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Kinetic friction.

  1. Jul 16, 2009 #1
    A sufficiently large push is given to start the block moving. After that, what magnitude Fpush must be applied to make the block accelerate to the right at 5.71 m/s2

    From a previous problem we concluded that coefficient of kinetic friction is .34 and static is .43

    I do not understand how to determine the blocks mass to be able to determine the Fpush needed. any tips
     
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  3. Jul 16, 2009 #2

    cepheid

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    Neither do I.

    Was this the exact statement of the problem? If not, can you please use the proper template for posting homework questions so that we are sure all of the pertinent information has been posted?
     
  4. Jul 16, 2009 #3
    Mag of frictional force from the previous question was determined to be 7.4. and then the coefficients were determined. thats all the information i was given to be able to answer the question of

    A sufficiently large push is given to start the block moving. After that, what magnitude Fpush must be applied to make the block accelerate to the right at 5.71 m/s2

    But i think the 7.4 is from a different question and is not related to this one or maybe it is?
     
  5. Jul 16, 2009 #4

    cepheid

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    The fact that it says "the" block and not "a" block, suggests that it is referring to a specific block that has already been mentioned or that you already know about. Is this the next SUB-part of a one long question? If so, can you just post the whole question? Can you post it word for word, please?

    If it is not a sub-part of a larger problem, then I find it very bizarre that a *stand-alone* problem would be worded in this way. What you have posted here is not enough info to determine Fpush.
     
  6. Jul 16, 2009 #5
    you were right i did not see the information listed above the block, only thing missing is the blocks mass which is 5KG
     
  7. Jul 16, 2009 #6
    and it also wants to know the direction of the overall force of the table onto the block.

    The block is at rest on a table.
     
  8. Jul 16, 2009 #7

    cepheid

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    Soooo....can you now answer the problem?
     
  9. Jul 16, 2009 #8
    -fk = ma correct?

    so 5.71x5 kg = 28.55 is the force required to accelerate it at 5.71
     
  10. Jul 16, 2009 #9
    and the table overall force is up (perpendicular to the table correct)?
     
  11. Jul 16, 2009 #10

    cepheid

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    Not really. The expression ma gives the NET force, which is the sum of two opposing forces, the applied force (which you are trying to calculate) and the frictional force (which you know).

    You know that 28.55 N can't be the applied force, because if you subtract the frictional force from it to get the net force, and then divide this by 5kg, you don't get an acceleration of 5.71.
     
  12. Jul 16, 2009 #11

    cepheid

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    No, the table exerts two forces on the block. The overall force will be the sum of these.
     
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