1. Limited time only! Sign up for a free 30min personal 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!

Work done by friction in a system

  1. Apr 10, 2009 #1
    1. The problem statement, all variables and given/known data

    work done by friction=? as big M slides through a distance of 3 meters.
    what is the speed of small m when it hits the ground?
    M=5kg m=2kg m is 3meters above ground muk is.25

    I can't figure out how to post a drawing (first post)Big M is horizonatal and frictionless b4 muk attached by a string over a pulleyattached by same string to m which is the 3 meters above ground.
    2.
    Relevant equations


    wf = -fkd wf=e2-e1

    3. The attempt at a solution
    do i use cos180 since it is horizontal and I am lost on part 2
     

    Attached Files:

    Last edited: Apr 10, 2009
  2. jcsd
  3. Apr 10, 2009 #2

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

    A picture would be helpful. One could created a gif or bmp file and attach it, or put the image on a site like imageshack or flickr and use tags (roll the cursor over the buttons just above the posting box).


    So M is sliding on the horizontal in one direction, and m is pulled over it in the opposite direction at some point. They are connected by a string which passes through/around a frictionless pulley. Is M sliding on a frictionless surface? So then the only friction interaction is between M and m?

    If the only friction is between M and m, then one only needs to consider the friction while they are in contact. What is the length of M, i.e. how far does m travel over M?
     
  4. Apr 10, 2009 #3
    Ok so far i have work done by friction -36.75 which the cos 180 just gives me the negative.

    and the speed of small m=6.1 m/s
    using the eq-----mk*mgd=mgh2+1/2mv^2-mgh1
    am I on the right track?
    I feel like i have missed some steps?
     
  5. Apr 11, 2009 #4

    Astronuc

    User Avatar
    Staff Emeritus
    Science Advisor

    The work done by friction is correct. 36.75 J is lost due to friction.

    The velocity seems a bit high.

    See if this helps - http://hyperphysics.phy-astr.gsu.edu/hbase/hpul2.html#c1

    In using the eq-----mk*mgd=mgh2+1/2mv^2-mgh1, one must include the kinetic energy of M, and don't forget to distinguish between m and M.
     
    Last edited: Apr 11, 2009
  6. Apr 11, 2009 #5
    Thank you-I like the link you sent. Looks like I need to still take the square root of the 6.3 which makes the correct answer...2.51m/s.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work done by friction in a system
  1. Work Done by Friction (Replies: 11)

Loading...