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 of friction quick question

  1. Feb 10, 2005 #1
    a block of mass 10kg is attached to a spring with a constant of 1850 N/m. the coefficient of sliding friction between the block and the surface is .25
    the block is pulled back 11 cm to the right and released.

    the question asks the speed at different points, and how far the distance after equilibrium it travels.

    THe problem i have is this.... i am fine on all of this except for how to figure the friction into the equation? i know the velocity neglecting friction, and i know the force of friction acting on the box. i need something to put the two together. can someone please lend me a small hand?
     
  2. jcsd
  3. Feb 10, 2005 #2

    Doc Al

    User Avatar

    Staff: Mentor

    The stretched spring has a certain PE. If there were no friction, as the spring contracted [itex]-\Delta \texrm(PE) = \Delta \texrm(KE)[/itex]. But with friction, some of that PE is transformed to heat via work done against friction. So: [itex]-\Delta \texrm(PE) = \Delta \texrm(KE) + W_\texrm{(friction)}[/itex]. What's the work done against friction?
     
  4. Feb 10, 2005 #3

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    How about considering the energy?

    [tex]\Delta K + \Delta U = -fd [/tex]

    where f is the friction force, d is the distance travelled, and K and U are kinetic and potential energy, respectively. This is a statement that says the change in total energy of the system is equal to the work done on the system (the work done on the system in this case is negative (frictional loss)).

    Try take it from there; we can help more if you need it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Force of friction quick question
Loading...