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Homework Help: Question about 1D Kinematics

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data

    A 55 kg woman jumps to the floor from a height of 1.5 m.

    If her body comes to rest during a time interval of 8.00 * 10^-3 seconds, what is the force of the floor on her feet?

    2. Relevant equations
    Unit conversion and
    g = 9.8m.s^2 ~ 10 m/s^2

    3. The attempt at a solution

    The way I understand it, she is at rest, so

    Fnet = 0

    Fnormal - Fg = 0

    Fnormal = Fg = mg = 550 N

    This is wrong, and the answer is around 300 N. Why is this?
  2. jcsd
  3. Feb 28, 2013 #2
    Intuitively, does your result make sense to you?

    You just said the answer is mg, so the force on her feet is independent of the height from which she jumps. Tell me--if you jump from a step to the ground, do you think you feel the same force as if jumping out of a skyscraper?

    What she's really doing is colliding with the ground. During that time she experiences an average force. Can you figure out what this average force is?
  4. Feb 28, 2013 #3
    Also, are you sure that the answer is 300 N?
  5. Feb 28, 2013 #4
    Sorry, 3.8 * 10^4 N

    Right, so her impact force would be something like

    Fi = ma = m * v / t

    v at 8 * 10^-3 would be

    Vf = Vi + at = 0 - 9.81m/s^2 * 8 * 10^-3 s = 0.07848 m / s

    F = (0.078 m/s / 8*10^-3 s) * 55kg = 0.00053625 N

    Which would be wrong, considering it would have to be opposite of this force and this would be too small

    I know that F= ma, so her Fg - Fi has to = 0
  6. Feb 28, 2013 #5
    Look at this carefully and tell me what you did wrong. Don't blindly use equations; think!
  7. Feb 28, 2013 #6
    Vf = Vi + at = 0 - 9.81m/s^2 * 8 * 10^-3 s = 0.07848 m / s

    All of the units cancel out well, to provide me with (m/s). I'm assuming that initial velocity is zero (when she steps off the object, her initial velocity will be zero). This should get me her velocity at time t. I don't see any error, except for the sign of the answer (which would be negative in this case due to gravity, going down)

    im going to disregard my old attempt and retry another.

    At t = 0.008, she should still be in the air, considering she's in the air for...

    d = vi * t - g/2 t^2
    d = -4.9m/s^2 * 0.008^2 s ^2

    At t = 0.008 seconds, she's still in the air (only fallen -0.3136 millimeters). It seems impossible for her to be at rest, because she's not touching the floor yet
    Last edited: Feb 28, 2013
  8. Feb 28, 2013 #7
    8.00 * 10^-3 seconds is not the time that she drops for--it's the time period of her collision with the ground (i.e. the time period where the force from ground is still accelerating her up). So her final velocity has nothing to do with 8.00 * 10^-3 seconds.
  9. Feb 28, 2013 #8
    Okay, I see the confusion. The question is poorly worded and assumes familiarity with the type of problem it's asking.

    It's really saying "The woman's body comes to rest within 8.00 * 10^-3 seconds after contacting the floor". So, the floor is accelerating her up for 8.00 * 10^-3 seconds AFTER she contacts the floor.
  10. Feb 28, 2013 #9
    Okay, so her final velocity is 5.4 m/s [down], and she's in the air for 0.15 seconds.

    Thank you for your help and for clearing up the confusion.

    (5.4 m/s)/(8.00 * 10^-3 s) * 55 kg = 37 125 newtons
  11. Feb 28, 2013 #10
    Correct (you have a rounding error, the velocity rounded should be 5.5 m/s--should get you closer to 3.8 * 10^(4) N).
  12. Mar 1, 2013 #11
    This calculation seems plausible. Can someone explain why it's wrong?
  13. Mar 1, 2013 #12
    Because the time he used is a measure of how long it takes the floor to accelerate the woman in the opposite direction once she's impacted the form. It is the duration during which the floor is accelerating her (and, physically, it stems from the impact of the woman deforming the surface of the floor).
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