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Finding average force and time

  1. Feb 28, 2016 #1
    1. The problem statement, all variables and given/known data

    When an 98.0-g piece of toast is inserted into a toaster, the toaster's ejection spring is compressed 7.20 cm. When the toaster ejects the toasted slice, the slice reaches a height of 3.4 cm above it's starting position. What is the average force that the ejection spring exerts on the toast? What is the time over which the ejection spring pushes on the toast. Assume that throughout the ejection process the toast experiences a constant acceleration.

    2. Relevant equations
    Fs=-k/\x
    weight=mg
    PE=mgh

    3. The attempt at a solution
    I'm not entirely sure how to get the spring constant
    I've tried equating PE=Fs...
    mt=98.0g
    Sc=.0720m
    Ht=.034m

    Thus, mgh = -k/\x
    (98.0)(.034)= -k(.034-.0720)
    k=89.76

    but this seemed rather large, so then I assumed the mass should be in kg:
    k=.0877
    which made a little more sense.
    Do I now just plug those values in Fs=-kx ?
    Fs=-(.0877)(.034-.0720)= .00333 ?
     
  2. jcsd
  3. Feb 28, 2016 #2

    jbriggs444

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    This is not an easy question to interpret. On the one hand, it asks for "average force". On the other hand it specifies that acceleration is to be treated as a constant. The way that I reconcile those two clauses is to imagine that the toaster spring is not free to rise all the way to its relaxed length. Instead, it is held captive in a track or other arrangement. That means that the upward force that it exerts on the slice of bread is nearly constant throughout its 7.2 cm stroke and the spring constant is irrelevant. The "average force" clause acknowledges that the force is not constant. The "constant acceleration" clause gives us permission to ignore that lack of constancy. The 3.4 cm is the height to which the toast keeps flying upward after the spring hits the top of its track and stops.

    For many toasters, the assumption of a captive spring arrangement and flying toast is realistic.

    It was not easy to read your response.
    You declared a variable "mt" which presumably represents the mass of the toast. But then when you went to write down a formula you ignored that variable and used "m" instead. You declared a variable "Sc" which presumably represents the Spring compression. But again, instead of using that variable, you used "x" instead. And you used "k" to denote some sort of spring constant.

    But it is not at all clear what "-k/\x" is supposed to mean. Is that multiplication, exponentiation, squaring, extracting a square root or what?

    And I have no idea what significance there is to .034 - .0720.
     
  4. Feb 28, 2016 #3
    When you calculate that the force is equal to K*d you are basically getting the force at that exact moment.
    Remember energy conservation, If it made the toast go up to that height then it must did work to it this work is equal to its potential energy, If you want to get the k you can get it by that. But you dont need it to solve this question.

    Remember the work formula: F*d
    F here is the average force that if times to d gives me the work that I have done. So basically once you get numeral value of work make it equal to the formula of the work and the d here should be??? The distance where the springs pushes the toast. Think about it a little bit.
     
  5. Feb 28, 2016 #4

    haruspex

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    I'm afraid that's an all-too-common misconception. Average force is ##\vec F_{avg}=\Delta \vec p/\Delta t##, where ##\vec p## is momentum. If the force is not constant then this is very unlikely to satisfy ##|\vec F_{avg}|=\Delta E/|\Delta \vec s|##.
     
    Last edited: Feb 28, 2016
  6. Feb 28, 2016 #5

    jbriggs444

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    Agreed in general. But in this case there is a stipulation that the force may be treated as a constant.
     
  7. Feb 28, 2016 #6

    haruspex

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    Yes, but Biker's post omitted that, so made it seem that it could be applied in general.
     
  8. Feb 28, 2016 #7
    Yeah, i apologize. i was trying to say /\x meaning delta x or displacement rather.

    .034 - .0720 is what I assumed to be the displacement.

    I am lost as how to go about to solve this honestly.
     
  9. Feb 28, 2016 #8
    How do I go about solving this without the k constant?

    for displacement would it be (.034 - .0720) ? if W = Fd how would I go about finding the work without the force and merely the displacement? should I first find the constant acceleration and re-equate it: W = (ma) * d ?
     
  10. Feb 28, 2016 #9

    haruspex

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    What may be getting in your way is that you are given the mass. Dimensional analysis shows this to be irrelevant. There are no other givens which involve a mass dimension, just two distances and the acceleration g when airborne. So knowing the mass of the toast cannot help in finding the other acceleration.
    Just treat it as a SUVAT problem in two stages, each with its own constant acceleration.
     
  11. Feb 28, 2016 #10
    Ah I see. So I'm left with the two distances and gravity.

    Forgive my ignorance, but what is a SUVAT problem?
     
  12. Feb 28, 2016 #11
    Also, from what I gathered is my initial starting point -.0720 since it's below its equilibrium, d = (0-.0720) ?
     
  13. Feb 28, 2016 #12

    haruspex

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    The SUVAT equations involve five variables, s, u, v, a, t. They relate distance, initial speed, final speed, acceleration and time. They are only valid for constant acceleration. You probably know them, but perhaps not by that name.
    It is a little confusing, but the spring mechanism raises the toast by .072m, then the toast continues in flight another .034m. Each stage has constant acceleration, so you can apply the SUVAT equations to each stage separately. One variable links the two: which is it?
     
  14. Feb 28, 2016 #13
    Distance? Or would it be time?
     
  15. Feb 28, 2016 #14

    haruspex

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    Each has its own distance, each takes its own time. Think again.
     
  16. Feb 28, 2016 #15
    Speed then right?
     
  17. Feb 28, 2016 #16

    haruspex

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    Yes. Each SUVAT equation involves four of the five variables, so the trick is to figure out which one you do not care about and use the equation that omits it. In the first phase, which variable is uninteresting?
     
  18. Feb 28, 2016 #17
    I see, that make sense.
    Acceleration is constant so that isn't needed.
     
  19. Feb 28, 2016 #18

    haruspex

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    That's the one. You are not told the acceleration, you are not asked to find it, and you do not need it for the second phase. You have enough information to find it, but that's just extra work.
     
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