1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar

    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

    User Avatar

    Staff: Mentor

    That would be my answer. How'd we do?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?