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Finding Initial Speed given Kinetic Friction, Mass, and more

  1. Nov 21, 2014 #1
    Hi, I'm new here and I'm having trouble with a question.

    1. The problem statement, all variables and given/known data


    A 20.0-kg curling stone is released at the hog line and moves 28.35 m [W] to sit on the button and score a point. If the coefficient of kinetic friction between the stone and the ice is 0.00200, what was the initial speed of the stone when it was released?

    So what I know is Vf = 0 m/s (the stone stops) , m = 20.0 kg , d = 28.35 m , μk = 0.00200

    Vi = ?

    2. Relevant equations
    I'm not actually sure what to use...
    FN = ma
    Ffk = μkFN

    3. The attempt at a solution
    I looked around at other questions and tried to do the question based on another similar question. But most were looking for displacement or final velocity.

    Ffk = μkFN
    = (0.00200) (20.0 x ?)
    And then I got stuck
     
    Last edited: Nov 21, 2014
  2. jcsd
  3. Nov 21, 2014 #2
    Have you drawn a free body diagram that includes the forces acting on the stone? Have you written out the force balances based on the information in the free body diagram?

    Chet
     
  4. Nov 21, 2014 #3
    I have not written out the force balances. If I may ask, what are those? I may have learned them but referred to under different terms.
     
  5. Nov 21, 2014 #4
    So have you drawn a free body diagram? A force balance is (sum of forces = ma).
     
  6. Nov 21, 2014 #5
    Yes, I have.
     
  7. Nov 21, 2014 #6
    So let's see your force balance equations in the horizontal and vertical directions.

    Chet
     
  8. Nov 21, 2014 #7
    Would that mean:
    FN = ma
    Fg = mg
    Ffk = μkFN
     
  9. Nov 21, 2014 #8
    What is FN defined as? In the first equation, is a the acceleration in the vertical direction or the horizontal direction? I don't see a sum of forces in any of these equations. What is the sum of the forces acting on the stone in the vertical direction? What is the sum of the forces acting on the stone in the horizontal direction?

    Chet
     
  10. Nov 21, 2014 #9
    Ohhhh, well the acceleration will be in the horizontal or x direction as the rock isn't flying up in the air. So the sum of the forces in the vertical direction would be 0 N. I'm not actually sure how to determine the sum of forces in the horizontal direction, given the values. I'm sorry :(

    Edit: Maybe Fnet = Fapp + Fg(parallel) + Ffk
     
  11. Nov 22, 2014 #10
    I think you're beginning to get the right idea. Next, I want you to make a list of the forces acting on the stone in the vertical direction and in the horizontal direction (in words, not symbols).

    Forces acting on stone in vertical direction are: (complete this sentence)

    Forces acting on stone in horizontal direction are: (complete this sentence)

    Hint: You should be able to see these forces as arrows in your free body diagram.

    Chet
     
  12. Nov 22, 2014 #11
    Forces acting on the stone in the vertical direction are: Normal Force and the Force of Gravity. Forces acting on the stone in the horizontal direction are Applied Force and the Kinetic Friction. The net force in the horizontal direction should be equal to the Force Applied plus Force of Kinetic Friction and the Force of Gravity in the Parallel direction.
     
  13. Nov 22, 2014 #12
    NICE JOB.
    Very nicely done. The only correction I would make would be that the Applied Force is no longer present after the stone is released. So, in the horizontal direction, the only force acting on the stone is the Kinetic Friction.

    Now, let's translate all this into the language of mathematics. Let
    N = normal force from the ground
    F = kinetic friction force from the ground
    W = mg = weight of stone
    v0= horizontal velocity of stone after applied force has been released

    Now please write the force balance on the stone in (a) the vertical direction and (b) the horizontal direction, based on your identification of the forces involved. For the horizontal direction, don't forget to include ma.

    Chet
     
  14. Nov 22, 2014 #13
    Can you help me with this? I do not know what "force balance" is.
     
  15. Nov 22, 2014 #14
    Newton's second law says that you add up the forces in each direction, and then set the sum equal to the mass times the acceleration in that direction. Make sure that you have the right signs when you add up the forces. Have they covered this in your course yet?

    Chet
     
  16. Nov 22, 2014 #15
    Yes! I think we may have referred to this in different terms though?

    So for the vertical direction: Fnet = FNormal + Fg. This cancels out giving zero.

    For the Horizontal Direction we have Fnet = ma
    Ffk = μkFN
     
  17. Nov 22, 2014 #16
    I'm going to express your results in a little different form, using the symbols I mentioned in my previous post:

    Vertical:

    N - mg = 0

    Horizontal:

    - F = ma

    As you already indicated, the friction force F is related to the normal force N by:

    F = μkN

    Now solve the Vertical equation for N in terms of mg, and then substitute that into the equation for the frictional force F. Then substitute this into the Horizontal equation. What do you end up with?

    Chet
     
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