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The second question: Kinetic Friction

  1. May 27, 2005 #1
    A 65.0 kg person, running horizontally with a velocity of +4.06 m/s, jumps onto a 10.4 kg sled that is initially at rest.
    (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away.

    Which I figured out was 3.5 m/s...now my question:

    (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

    Any suggestions on how to go about solving this?
     
  2. jcsd
  3. May 27, 2005 #2
    The work done by friction must be equal in magnitude to the kinetic energy of the sled-person system but opposite in sign for the object to be at rest after displacing 30m.

    Knowing work is equal to [tex]Force \times Displacement[/tex] you can then solve for friction which then allows you to obtain the coefficent of kinetic friction.
     
  4. May 27, 2005 #3
    How should I calculate the horizontal force, to be able to calculate F=ma? What is the acceleration?
     
  5. May 27, 2005 #4
    The only horizontal force should be friction unless someone is pulling the sled-person system. Is it asking for an acceleration?
     
  6. May 27, 2005 #5
    Not directly...you can see I'm lost here. I do well visualizing things with equations...maybe that'd help..?
     
  7. May 27, 2005 #6
    Thats exactly the question u should ask yourself.
    Bring the three kinematic equations into your mind at this point.

    -- AI
     
  8. May 27, 2005 #7
    Don't worry about accelerations. You can if you want but it will make the problem more confusing to you I imagine. It is just an energy problem.

    The thing to remember is that the [tex]f = (W \times {\Delta x}) = {\Delta KE}[/tex]

    Also, remember [tex]f=\mu{mg}[/tex].
     
  9. May 27, 2005 #8
    Ahh, there's my hangup...how do I calculate work, and is the quantity of dX just 30m?
     
  10. May 27, 2005 #9
    gokugreene,
    i too like nutster cant see what you are actually trying here?

    Btw, nutster
    you can use v^2 = u^2 + 2as to get acceleration since u have v,u and s
    I suppose once u have 'a' , u know what to do

    -- AI
     
  11. May 27, 2005 #10
    Yes, the change in x is 30m.

    Tenali, he is looking for the coefficient of kinetic friction.

    You can find the coefficient of kinetic friction using the kinematic equations as Tenali suggested, but that isn't the way the problem is meant to be worked, but it will work.
     
    Last edited: May 27, 2005
  12. May 27, 2005 #11

    OlderDan

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    We use dX to represent a very small displacement. Work is the integral of the component of force in the direction of dX. In this problem, the force is the frictional force, and it is constant. That makes the integral very simple. The work is the constant force times the total distance the sled moves against that force, resulting in negative work done by friction on the sled.
     
  13. May 27, 2005 #12

    OlderDan

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    You of course meant to say

    [tex]W = (f \times {\Delta x}) = {\Delta KE}[/tex]
     
  14. May 27, 2005 #13
    Yes, thanks for the correction!
     
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