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How to use the work-energy theorem to find mu?

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data
    please check this link out to make it simpler to understand...

    http://www.scribd.com/doc/62694749/MC-Collision1D2

    there is a friction block and a spring scale... it gives two measurements (the vertical force and the horizontal force)
    Fvertical=88.2 N and
    Fhorizontal=75.8 N
    now i'm suppose to find the coefficient of friction...
    and also, i'm suppose to find out if the driver exceeded the 70 km/h speed limit
    please help me out!


    2. Relevant equations
    m1vi1+m2vi2 = (m1+m2)vf

    KE=1/2(m)(v2)




    3. The attempt at a solution
    for the mu:
    at first, i tried to use the "elementary" method of summation of forces along the axes but since we're talking about collisions, i think i should use the work-energy theorem... but now i don't know how to put mu into the equation: KE=1/2(m)(v2)
    ...

    for the speed limit:
    well, it's an inelastic collision obviously... one car is at rest but they didn't give out the vi1 of the yellow car...

    please help! i ran out of ideas on how to solve this!
     
  2. jcsd
  3. Nov 5, 2011 #2

    Redbelly98

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    Welcome to Physics Forums. I'm going to copy-and-paste the problem statement, just because that makes it easier to discuss here.

    Okay. What is an equation that involves friction force and coefficient of friction?
    Good, that equation is the basis for solving problems like this.

    They don't give vi1 because that is what you are supposed to find out.

    For starters, I would:

    1. Figure out what μ, the coef. of friction, is. You can use the measurements Fvertical=88.2 N and Fhorizontal=75.8 N, along with the usual equation that involves friction force and μ.

    2. Figure out how fast the two cars were going after the collision. You can use the 12.25 stopping distance, and what the acceleration must have been given the μ calculated previously.
     
  4. Nov 7, 2011 #3
    thanks for confirming and killing any doubts i have.

    for the mu, i used the equation for the spring scale and friction block


    µ = Fspring /Fnormal
    µ =Fspring /(mblock •g )

    so i assumed that the horizontal force is from the spring scale and the vertical force is from the friction block...

    so i got:

    = 75.8/88.2
    µ = 0.86

    i don't think i'm right... my physics professor said that i should use the work and KE theorem to solve for mu... how am i suppose to do that?
     
  5. Nov 7, 2011 #4

    Redbelly98

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    You are correct on µ. It sounds like your professor misspoke (or, possibly, was misheard? :smile:). Work and KE theorem can be used to find velocities that are relevant to this problem, but that involves knowing µ already -- which now you do.

    What are the unknown quantities here? The next step is to work on finding their values, and the KE theorem will be useful.
     
  6. Nov 8, 2011 #5
    hahah... tnx! I knew i was right... one way or another.
    with that out of the way...

    so now i got
    m1=2.674 kg
    m2=1.100 kg
    vi2=0
    and with two unknowns...

    vi1=?
    vf =?

    and in order to fine them i got these two equations...
    vi1=(m1+m2)vf/m1 and
    vf= m1vi1/(m1+m2)

    hmm... would i be wrong if i substituted 70m/s to via? if i am... is there any way to find via and vf at the same time with putting mu and distance into these equations?
     
  7. Nov 8, 2011 #6

    Redbelly98

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    Yes, so if you can find vf then you'll have vi1.
    Yes, 70 m/s would be wrong, and for two reasons. (1) It's 70 km/hr, not m/s, but more importantly (2) that is the posted speed limit, and we can't assume the driver's velocity was at the speed limit -- we'd already have our answer then, wouldn't we?

    Instead, apply the work-energy theorem to find vf.

    Work-energy theorem: W = Fnet·d = ΔKE

    Apply that to the final 12.25 m distance that the two vehicles travel before coming to a stop.
     
  8. Nov 8, 2011 #7
    oh my God, thank you!
    i just had that Eureka moment with your help!!!
    XD
     
  9. Nov 8, 2011 #8
    ok!
    So this is what i did...

    well first you need to find Fnet so...
    its
    Ʃfx
    Fa-Ff=+ma

    since the velocity is constant... i think that acceleration is zero... so i get

    75.8-Ff=0
    ________________________________
    and next...
    ƩFy
    FN-Fw=0
    FN=Fw
    FN=88.2

    and if we apply the coefficient of friction to the Ʃfx... we still get zero so it checks...
    _____________________
    so our Fnet=88.2

    using that we get...
    W = Fnet·d
    W = 88.2(12.25)
    W = 1080.45
    ________________________
    then applying the energy-work theorem wherein W=ΔKE
    ΔKE=1/2mv2
    1080.45=1/2(2.674)(v2)
    1080/1.337 = v2
    808.12=v2
    v= 28.43

    Right?

    Thank you for not giving up on me and for not giving me the answers directly and letting my brain do some work...

    so all in all... i won't be pressing charges because he did not exceed the speed limit... it's the red car's fault for stopping anyway... heheh
     
  10. Nov 8, 2011 #9
    but before doing that... i should watch the units more carefully next time...

    yeah... i will have to charge the yellow car with over speeding... i should really know my units...

    so i converted 70 km/hr to m/s

    and i get 19.44m/s
    and the driver did... get over that limit...

    sorry, sometimes i'm a klutz...
     
  11. Nov 8, 2011 #10

    Redbelly98

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    Let me stop you here.

    What is this force Fa? Is it a force that acts on the cars? If you haven't already, draw yourself a force diagram that describes the cars skidding 12.25 m after the collision.

    After the collision, the cars are going at speed vf, then the come to a stop (v=0) in 12.25 m. So the speed changes, it is not constant.

    88.2 N is the weight of the friction block. The cars are different than that, right?
     
    Last edited: Nov 8, 2011
  12. Nov 10, 2011 #11
    doh! me and my hastiness...
    i will work on this!
     
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