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Mass of block given μK and Force?

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A student pushes a block with 240 N of force across a horizontal surface with a coefficient of kinetic friction of 0.4. The block accelerates at a rate of 0.88 m/s2. Find the mass of the block.

    2. Relevant equations

    μK = FK / FN
    F=ma

    3. The attempt at a solution

    My first attempt was to figure out mass using F=ma

    m = f/a
    = 240 N / 0.88 m/s2
    = 272.7 kg​

    but this is not taking into account the kinetic friction...

    So what I did was

    FK = μK x FN
    = 0.4 x (272.7 kg x 9.8 m/s2
    = 1069.1 N​

    I know I'm doing this wrong and I need to know how to derive at net force to find acceleration
     
  2. jcsd
  3. Nov 15, 2012 #2

    Doc Al

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    To use ƩF = ma you must include all the forces.

    You are using the incorrect answer from your first attempt.

    Instead, just call the mass "m" and set up your equation using the net force. Then you can solve for the mass. What's the friction force in terms of "m"?
     
  4. Nov 15, 2012 #3
    μK = FK / FN
    FN = FK / μK
    mg = FK / μK
    m = (FK / μK) / g

    Like this?
     
  5. Nov 15, 2012 #4

    Doc Al

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    No. (That won't work, since FK is proportional to the mass, so your equation ends up being m = m.)

    Use ƩF = ma. First work on the forces. There are two forces acting. What are they?
     
  6. Nov 15, 2012 #5
    force applied of 240 N and kinetic friction

    I can't find magnitude of kinetic friction because I cannot find Force normal.
     
  7. Nov 15, 2012 #6

    Doc Al

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    Good.

    If the mass is "m", what's the weight?
     
  8. Nov 15, 2012 #7
    weight = 9.8m
     
  9. Nov 15, 2012 #8

    Doc Al

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    Good. So now you have the normal force. Keep going.
     
  10. Nov 15, 2012 #9
    μK = FK / FN
    FK = μK × FN
    FK= 0.4 x 9.8m

    I have 2 unknowns?
     
  11. Nov 16, 2012 #10

    Doc Al

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    Good.

    The only unknown is the mass, which is what you are trying to find.

    Keep going. What is ƩF?
     
  12. Nov 16, 2012 #11
    Fapplied = 0.88m


    just guessing here:
    Fnet = FA [fwd] + Ff [bwd]
    Fnet = FA [fwd] - Ff [fwd]
    Fnet = 0.88m - (0.4 x 9.8m)
     
  13. Nov 16, 2012 #12

    Doc Al

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    The applied force is given as 240 N.

    This is good.
    Fix this and then use Fnet = ma.
     
  14. Nov 16, 2012 #13
    oh wow my bad...

    Fnet = 240 - (0.4 x 9.8m)
    = 240 - 2.92m​

    Still don't get what I need to do next?
     
  15. Nov 16, 2012 #14

    Doc Al

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    Apply Newton's 2nd law: Fnet = ma
     
  16. Nov 16, 2012 #15
    240 - 2.92m = 0.88m
    240 = 0.88m + 2.92m
    240 = 3.8m
    m = 63 kg
     
  17. Nov 16, 2012 #16

    Doc Al

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    That's the right idea. But I think you made a mistake earlier (probably a typo):

    Fix that, resolve your equation, and you'll have the correct mass.
     
  18. Nov 16, 2012 #17
    Fnet = 240 - (0.4 x 9.8m)
    = 240 - 3.92m

    240 - 3.92m = 0.88m
    240 = 0.88m + 3.92m
    240 = 4.8m
    m = 50 kg

    wohooooooo thanks a lot!:biggrin:
     
  19. Nov 16, 2012 #18

    Doc Al

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    Yay! :approve:
     
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