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Friction and horizontal movement

  1. Nov 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A block of mass m lies on a horizontal table.

    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 of the block after it begins to move.

    2. Relevant equations

    friction force=1/2mass*mu_s*g

    3. The attempt at a solution

  2. jcsd
  3. Nov 1, 2007 #2


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    The friction force is proportional to the weigth (mg) by some factor [itex]\mu[/itex], which is the coefficient of friction. In general the static coefficient of friction [itex]\mu_s[/itex] is greater than the coefficient of kinetic friction [itex]\mu_k[/itex], so after the block moves there would be an excess force that would accelerate the block.

    If one starts with [itex]\mu_s[/itex]mg applied to move the block, then continues applying that force against the kinetic friction [itex]\mu_k[/itex]mg, then the net force applied is simply the difference.

    F = [itex]\mu_s[/itex]mg - [itex]\mu_k[/itex]mg or F = ([itex]\mu_s[/itex] - [itex]\mu_k[/itex])mg.

    Then applying F = ma => a = F/m, because one still has to accelerate the entire mass m.
  4. Nov 1, 2007 #3
    I can't quite figure it out, because when I submit answers it tells me that m is not part of the correct equation. So it must be done without terms of mass
  5. Nov 1, 2007 #4
    yes m wont be a part of the equation, when you equate the equation (Ms-Mk)mg to ma, the m cancels out.
  6. Nov 1, 2007 #5
    Is this an online quiz you are attempting. I would like some practice too. Do you mind sharing the site with me?
  7. Nov 1, 2007 #6
    It's my homework, which is all done online, which means I get instant feedback on the answers I submit. Unfortunately that means when I get it wrong I can't proceed until I get it right.
  8. Nov 1, 2007 #7
    oh ok..so obviously I can't access it. If you come across any sites with problems related to Motion and Dynamics, could you let me know? Also, are you clear on this question?
  9. Nov 1, 2007 #8
    Alright :)

    I'm not clear on this question unfortunately. I can't figure out how to get this equation without using the term m.
  10. Nov 1, 2007 #9
    ok..F net = (Ms -Mk )mg..clear on how did Astronuc get tht?
  11. Nov 1, 2007 #10
    Just incase you get confused, Ms is the coefficient of static friction, and Mk is the coefficient of kinetic friction. They are the same things as [tex]\mu[/tex]s and [tex]\mu[/tex]k..
  12. Nov 1, 2007 #11
    ok, i'll just finish this off. so we know that ([tex]\mu[/tex]s-[tex]\mu[/tex]k)mg= Fnet
    Now we also know that Fnet=ma (newtons 2nd law, where m is the mass)
    So we can equate these equations
    ([tex]\mu[/tex]s-[tex]\mu[/tex]k)mg= ma

    The "m's" cancel out, leaving you with
    ([tex]\mu[/tex]s- [tex]\mu[/tex]k)g= a
    Solve for a! Hope that helps..
  13. Nov 1, 2007 #12
    oh wait, you dont are not given numbers, then this is your answer..:)
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