You are testing a new amusement park roller coaster with an empty car with mass 127 kg. One part of the track is a vertical loop with radius

Question

You are testing a new amusement park roller coaster with an empty car with mass 127 kg. One part of the track is a vertical loop with radius 11.8 m. At the bottom of the loop (point A) the car has speed 27.1 m/s, and at the top of the loop (point B) it has speed 8.9 m/s. As the car rolls from point A to point B, how much work is done by the friction

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Maris 4 months 2021-09-05T10:22:46+00:00 1 Answers 2 views 0

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    2021-09-05T10:24:10+00:00

    Answer:

    Wf = 7876.5 J

    Explanation:

    To find the work done by the friction force you use the following formula, in which the total work is given by the change in the kinetic energy of the car:

    W_T=\Delta E_k=\frac{1}{2}m(v_B^2-v_A^2)  (1)

    m: mass of the car = 127 kg

    vB: speed at the top of the loop = 8.9 m/s

    vA: speed at the bottom of the loop = 27.1 m/s

    The total work is the contribution of the work done by the gravitational force and also the work done by the friction force, that is:

    W_T=-mgh-W_f  (2)

    g: gravitational acceleration = 9.8m/s^2

    h: height of the car to the ground at point B = 2R = 2(11.8m) = 23.6 m

    Then, you first equal the equations (1) and (2). Next, you replace the values of vB, vA, m, g and h, in order to obtain the work done by the friction force:

    W_f=-mgh-\frac{1}{2}m(v_B^2-v_A^2)\\\\W_f=-(127kg)(9.8m/s^2)(27.1m)-\frac{1}{2}(127kg)((8.9m/s)^2-(27.1m/s)^2)\\\\W_f=7876.5 \ J

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