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Equations for fk, Wfk, and WF as an expression

  1. Oct 18, 2012 #1
    1. The problem statement, all variables and given/known data

    (a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force vector F at an angle θ with respect to the horizontal. The coefficient of kinetic friction between block and table is μk. Is the force exerted by friction equal to μkmg?
    ANSWER = No (Correct )

    Enter the expression that describes only the magnitude of the force exerted by friction. (Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)

    fk = ?

    (b) How much work is done by the friction force and by vector F ? (Don't forget the signs. Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)
    Wfk = ?
    WF =?



    3. The attempt at a solution

    fk= μk (mg-Fsin(θ))
    Wfk=-(cos(θ))d
    WF=mg(sin(θ))d

    *d= distance

    I'm struggling coming up with the correct equations. Any help is appreciated.
     
  2. jcsd
  3. Oct 18, 2012 #2

    tiny-tim

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    welcome to pf!

    hi svfetz! welcome to pf! :smile:
    your fk is correct :smile:

    but i don't understand what formula you're using for work done :confused:
     
  4. Oct 18, 2012 #3
    Hi :smile:

    For my submission for fk webassign didn't like my answer for it so I have no idea what I am doing wrong because you agree its right.

    For my other two equations I used W= FΔh and used cos and sin because it is getting pulled at an angle. So w=(Fcosθ)Δh?

    Ahhh, I'm so confused right now that I don't even know what I did anymore:confused:

    **For my Homework webassign hates me. (this is just the full problem)

    (a) A block with a mass m is pulled along a horizontal surface for a distance x by a constant force vector F at an angle θ with respect to the horizontal. The coefficient of kinetic friction between block and table is μk. Is the force exerted by friction equal to μkmg?
    No

    Enter the expression that describes only the magnitude of the force exerted by friction. (Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)
    fk =

    (b) How much work is done by the friction force and by vector F ? (Don't forget the signs. Use any variable or symbol stated above along with the following as necessary: g and F for the magnitude of vector F .)
    Wfk =
    WF =

    (c) Identify all the forces that do no work on the block.
    1. friction vector f k 2.vector n 3.vector F sin(θ) 4. mvector g 5.vector F cos(θ)

    (d) Let m = 2.00 kg, x = 4.00 m, θ = 41.0°, F = 17.0 N, and μk = 0.400, and find the answers to parts (a) and (b). (Include appropriate signs.)

    fk = -12.3 NThe response you submitted has the wrong sign.
    Wfk = -4.0 J Wrong
    WF = 68.15 J Wrong
     
  5. Oct 19, 2012 #4

    tiny-tim

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    hi svfetz! :smile:

    (just got up :zzz:)
    yes :smile:
    sorry, i don't follow any of these :confused:

    how did you get WF ? ​
     
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