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Conservation Of Energy

  1. Oct 27, 2008 #1
    1. The problem statement, all variables and given/known data
    A projectile of mass 1.350 kg is shot straight up with an initial speed of 13.0 m/s.

    (a) How high would it go if there were no air friction?
    I found this to be 8.622 m.

    (b) If the projectile rises to a maximum height of only 7.4 m, determine the magnitude of the average force due to air resistance.



    2. Relevant equations
    u + k = u + k

    3. The attempt at a solution
    I found the difference in potential energy at m = 7.4 and m = 8.622, which equals the work done. I then solved for F in the formula W = F D cos theta. What did I do wrong?
     
  2. jcsd
  3. Oct 27, 2008 #2

    mgb_phys

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    You did the first part by energy conservation ?
    If you apply the same thign to the second part how much energy is lost?
    Whats the relationship between force / distance /energy?

    ps. where does cos theta come from?
     
  4. Oct 27, 2008 #3

    LowlyPion

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    Show your work for b). I think your method is ok, you must have made a simple error.
     
  5. Oct 27, 2008 #4
    I'm not too sure what you mean, although I do believe I did energy conversion for part a.
    (8.622)(9.8)(1.350)-(7.4)(9.8)(1.350) = 16.16706

    W=FD cos theta
    16.16706 = F(8.622 - 7.4) cos 180
    F = -13.23
    It asks for magnitude so I used 13.23.
     
  6. Oct 27, 2008 #5
    What did I do wrong?
     
  7. Oct 27, 2008 #6

    LowlyPion

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    You found the work from the air resistance, but your calculation is incorrect.

    16.17 N-m = F * D = F * 7.4

    F = 16.17/7.4 N
     
  8. Oct 27, 2008 #7
    Which calculation, and why are you using 7.4 for your D?

    By the way, thank you for all of your help.
     
  9. Oct 27, 2008 #8

    LowlyPion

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    7.4m is the distance over which the force acts.

    The difference in distance is really not relevant except insofar as it helps you determine what the total work of the resistance was.
     
  10. Oct 27, 2008 #9
    Question (b) is ambiguous: the "average" of the friction force could refer to an average over distance but also to an average over time. Since the velocity of the projectile is changing, the two averages are not the same.
     
  11. Oct 27, 2008 #10

    LowlyPion

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    I don't think it is all that ambiguous. The difference in resulting Max potential energy and Max theoretical PE gives you the total for the integral of the variable Force acting over that distance. Using then the total distance to determine the average looks appropriate to me.
     
  12. Oct 27, 2008 #11
    I know that there is one obvious solution, but that does not mean the question isn't ambiguous!

    True: you would have to know more in order to calculate the time-averaged force: for instance, that the magnitude of the air friction force is proportional to the speed of the projectile.
     
    Last edited: Oct 27, 2008
  13. Oct 28, 2008 #12

    alphysicist

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    Hi borgwal,

    I don't agree; I don't think the time-average force would be valid for this problem. I do agree that the time average and displacement averaged forces are different here:

    [tex]
    F_{\rm (t-ave)} = m g - \frac{m v_i}{\Delta t}
    [/tex]

    [tex]
    F_{\rm (y-ave)} = m g - \frac{1}{2} m \frac{v_i^2}{\Delta y}
    [/tex]

    But the real question is what do we really want when we are asked for an average force? We want a constant force, such that if we replaced the real force with that constant force, the result would be the same.


    So in the problem from the original post, with the "real" air resistance force the object rises to a height of 7.4m, and it takes some unknown time [itex]T[/itex] to reach that height.


    If we replace the real force of air resistance with the constant time-averaged force, it would take the same amount of time [itex]T[/itex] to reach its maximum height, but it would reach a different maximum height (not 7.4m).

    If we replace the real force of air resistance with the constant y-averaged force, it would reach a maximum height of 7.4m, but would take a different amount of time (not equal to [itex]T[/itex]) to reach that height.


    Since in this problem they give us the height as a constraint, using the time-averaged force would not fit the original problem.
     
    Last edited: Oct 28, 2008
  14. Oct 28, 2008 #13
    Yes..I know I'm only nitpicking for the fun of it, sorry! I know what the natural and obvious meaning of the question is, and I'd be surprised if a good student would assume anything else. Nevertheless, I stick with my point:

    1) when quoting an average, one should always specify an average over what is taken, unless the average is *always* defined to be over the same variable (but for forces, we average sometimes over time, and sometimes over displacement).


    2) the friction force is not a function of position, nor of time, but of velocity (to a good approximation). In this way, too, I see no difference between time- and position averages, and might as well argue for a velocity-average!

    3) if I'd give you *more* information, namely that the projectile reaches a height h in a time t, then your method of replacing the actual force by a constant force would not work. I'd say the additional information does not *make* the question ambiguous, it just brings out the ambiguity of the original question.


    Again, I'm just having fun...
     
  15. Oct 28, 2008 #14

    alphysicist

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    Same here! I always fear people will get defensive and take my comments somehow personally; but I just like discussing these things.

    This is what I would completely disagree with! There are all types of averages: functional averages over different variables, arithmetic averages, geometric averages, harmonic, etc. If the type of average is specified then it is just a mechanical process of filling in the formula.

    But we don't take an average just to be taking an average. We take it for a purpose--and the purpose here is to find a "simple" force (constant force) that can be used to replace the real, complicated force.



    Let me give two trivial problems with simpler averages:

    1. Albert can make 10 sandwiches per hour, and then he packages them at a rate of 20 sandwiches per hour. What is the average rate?

    2. Bob makes 10 pizzas per hour, and Charles makes 20 pizzas per hour. What is the average rate?


    In each case we have rates of 10 and 20. Now if the problem had specified arithmetic average rate (which would give 15), or geometric average rate (giving 14), or harmonic average rate (giving 13), you would just fill in the formulas and get those answers.

    But like I said, you take an average for a purpose. In #1, we would want to find at what single rate would Albert have to work at making sandwiches and also packaging them so that he would have the same output. In that case the harmonic average of 10 and 20 gives the answer. If he both makes and packages at 13 sandwiches/hour (about 13.3), his output is the same.

    For #2, we want to know at what single rate could Bob and Charles both work at to give the same output. In this case is is the arithmetic average. If Bob and Charles both make pizzas at 15 pizzas/hour, their total output is the same.


    In both cases, we don't need (or want!) to specify which average to take--the point of the average is to replace the variable rates with a uniform rate to give the same outcome, and that is what specifies which average to use.


    I would claim that the same type of reasoning is important for the problem in this thread. We have an input (mass=1.35kg, initial speed=13m/s), and an output (max. height=7.4m), and the purpose of the average is to find a uniform force that could replace the non-uniform resistive force and still give the same output. The time-average of the resistive force will not do that; the position average of the resistive force will.





    That's what I disagree with you about: you are not arguing at all for a particular type of average. Do you have an argument why the velocity-average might be better?

    Of course I realize you brought up the velocity-average as just a rhetorical point. But when you say there is no difference between the time and position averages, I would say there is a huge difference--the position average force gives back the measurements in the problem, the time average force does not.


    I would definitely say that it would make the question ambiguous if the time were given.


    Here's another trivial example, built up in 4 parts:


    A: A rectangle has sides of length 1 and 4. We want to replace these non-uniform sides by sides of constant length (in other words, make a square!). What will be the length of each side of the square?-- This is completely ambiguous and confusing.

    B: A rectangle has sides of length 1 and 4. We want to replace these non-uniform sides by sides of constant length so that the perimeter is the same. What will be the length of each side of the square?-- Input is sides of 1 and 4; output is perimeter=10. There is no ambiguity here. The arithmetic average of 1 and 4 will give the length of each side of the square.

    C: A rectangle has sides of length 1 and 4. We want to replace these non-uniform sides by sides of constant length so that the area is the same. What will be the length of each side of the square?-- Input is sides of 1 and 4; output is area=4. There is no ambiguity here. The geometric average of 1 and 4 will give the length of each side of the square.

    D: A rectangle has sides of length 1 and 4. We want to replace these non-uniform sides by sides of constant length so that the area and perimeter is the same. What will be the length of each side of the square?-- Input is sides of 1 and 4; output is perimeter=10 and area=4. This is not just ambiguous--there is no answer, because we have added more requirements than an averaged (uniform) side length can satisfy.



    At each point, a particular average for parts A through D could have been specified, and an answer gotten just by filling in the blanks in the formula. But that would miss the point!



    The same thing applies to the problem in the thread. If the max height, and the time to reach the maximum height were both given, there is no average force that could recreate the output given--so what would be the point? It would just be an exercise in filling in a formula.


    Me too! (Typing this was much more enjoyable than what I should be doing!)


    Since this post is rather long, let me just summarize my argument. An average is not some arbitrary quantity to calculate, but it is an answer to this question: What uniform function (or single value) can we replace a non-uniform function (or set of values) and still get the same output? With that point of view, the decision of what average to use is an essential part of understanding the problem, and so should not be specified in a problem.
     
  16. Oct 28, 2008 #15
    This was fun! I appreciate your point: if there is one thing I dislike in problems, it is blindly plugging in some values into a prescribed equation.

    Now, writing this post took an average of 20 words per minute: the first sentence took an average of 28 words per minute, this one took 12 [just illustrating taking averages in creative ways :-) ]
     
  17. Oct 28, 2008 #16

    alphysicist

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    I enjoyed our discussion too, borgwal. See you around!
     
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