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Work and Energy Theorem

  1. Jun 15, 2009 #1
    1. The problem statement, all variables and given/known data

    A Hot Wheels car, whose mass is 200g is hauling a mass block of 150g along a flat table surface
    with a string as shown in the graph below. Assume the car and the attached mass is moving with
    an initial velocity of 1.7 m/s. A student predicts that the car will travel a maximum distance of
    42 cm. Is this a reasonable prediction? Explain fully.


    Same

    2. Relevant equations

    KE = 1/2mV2
    [tex]\Sigma[/tex]F = mgcos[tex]\theta[/tex]s

    3. The attempt at a solution

    I know that the formulas listed above are needed to solve this. After plugging in:

    m = .350, v = 1.7, I get KE = .505 J
    But where do I go from here to get distance? What would be theta be in this instance?

    I have this formula KEf = KF0 + [tex]\Sigma[/tex]F = mgcos[tex]\theta[/tex]s, but I'm not sure what to use for theta?

    I've attached the image to this post

    Thanks for any hints!
     

    Attached Files:

    Last edited: Jun 15, 2009
  2. jcsd
  3. Jun 15, 2009 #2

    cepheid

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    Staff Emeritus
    Science Advisor
    Gold Member

    We can't see your attachment yet, because it is pending approval. So it's a bit hard to help you with your problem.

    Theta typically refers to an angle. There is no angle in this problem, because, as it says in the description, the table surface is flat.

    What was the the information source that gave you the idea that you needed to use those formulae? Have you considered that perhaps they just may not be relevant to this problem? Also, what is the letter s in your sum of forces equation? mgcos(theta) already has units of force. So, if s is distance, then that equation makes no sense.
     
  4. Jun 15, 2009 #3
    The s stands for distance in the equation. This equation was what we discussed in class, so i assume it is what we should use, since it is the only one I've seen with all the variables:

    velocity, distance, and mass.

    The picture is something like this, where the 200g car is hauling that 150g block off the side of the table:

    | (pulley)--------rope------[200g car]
    | ________table_________________
    |
    r
    o
    p
    e
    |
    |
    |
    [150g block]
     

    Attached Files:

  5. Jun 15, 2009 #4

    LowlyPion

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    Homework Helper

    Welcome to PF.

    The block then - all m*g of it - is retarding the forward motion?

    Initially the car and the block are moving, as a system at 1.7 m/s positive x?

    That would seem to suggest that the retarding deceleration of the system is 3/7*g?
     
  6. Jun 15, 2009 #5

    cepheid

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    Staff Emeritus
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    Gold Member

    First of all can you see that

    [tex] \sum{F} = mg\cos \theta s [/tex]

    makes no sense?

    By a dimensional analysis, your left hand side and your right hand side are not consistent. You have

    [force] = [force] * [distance}

    This makes no sense. The left hand side should be the work done i.e. [itex] W = (\sum{F})s [/itex]

    In words, you are trying to invoke the work-energy theorem, which states that

    Work done = net force*distance = change in kinetic energy = difference between final and inital KE

    or in symbols:

    [tex] W = (\sum{F})s = \Delta KE = KE_f - KE_i [/tex]

    Second of all, can you see that the equations you have written are not necessarily relevant to the problem (esp. the factor of cosine theta, which has nothing to do with anything here!).

    Well, you would be much better off approaching problems using your understanding of physics, rather than using blind matching of formulae to variables! The whole point of doing these labs is to apply your knowledge of physics, rather than to go through a mechanical exercise in applying formulae that is not accompanied by any conceptual understanding of what is going on.

    You would be well-advised to look at LowlyPion's hint, which can be bit cryptic, unless if you realise that the word "that" in his third sentence refers specifically to the student's estimate of a distance of 42 cm.
     
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