A boy in a wheelchair (total mass 56.5 kg) has speed 1.40 m/s at the crest of a slope 2.05 m high and 12.4 m long. At the bottom of the slop

Question

A boy in a wheelchair (total mass 56.5 kg) has speed 1.40 m/s at the crest of a slope 2.05 m high and 12.4 m long. At the bottom of the slope his speed is 6.30 m/s. Assume air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N. Find the work he did in pushing forward on his wheels during the downhill ride.

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Kim Chi 4 years 2021-08-04T23:20:57+00:00 1 Answers 141 views 0

Answers ( )

  1. thanhcong
    0
    2021-08-04T23:22:00+00:00
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    Answer: 439.01 J

    Explanation:

    The Kinetic Energy at the top + Potential Energy at the top + Work Done = Kinetic Energy at the bottom + Friction loss

    1/2mv² + mgh + W = 1/2 mv² + Fd

    1/2 * 56.5 * (1.4)² + 56.5 * 9.8 * 2.05 + W = 1/2 * 56.5 * (6.3)² + 41 * 12.4

    1/2 * 56.5 * 1.96 + 56.5 * 9.8 * 2.05 + W = 1/2 * 56.5 * 39.69 + 41 * 12.4

    55.47 + 1135.085 + W = 1121.243 + 508.4

    1190.555 + W = 1629.643

    W = 1629.643 – 1190.555

    W = 439.088

    W = 439.09 J

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