In short-track speed skating, the track has straight sections and semi-circles 16 m in diameter. Assume that a 66 kg skater goes around the

Question

In short-track speed skating, the track has straight sections and semi-circles 16 m in diameter. Assume that a 66 kg skater goes around the turn at a constant 12 m/s.
A. What is the horizontal force on the skater?
B. What is the ratio of this force to the skater’s weight?

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Hải Đăng 5 years 2021-08-16T18:40:14+00:00 1 Answers 522 views 1

Answers ( )

  1. nguyetanh
    0
    2021-08-16T18:41:44+00:00
    Reply

    Answer:

    a. 1,188 N

    b. 1.836

    Explanation:

    The computation is shown below:

    a. For horizontal force, first we need to find out the circular path radius which is shown below:

    As we know that

    r = \frac{d}{2}

    = \frac{16}{2}

    = 8m

    Now the horizontal force is

    F = \frac{m\times v^2}{r}

    where,

    m = 66 kg

    v = 12 m/s

    and r = 8 m

    Now placing these values to the above formula

    So, the horizontal force is

    F = \frac{66\times 12^2}{8}

    = 1,188 N

    b. Now the ratio of force to the weight of skater is

    = \frac{1,188 N}{66\ kg \times 9.8m/s^{2}}

    = 1.836

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