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!

Force/net work problem

  1. Sep 25, 2004 #1
    Hello! I need a little help here. I have been trying to figure out this problem, but I just can't and I don't know where I am going wrong.

    A 116 -kg crate is being pulled across a horizontal floor by a force P that makes an angle of 35.2 ° above the horizontal. The coefficient of kinetic friction is 0.212 . What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?

    now I know what W=FdCos(theta) and that fs=uk*F....but I just don't know where to go from there and I don't know what the end of the question means..."so that hte net work done by it and the kinetic frictional force is zero?"...does that mean I just set the net work and frictional force equal and solve for the force?
     
  2. jcsd
  3. Sep 26, 2004 #2
    Any moving object on a surface with friction will have kinetic friction.

    Ah I am starting to see now. The question intends to ask "what should P be if the work done by you and the work done by kinetic friction cancel each other out?" So yeah you're right - almost. You can't set work equal to force because they are not of the same unit. Set your horizontal component of P equal to the kinetic friction experienced because if both forces cancel each other out, no work can be done in moving the box even if it's moving at a constant speed.
     
    Last edited by a moderator: Sep 26, 2004
  4. Sep 26, 2004 #3
    Thanks! That's exactly what someone said when I asked them about it today and it worked.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Force/net work problem
  1. Net force problem (Replies: 4)

  2. Net Force and Net Work (Replies: 1)

Loading...