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Pushing a block question (Friction)

  1. Oct 14, 2009 #1
    1. The problem statement, all variables and given/known data


    A block of mass m lies on a horizontal table. The coefficient of static friction between the block and the table is [tex]\mu[/tex]s. The coefficient of kinetic friction is [tex]\mu[/tex]k, with [tex]\mu[/tex]k < [tex]\mu[/tex]s.

    Part A
    If the block is at rest (and the only forces acting on the block are the force due to gravity and the normal force from the table), what is the magnitude of the force due to friction?

    Part B
    Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force F must you be pushing the block just before the block begins to move?
    Express the magnitude of F in terms of some or all the variables mu_s, mu_k, and m, as well as the acceleration due to gravity g.

    Part C
    Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force?
    Express your answer in terms of some or all of the variables mu_s, mu_k, and m, as well as the acceleration due to gravity g.

    Part D
    Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration a of the block after it begins to move.



    3. The attempt at a solution

    A: 0

    B: [tex]\mu[/tex]s * mg

    C: I know the magnitude of the frictional force is equal to the\horizonatal push but I don't know how to express that in variables

    D: I am clueless on this one
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 14, 2009 #2

    rl.bhat

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    The frictional force is a self adjusting force. When it pushed with half the maximum frictional force, the body will not move. So the frictional force = ....?
    For last part use μk.
     
  4. Oct 14, 2009 #3
    1/2[tex]\mu[/tex]s * mg ?
     
  5. Oct 14, 2009 #4

    rl.bhat

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  6. Oct 14, 2009 #5
    For part d, it is just [tex]\mu[/tex]k?
     
  7. Oct 14, 2009 #6

    rl.bhat

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    No.
    a = frictional force/mass.
     
  8. Oct 15, 2009 #7
    So, it is [tex]\mu[/tex]k/m since after you start moving the block, you are using [tex]\mu[/tex]k and no longer [tex]\mu[/tex]s
     
  9. Oct 15, 2009 #8

    rl.bhat

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    Frictional force is μk*mg
     
  10. Oct 15, 2009 #9
    Mastering Physics says The correct answer does not depend on the variables: m, m, [tex]\mu[/tex]k.
     
  11. Oct 15, 2009 #10

    rl.bhat

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    Yes.
    a = frictional force/mass.
    See what you get.
     
  12. Oct 15, 2009 #11
    I did [tex]\mu[/tex]k * mg/m = a but it said The correct answer does not depend on the variables: m, m, [tex]\mu[/tex]k.
     
  13. Oct 15, 2009 #12

    rl.bhat

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    Acceleration is μk*g.
    You can it as (Fr/N)*g
     
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