mu*m*g. That's a difference.
     
  9. Jan 25, 2013 #8
    Ok yes I did that and I got 56kg as an answer. Still incorrect :/
     
  10. Jan 25, 2013 #9

    Dick

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    Put the numbers in and show how you got it. That's not what I get.
     
  11. Jan 25, 2013 #10
    Wait are you getting 37.485...kg? I tried agAin and that's what I got
     
  12. Jan 25, 2013 #11

    Dick

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    Yes, that's about right.
     
  13. Jan 25, 2013 #12

    tms

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    Yes, but you have too many significant figures.
     
  14. Jan 25, 2013 #13
    Yay I got it! :) thank you so much!!!
     
  15. Jan 25, 2013 #14
    With significant figures it is 37 right? I think the textbook just rounded wrong
     
  16. Jan 25, 2013 #15

    Dick

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    Depends exactly what you use for g. If you use 9.8m/s^2 you get 37.5kg. Round up, round down? Not really important.
     
  17. Jan 25, 2013 #16

    tms

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    You could probably get away with 37.5, since the acceleration has 3 figures, the the force has 3, too, arguably.
     
  18. Jan 25, 2013 #17
    The blocks acceleration is 2.53m/s^2 which is three significant figures so an answer of 37.5kg is correct. Just don't forget the correct units.
     
  19. Jan 25, 2013 #18
    Oh ok. I've learned significant figures differently. 0.15 (the coefficient of friction, cant scroll up and see but i think thats the value) has 2 significant figures and so does the force (150N). The acceleration has 3 but you go by the least significant term so it has 2 in the end
     
  20. Jan 25, 2013 #19

    Dick

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    You said the book gave 38kg. So they must be rounding 37.5kg up to 38kg. This is really all much less important than getting the physics right.
     
  21. Jan 25, 2013 #20
    The force thought with 150N has three significant figures. Leading zeros are not included in significant figures but trailing zeros are. The right answer was found but on a test it is always best to account for significant figures and I have always been expected too. Here is a link on a brief overview of significant figures.
    http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/sigfigs.html
     
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