physics help?

I don't know how to answer these two questions. Thanks in advanced?

1. A sumo wrestler originally jumps 10 cm above a trampoline but has the same total energy of a small ball that was 200 cm higher than the trampoline. What causes this equivalency?

would it be caused by the difference in mass and height since potential energy=mgh

2. If a 40 kg gymnast and a 400 kg sumo wrestler each dropped from 1 m above the trampoline, find the final position of each athlete. Assume the trampoline is a simple spring obeying Hooke's law with a k value of 12 000 N/m.

How would I answer this?

3 Answers

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  • 6 years ago
    Favorite Answer

    You're right for question 1. The sumo wrestler has a larger mass than the ball, but has a height 20 times less than the ball. This means that the mass of the sumo is 20 times more than the ball. But If you think about it, a small ball is probably within 2kg. If this is the case the sumo is only 40kg, which is not heavy or realistic at all. Because of this, i assume that the sumo is at the peak of his jump, whereas the ball is still moving. This means that the sumo only has gravitational potential energy, whereas the ball has an additional kinetic energy.

    For question 2:

    gravitational potential is equal to elastic potential.

    Position of the sumo wrestler:

    mgh=1/2kx²

    x=√2mgh/k

    =√(2)(400)(9.81)/12000

    =0.808m

    The final position of the sumo wrestler is 0.808 meters below the resting level of the trampoline.

    position for the gymnast:

    mgh=1/2kx²

    x=√2mgh/k

    =√(2)(40)(9.81)/12000

    =0.256m

    The final position of the gymnast is 0.256 meters below the resting level of the trampoline.

    This makes sense because the sumo wrestler is much heavier than the gymnast, and would therefore stretch the trampoline to a greater distance.

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  • Jim
    Lv 7
    6 years ago

    1. Yes, as "g" is constant, it is the PRODUCT of m & h that makes them equivalent.

    2. "final position" is too poorly worded IMHO - I believe the question is better asked this way::

    After each person drops from 1 m onto the trampoline what are the max trampoline deflections of each?

    Solution: convert the max GPE of each person to SPE of trampoline, solve for x {trampoline deflection}.

    gymnast:

    GPE = SPE

    mgh = 1/2kx²

    40(9.8)(1) = (0.5)(12,000)x²

    x² = 0.06533333

    x ≈ 0.26 m ANS

    Sumo wrestler:

    GPE = SPE

    mgh = 1/2kx²

    400(9.8)(1) = (0.5)(12,000)x²

    x² = 0.653333

    x = 0.81 m ANS

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