Ask Question
6 November, 09:58

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.

+2
Answers (1)
  1. 6 November, 12:18
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers