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Homework Help: Friction Problems

  1. Oct 16, 2006 #1
    1)If the coefficient of kinetic friction between a 35-kg crate and the floor is 0.3, what horizontal force is required to move the crate at a steady speed across the floor?

    FN=mg=(35 kg)(9.8 m/s2)=343 N
    Ffr= kF N=(0.30)(343 N)=102.9 N




    2)A 1200-N crate rests on the floor. How much work is required to move it at constant speed
    a) 4 m along the floor against a friction force of 230 N
    b) 4 m vertically?

    a)
    W G = mgxcos90°=0
    W N = FNxcos90°=0
    W net = (Fnet)xx=( Fnet – Ffr)x= (1200 N – 230 N)(4 m)= 3880 J

    b)
    W G = mgxcos90°=0
    W N = FNxcos90°=0
    W net = (Fnet)x=( Fnet – Ffr)x= (1200 N – 0)(4 m)= 4800 J

    Just wanting to make sure I am doing everything correctly.
     
  2. jcsd
  3. Oct 16, 2006 #2

    berkeman

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    Staff: Mentor

    -1- looks good

    -2a- They give you the force that is opposing your push. Work is force x distance.

    -2b- The force opposing the lift is just the weight, so you have the correct answer. But I don't really understand what you are trying to say with the two equations =0 part. The work done is just the change in potential energy for the object between the two different heights.
     
  4. Oct 16, 2006 #3
    2a)

    Would it just be
    W=Fd
    W=230N*4=920 J
     
  5. Oct 16, 2006 #4

    berkeman

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    Staff: Mentor

    That would be my answer. How'd we do?
     
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