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!

Homework Help: Friction on a incline

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A block of mass = 4kg is pushed down an incline with a force of 4 N. The coefficient of kinetic friction of the incline = 0.11 . How far will the block move on the incline in the first 2 seconds after starting from rest ?
    Given that angle of incline = 30 degrees.

    2. Relevant equations

    3. The attempt at a solution

    I am having trouble understanding the role of the PUSH. since the force due to the push acts only for a moment, how do I account for it in the equations of motion ?
  2. jcsd
  3. May 8, 2007 #2

    Chi Meson

    User Avatar
    Science Advisor
    Homework Helper

    mgsinx represents one of the forces parallel to the surface of the plane.

    umgcosx represents another (do you know which one?)

    THere is a third force (the pushing force). How does this third force join the first two to give you the net force?
  4. May 8, 2007 #3
    umgcosx is the force of friction.

    The pushing force is instantaneous...so it will be there in my eqn. at one instant but not after that ? I feel its P+mgsinx-umgcosx=ma , where P=pushing force.
  5. May 8, 2007 #4
    i think P+mgsinx-umgcosx=ma so..
    it's nothing ; enhance force:rolleyes:
  6. May 8, 2007 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Somehow you got it into your head that the push acts just for an instant, but that's NOT what the problem says. Assume the push is maintained as the block goes down the incline.
  7. May 9, 2007 #6
    Yeah this gives me the correct answer which is 10 m.
    But i have a query,how would you solve this if the force was only instantaneous ?

    Thx for the help ChiMeson and DocAl
  8. May 9, 2007 #7
    In the case of "instantaneous", you still need to know the time the force acts on the block in order to calculate how much momentum the block gains from the force.
  9. May 9, 2007 #8

    Doc Al

    User Avatar

    Staff: Mentor

    As Weimin says, you need to think in terms of the impulse that the force exerts, which produces a change in momentum of the block. The impulse = force*time that the force acts. (Even if it seems "instantaneous", to have an effect the force must act for some nonzero amount of time.) Once you have the change in momentum, you can calculate the speed of the block after the impact. Then it's just another force/acceleration problem, only now the block has some speed instead of starting from rest.
  10. May 9, 2007 #9
    Ahh.. fine. Thx for the explanation :smile:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook