Physics

A child bounces a 48 g superball on the sidewalk.
The velocity change of the superball is
from 22 m/s downward to 19 m/s upward.
If the contact time with the sidewalk is 1
800
s, what is the magnitude of the average force
exerted on the superball by the sidewalk?
Answer in units of N

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  2. 👎 0
  3. 👁 67
asked by famin
  1. t = (1/800)s.
    V = Vo + a*t = 19.
    22 + a*(1/800) = 19,
    a/800 = -3,
    a = -2400 m/s^2.

    F = M*a = 0.048 * (-2400) = -115.2 N.
    The negative sign means the force opposes the motion.

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    posted by Henry2,
  2. Correction: V = Vo + a*t = 19 - (-22).
    -22 + a*(1/800) = 41.
    a/800 = 63,
    a = 50,400 m/s^2.

    F = M*a = 0.048 * (50,400) = 2419 N.

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    2. 👎 0
    posted by Henry2,

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