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Power Problem at inclined angle

  1. Oct 14, 2008 #1
    A motorcycle (mass of cycle plus rider = 232 kg) is traveling at a steady speed of 21 m/s. The force of air resistance acting on the cycle and rider is 205 N. Find the power necessary to sustain this speed if (a) the road is level and (b) the road is sloped upward at 29.2° with respect to the horizontal.

    I got through part a..
    where P=Fv=205N * 21 m/s= 4305 W

    but I'm not sure how to handle the opposing force of the motorcycle's weight down the slope...
    I THINK you can find it with
    W= F * cos29.2 s
    = 232kg * 9.8 m/s^2 * cos29.2 * 21
    (where 21=s given the velocity..ie distance per unit time?... or would it be 1/21 s in order to get rid of one of the 1/s)
    THEN with this amount..
    would you add it to part a?.. do you have to reconsider the air resistance since the bike is no longer traveling along the horizontal? (so.. 205N * sin29.2(?.. not cos because force is in opp direction) *21 m/s)..

    anyway.. a bit confused.. any insight would be very appreciated I have to turn this in tonight. Thanks!
  2. jcsd
  3. Oct 14, 2008 #2
    At first glance I believe you are correct in assuming W= F * cos29.2 s
    = 232kg * 9.8 m/s^2 * cos29.2 * 21 m/s then adding this value to the original value calculated in part A. As for your question about units they do work out. Remember that a newton is a kg m/s^2 so you end up with N*m/s or a j/s which is known as the watt!

    Also there should be no need to take the sin component of the force due to air resistance because the air resistance would be a result of the movement of the motorcycle which means the force directly opposes the rider the same way that friction does... unless of course the problem is referring to a windy day.

    edit: at second look taking the cos of 29.2 will give the wrong component of gravitational force that you are looking for
    Last edited: Oct 14, 2008
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