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Homework Help: Figuring acceleration using time and the reading of a weight scale help please!

  1. Oct 14, 2004 #1
    I'm new to this stuff, so the question may be really simple, but I just can't work it right...

    Alright, so the question I'm trying to figure is this:

    The student's normal weight is 500 N. He stands on a scale in an elevator and records the scale reading as a function of time.

    So for the first 5 seconds the scale is at 500 N, thus the acceleration is 0 m/s/s. Then the next 5 seconds the scale reading is at 700. Now how do I go about figuring the acceleration?

    I know F=ma. The only way I could think of solving this problem is putting the difference in for the force variable, the students weight in for the mass variable and then solving for acceleration.

    So for example for the second 5 seconds I used 200=500a and I got 0.4 m/s/s. This doesn't seem like the right answer so I must be missing something or just going about the problem completely the wrong way. If someone could point me in the right direction I'd really appreciate it!
  2. jcsd
  3. Oct 14, 2004 #2
    You have it mostly right, the 'm' in the equation is the student's mass, not weight. weight = mass*gravity.
  4. Oct 14, 2004 #3


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    500/9.8 = 700/x

    Cross multiply, solve for x, subtract g, and that is your acceleration upward.
  5. Oct 14, 2004 #4
    If I did it this way I do 500/9.8=700/x and x=13.72m/s/s. But then the last 5 seconds of the boys motion is downward and the scale reads 200N (300N0 lighter (when he was going up in the second 5 seconds the scale was 200N heavier [700N]). When I do 500/9.8=300/x I get x=5.88m/s/s. That doesn't seem right... So I assume I add 5.88m/s/s to gravity, then I get 15.68m/s/s.

    ...is that correct?
  6. Oct 14, 2004 #5
    No. You are using the formula [itex]F_{net}=ma[/itex], meaning the net force is accounted for (including gravity). You should really be starting all these problems with a free-body diagram, BTW.

    Review your work with this in mind and remember that your final answer should be negative because you defined the upward direction as positive.
  7. Oct 14, 2004 #6


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    Your first part is right. 13.72 - 9.8 is the acceleration.

    For the second part, since it's the opposite direction, you subtract that number from 9.8, instead of the other way around, to get the acceleration downwards. In this case it would be 9.8-5.88=3.92 m/s^2.
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