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Rock Problem

  1. Jan 25, 2006 #1
    I have this one problem solving question that confuses me. Someone pushes a 50kg rock a distance of 10m in a recent competition of strength. The coefficient of friction between the rock and the ground is 0.8 and the person exerts a force of 400 N. I've got my free body digram with my applied force as the force going foward. I know i have to use Fnet to find the amount of newtons but what do i use to find fnet. 400N is suppose to minus with something. I'd appreciate it if someone can bring me a bit further into the question so i can get to the answer. I hope this is enough work, i'm trying to get some review before my exams.
     
  2. jcsd
  3. Jan 25, 2006 #2
    you'll need to subtract the force opposing the push, ie the friction.
     
  4. Jan 25, 2006 #3
    so do i multiply 0.8 (coefficient of friction) by 400 N?
     
    Last edited: Jan 25, 2006
  5. Jan 25, 2006 #4
    ok i got 320N as my force of friction (from ff=uFn ff= 0.8x400N)
    then i use fnet= fapplied-ffriction to get 400-320=80 N
    use fnet in equation fnet=ma
    switch it around a= fnet/m a= 80N/50kg which equals 1.6 m/s2
    am i right? correct me if i'm wrong please, appreciate it.
     
  6. Jan 25, 2006 #5
    i got the answer now. its 0.16 m/s2 after correcting it.
     
  7. Jan 25, 2006 #6
    the Fnormal would be the force opposing the weight.
    which, by the way is equal to mg
     
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