A physics teacher (mass = 80.4 kg) is jogging through the woods and runs straight into a large oak tree at 4.9 m/s. Rebound speed is measure

Question

A physics teacher (mass = 80.4 kg) is jogging through the woods and runs straight into a large oak tree at 4.9 m/s. Rebound speed is measured at 4.0 m/s in the opposite direction. If the time of contact with the tree is 56 milliseconds, what is the magnitude of the force that the tree exerts on the teacher?

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Amity 5 years 2021-08-09T10:25:01+00:00 1 Answers 16 views 0

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  1. diemthu
    0
    2021-08-09T10:26:59+00:00
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    Answer:

    12777.86N

    Explanation:

    This problem is approached by understanding that the impulse felt by the physics teacher is equal to his change in momentum. This is mathematically represented as follows;

    Ft=mv-(-mu)................(1)

    where m is his mass, v is his final velocity, u is his initial velocity, t is the time for which the force F acts on him.

    It should also be noted that while he rebounds there is a change in his direction and that is why the mu carries a negative sign.

    Equation (1) can therefore be written as,

    Ft=mv+mu\\F=\frac{m(v+u)}{t}.............(2)

    equation (2) is a statement of Newton’s second law of motion.

    Given;

    m = 80.4kg

    u = 4.9m/s

    v = 4.0m/s

    t = 56 milliseconds = 0.056s

    Therefore;

    F=\frac{80.4(4.9+4.0)}{0.056}\\F=\frac{80.4*8.9}{0.056}\\F=12777.86N

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