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Homework Problem about the Work-Energy Theorem

  1. Nov 22, 2007 #1
    [SOLVED] Homework Problem about the Work-Energy Theorem

    Please help! This is my first year of taking physics and Im really confused....

    A puck on a shuffleboard decreases in speed from 1.0 m/s to .20 m/s while traveling 4.0 m. What is the coeffecient of kinetic friction between the puck and the board?



    I think that the equations we have to use to solve this would be....


    w= [tex]\frac{1}{2}[/tex]m[tex]V_{1}^{2}[/tex]-[tex]\frac{1}{2}[/tex]m[tex]V_{o}^{2}[/tex]

    w=Fd

    [tex]F_{f}[/tex]=[tex]\mu[/tex][tex]F_{N}[/tex]

    w = [tex]\frac{1}{2}[/tex]k[tex]d^{2}[/tex]

    k= [tex]\frac{1}{2}[/tex]m[tex]V^{2}[/tex]


    First : I was thinking about using the equation w= [tex]\frac{1}{2}[/tex]m[tex]V_{1}^{2}[/tex]-[tex]\frac{1}{2}[/tex]m[tex]V_{o}^{2}[/tex] to solve for work, then substitute the work value in for w=Fd then somehow put it in the equation [tex]F_{f}[/tex]=[tex]\mu[/tex][tex]F_{N}[/tex] . Then I got extremely confused...and I realized that I don't have the mass of the puck to put into the first equation.

    Second try: So, I decided to use w = [tex]\frac{1}{2}[/tex]k[tex]d^{2}[/tex] but in order to find out the k (spring constant) value, I need to use the equation k= [tex]\frac{1}{2}[/tex]m[tex]V^{2}[/tex]...which again, I dont have the mass for.

    Third try...I was going to use the equation F=kd to figure out F and then substitute it in for w in w=Fd and then solve for m in w= [tex]\frac{1}{2}[/tex]m[tex]V_{1}^{2}[/tex]-[tex]\frac{1}{2}[/tex]m[tex]V_{o}^{2}[/tex]
    . But I don't have the k again!!

    And now, I'm out of ideas...please help me...thank you! :) I hope this question makes sense....and my tries...if not, let me know and I'll try to fix them up! Thanks again!
     
  2. jcsd
  3. Nov 22, 2007 #2

    rl.bhat

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    Homework Helper

    Find the retardation of the body by using kinematic equatuon. Retarding force = m*a and normal reaction = mg. Plug it in the equation to get coefficient kinetic friction
     
  4. Nov 22, 2007 #3
    Try this

    Force of friction is given by:

    [tex]F_{f}[/tex] = [tex]\mu[/tex] N [1]

    Solving for [tex]\mu[/tex] we have

    [tex]\mu[/tex] = [tex]\frac{F_{f}}{N}[/tex] [2]

    The normal force N is given by:

    N = mg [3]

    Substituting [3] into [2] we have:

    [tex]\mu[/tex] = [tex]\frac{F_{f}}{mg}[/tex] [4]

    The force of friction is also defined by:

    [tex]F_{f}[/tex] = ma [5]

    where a is the deceleration and m is the mass of the puck. We also know given a distance and a constant acceleration (or deceleration in this case) that the acceleration is equal to the following (see: http://library.thinkquest.org/10796/ch3/ch3.htm ):

    a = [tex]\frac{(V_{f}^{2} - V_{i}^{2})}{2d}[/tex] [6]

    Substituting [6] into [5] we have:

    [tex]F_{f}[/tex] = [tex]\frac{m(V_{f}^{2} - V_{i}^{2})}{2d}[/tex] [7]

    and finally substituting [7] into [4] we have:

    [tex]\mu[/tex] = [tex]\frac{(V_{f}^{2} - V_{i}^{2})}{2dg}[/tex]
     
  5. Nov 23, 2007 #4
    photonsquared, I used your method to solve the problem....it made sense to me and this is the answer that I got [tex]\mu_{k}[/tex]= .0122. Is there anyway you could tell me if this is correct? Thank you very much for your help! :smile:
     
  6. Nov 23, 2007 #5
    That is the same value I calculate. The reason you don't need the mass of the puck is because the masses cancel out when you substitute eqn 7 into 4. Hope this helps.
     
  7. Nov 23, 2007 #6
    Ok thank you very much photonsquared! :) Thanks rl.bhat for your response!
     
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