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Friction help

  1. Mar 24, 2009 #1
    Friction help!!!

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

    It is sometimes claimed that friction forces always slow an object down, but this is not true. If you place a box of mass M on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed v of the belt. The coefficient of friction (greek letter mu "looks like a u") between box and belt is . Do not worry about italics. For example, if a variable g is used in the question, just type g and for (u) use mu.

    I have 2 questions that do not have numbers, just have to solve for an equation.

    (a) What is the distance d (relative to the floor) that the box moves before reaching the final speed v? (Use energy arguments to find this answer.)

    (b) How much time does it take for the box to reach its final speed?



    The attempt at a solution

    not sure how to start. i figured conservation of energy. also the block will experience -mg force, it will have normal force cancelling it out. so then only force is friction force due to belt. this is where i get stuck. also not sure if i have to factor in change of thermal energy or not. anything with friction with have change in thermal.
    any help would be great
    thanks
     
  2. jcsd
  3. Mar 24, 2009 #2

    LowlyPion

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    Re: Friction help!!!

    Until it comes to rest on the moving surface there is a force acting on M that is equal to the force of friction.

    F = μ*M*g

    This force will accelerate M until it is going at the speed of the belt.

    Since F = m*a

    then you can figure how fast it accelerates. Armed with acceleration you can write an equation from kinematics that relates this acceleration and the velocity of the belt to distance. Likewise the time based on velocity and acceleration.
     
  4. Mar 24, 2009 #3
    Re: Friction help!!!

    ok since F= u*M*g and F = m*a i can set them equal
    m*a = u*M*g i solved for a by dividing by m and get
    a = (u*M*g)/m would the (m) cancel? ther is a big M and a small m not sure

    then if this is a F = m*a
    so F = m*((u*M*g)/m)

    am i right so far? i feel like i messed up somewhere
     
  5. Mar 24, 2009 #4

    LowlyPion

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    Re: Friction help!!!

    Sorry, of course the m's cancel.
     
  6. Mar 24, 2009 #5

    LowlyPion

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    Re: Friction help!!!

    But what you are really interested in is a, isn't it?

    That way you can relate a, v, and x.

    And you can relate a, v and t.
     
  7. Mar 24, 2009 #6
    Re: Friction help!!!

    how are these related? i know if your integrate acceleration you get velocity. that is back to calc 1 stuff. but dont think that works for physics and i dont know where the x comes into play.
     
  8. Mar 24, 2009 #7

    LowlyPion

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    Re: Friction help!!!

    These are kinematic relationships. But never mind that. Do it as they suggest.

    What will the increase in kinetic energy be?

    And you have the force ... so how much work is done on getting the block to speed?

    W = F * d
     
  9. Mar 24, 2009 #8
    Re: Friction help!!!

    kinetic energy is = 1/2mv^2
    delta(E) = w+q i said q is 0
    so delta(E) = w delta(E)= KE (1/2mv^2)
    w=f*d so
    1/2mv^2 = (m(u*g)) * d
    solve for d i get
    d= v^2/2g
    where did i go wrong? this is the wrong answer
     
  10. Mar 24, 2009 #9

    LowlyPion

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    Re: Friction help!!!

    Well you have it, but you dropped a constant looks like.

    1/2*m*V2 = μ*m*g * d

    d = V2/(2*μ*g)
     
  11. Mar 24, 2009 #10
    Re: Friction help!!!

    yeah realized that
    i got the answers now

    thanks for all the help
     
  12. Mar 24, 2009 #11
    Re: Friction help!!!

    how did you get the time?
     
  13. Mar 24, 2009 #12

    LowlyPion

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    Re: Friction help!!!

    Can't you get it from distance and acceleration?
     
  14. Mar 24, 2009 #13
    Re: Friction help!!!

    yes that is what i did messed with some of the letters.
    eventually simplified down to

    t= v/(mu*g) where mu is just u not mass * u
     
  15. Mar 24, 2009 #14
    Re: Friction help!!!

    i tried:

    d/sqrt(2*d*mu*g)

    ...it was wrong
     
  16. Mar 24, 2009 #15

    LowlyPion

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    Re: Friction help!!!

    That's because your acceleration is μ*g

    V is a constant and is already given. So to find time

    V = a*t = μ*g*t

    t = v/(μ*g)
     
  17. Mar 24, 2009 #16
    Re: Friction help!!!

    thx a lot for u guys helping

    Now I get the answer
     
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