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Homework Help: Curling stone coming to rest - friction

  1. May 31, 2010 #1
    This is just a revision question for an exam i have in a couple weeks.

    I think i kow the answer but just want to double check.

    a curling stone is released with an instial speed of 3.1 m/s
    the coefficent of sliding friction between the stone and ice is 0.017
    What is the distance the stone will come to rest?
    A 14m B 19m C 29m D 58m E Impossible to tell without mass

    I believe it to be E

    IF i did have the mass what equation would i need?
    [tex]\mu[/tex][tex]_{}slide[/tex] * mg ??

    Many thanks
    Last edited: May 31, 2010
  2. jcsd
  3. May 31, 2010 #2
    Re: friction

    you sure ? If a ton of steel is on a ice and I push it to give it a momentum, will it stop at the same place than a curling stone of 10kg ?

    Think about it.
  4. May 31, 2010 #3
    Re: friction

    btw it was a joke. You had it almost right :P

    coefficient of dynamics friction * Force which equals Mass times Acceleration.
  5. May 31, 2010 #4
    Re: friction

    ok re-looked at it and have decided on D 58m as D=V2/the coefficent of sliding friction * gravity (9.8 m/s)

    That look correct?
  6. May 31, 2010 #5
    Re: friction

    How can you tell the acceleration ? (which is negative) you conclude it is the gravity ?
  7. May 31, 2010 #6
    Re: friction

    well the equation i have is D=V2/[tex]\mu[/tex]slide * g

    I just put in my values of 3.1 m/s2/ 0.017*9.8 m/s

    It gave me an answer of 57.6 which is the same as answer D 58m (2sf)
  8. May 31, 2010 #7
    Re: friction

    I already told you you were right.

    It's just my way to teach. To make sure you understood and looked at your response again to be sure.
  9. May 31, 2010 #8
    Re: friction

    ha ha oh right, ok thanks
    Do you know much about collisions in three dimensions between 2 objects? i.e (vbx,vby)?
  10. May 31, 2010 #9
    Re: friction

    It's the same thing then between two objects in 2 dimensions. You just add another dimension.

    You seperate the forces in x, y, z instead of only x and y. That's it.
  11. May 31, 2010 #10
    Re: friction

    sorry ive been an idiot, i meant in 2 dimensions.
    ill tell you the question....

    2 objects with masses A 0.60kg and B 0.45kg are travelling towards one another parallel and anti parallel at uAx 1.5m/s and uBx -2.0m/s
    the velocity of A after collision is (0,12)m/s what is (vBx,vBy) of B after collision?

    I just cant find in my text how to do it for (x,y)
  12. May 31, 2010 #11
    Re: friction

    Hint: 3rd law of Newton.
  13. May 31, 2010 #12
    Re: friction

    so if object A has an opposite but equal in magnitude force on object B when they collide then the force acting on B would be f=ma which would be 0.9N, if i rearrnged to find for v i get v=f/m which gives me 2m/s but as B is travelling at -2m/s it would reusult in 0m/s so B would come to a stand still and have (vBx,vBy) (0,0)m/s

    is that sort of the way to go?
  14. May 31, 2010 #13
    Re: friction

    You "can" do it that way. But the best way is to go with the 3rd law of newton: the conservation of energy.

    Or I may say, the momentum. m1 x v1 = m2 x v2
  15. May 31, 2010 #14
    Re: friction

    yer i was going to ask if there was a better way of doing it. That makes sense :) Thanks
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