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Spring question or is it?

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data

    When an 86.5-g piece of toast is inserted into a toaster, the toaster's ejection spring is compressed 7.85 cm. When the toaster ejects the toasted slice, the slice reaches a height of 3.9 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

    Not sure. I'm confused about 'assumptions' with the question. Seems broken to me. I approached it using different ways; 1. finding the spring constant of the spring, plugging that into the force formula [f=kx] OR ignoring the fact that the spring is a spring, then comparing forces [setting them equal to each other] and solving for some variable. The first method seems to work, but has given twice the correct answer. I'm not looking for the answer, I'm looking for the methodology.


    3. The attempt at a solution

    k= (2mg(3.9cm+7.85cm) / (7.85cm^2)) = 32.36

    f= k x

    f= (32.36) (.039 - .0785) = 1.27822 N


    correct answer is: 1.27 N
     
  2. jcsd
  3. Nov 20, 2011 #2

    PhanthomJay

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    The spring exerts a force on the toast only when it is in contact with it over its compressed and uncompressed length, a distance of 0.0785 m. . The max force delivered by the spring is kx. Since it delivers no force when x = 0, the average force of the spring is half the max force.
     
  4. Nov 20, 2011 #3
    Thanks!

    How would you find the time?

    question: 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.

    The way I attempted solving this problem was to use acceleration = force divided by mass then use conservation of energy to determine the exit velocity by equating .5kx^2 = .5mv^2. Next, I found time by dividing velocity by acceleration to find time from equation v = axt; however, this is not providing the correct result which is 0.18s; I am coming up with .104 seconds. Any help would be great because I am pretty sure my physics is right.
     
  5. Nov 20, 2011 #4

    PhanthomJay

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    the toast pops up vertically. so you must include its potential energy in your equation, which you neglected to do..
     
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