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Forces problem

  1. Oct 13, 2006 #1
    Hello, I am a junior in high school now taking AP Physics B. i'm stuck on a series of questions regarding force:

    1. Assume you are on a planet similar to earth where the acceleration of gravity is 10m/s^2 and the positive directions for displacement velocity and acceleration are upward. At t=0sec., an elevator is at a displacement of x=0 with a velocity of v=0m/s. A student whose weight is 400N stands on a scale in an elevator and records the scale reading as a function of time. The data is shown below:

    0-5s--------400N
    5-10s-------500N
    10-15s------600N
    15-20s-----0N

    a) Which direction is the force due to the scale?
    b) What is the mass of the student?
    c) Calculate the acceleration of the elevator during the 15-20s interval.
    d) Draw the a vs. t graph.
    e) What is the velocity of the elevator at t=20s?
    f) Draw the v vs. t graph.
    g) What is the displacement of the elevator above the starting point at the end of 20s?
    h) Draw the d vs. t graph.

    I was only completely sure about a and b, for a the direction is up and for b the mass is 40kg. I obviously don't want just answers, but steps and hopefully explainations to go along with it.

    Thank you.
     
  2. jcsd
  3. Oct 13, 2006 #2

    OlderDan

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    a) and b) are good. Show us what you tried to do for the other parts that have numerical answers. If you can post or at least describe the graphs, that would be good too.
     
  4. Oct 14, 2006 #3
    Well, since they all depend on each other (from accel find velocity from velocity find displacement) so I'll show you what I did for the acceleration problems:

    I looked at the readings, the weight force from the student is constant at 400N, the only thing that changes is the tension in the elevator string (the opposing force, i assume). So if the scale reads 500, as it does at 5-10sec, i assumed that to be the net force, and thus (since i assume it's going up) the tension in the string must be 900N. so using f=ma, i said:

    T-mg=ma
    t-400=40a
    T=40a+400

    So for example, at 5-10sec, i found the tension (T) to be 900N, but that gave me an acceleration of:

    900=40a-400
    1300=40a
    a=32.5

    which can't be right. but if i make it going down in the 5-10sec.:

    a=7.5

    I'm just not sure which is correct.
     
  5. Oct 14, 2006 #4

    OlderDan

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    900N is not correct. The student weight is 400N and constant. When he sees 400N on the scale it is because the scale is just supporting the weight, no more and no less. In this case the net force acting on him is zero. When s/he reads 500N it is because the scale is pushing up with a force greater than the 400N needed to support the weight. The 500N acts upward. The 400N weight is still downward. The net force on the student is 100N upward
     
  6. Oct 14, 2006 #5
    Okay, seems simple enough. I was thinking in terms of the elvator cable tension effecting the reading, not the force pushing upward from the scale. so at no net force on the student (400N reading) the scale pushes up 400n, on a 500N reading, the scale pushes up 500n, right? What about when the scale reads 0?

    my teacher never did a problem like this, so the whole scale thing is odd to me.
     
  7. Oct 14, 2006 #6

    OlderDan

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    The scale always reads the force it is applying to the student. When it reads 500, the net force on the student is 100N upward. When it reads zero, the only force acting on the student is gravity. So what happens to the student? Unfortunately, this last condition will end rather abruptly at some point.
     
  8. Oct 15, 2006 #7
    ok, i understand now. I've finished the problems and would like feedback:

    6. Calculate the acceleration of the elevator at the end of 20s: -10m/s2 (the g value given in the problem)

    7. Choose the correct a vs t graph: It looks exactly like the f vs t graph. Only the F vs T y-axis goes up by 100's and the a vs t graph goes up by 2.5's

    8. What is the velocity of the elevator at the end of 20s? Since acceleration doesn't remain constant, i split it as so using a=change in v/t:

    0=vf-0/5
    vf=0

    2.5=vf-0/5
    vf=12.5

    5=vf-12.5/5
    vf=37.5

    -10=vf-37.5/5
    vf=-12.5

    I took the acceleration and the v0 calculated by the vf in the last 5sec interval to solve. so, the answer i gave was -12.5m/s.

    9. Choose the correct v vs t graph: easy enough, i used my calculations from the previous problem.

    10. What is the displacement of the elevator above the starting point at the end of 20s: I used vf^2=v0^2+2ax to solve for the displacement. Since acceleration was not constant, i again split it up:

    0^2=0^2+2(0)(x)
    x=0m

    12.5^2=0^2+2(2.5)(x)
    x=31.25m

    37.5^2=12.5^2+2(5)(x)
    x=125m

    -12.5^2=37.5^2+2(-10)(x)
    x=-62.5 (i knew this had to be negative, even though the math doesn't say so)

    xtotal=0+31.25+125-62.5
    xtotal=93.75m

    11. Choose the correct d vs t graph: I had trouble with this one, none of them really matched my results. The closest one i had was at the end of 15s the displacement as about 90m. One does show the displacement at the end of 15sec being about 156m, but at the end of 20sec it reads the displacement as only about 210m, which doesn't make sense.
     
  9. Oct 15, 2006 #8

    OlderDan

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    What makes you think the last x has to be negative?

    I didnt see anything wrong with what came before that.
     
  10. Oct 15, 2006 #9
    I figured it was because it was accelerating downward.
     
  11. Oct 15, 2006 #10

    OlderDan

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    When you throw a ball in the air, as soon as you release it, it is accelerating downward, but it is still going upward. It takes time and distance to reverse the direction of motion.
     
  12. Oct 15, 2006 #11
    oh, i see. Thanks for your help
     
    Last edited: Oct 15, 2006
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