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Kinetic Energy and Work Energy Theorem

  1. Feb 26, 2008 #1
    A 65kg bicyclist rides his 8.8kg bicycle with a speed of 14m/s. (a) How much work must be done by the brakes to bring the bike and rider to a stop? (b) How far does the bicycle travel if it takes 4.0s to come to rest? (c) What is the magnitude of the braking force?

    So for part (a) the equation I would use is W=1/2mv_f^2 - 1/2mv_i^2 correct? Which then I get for an answer is 7232.4=W which means 7232.4 Newtons?

    For part (b) I know the work that it takes to stop the bicyclist and bicycle but what I'm confused on what equation I would need to use to figure that out.

  2. jcsd
  3. Feb 26, 2008 #2
    For part (a), work is the change in kinetic energy, which you did good...but what are the units for work? If it is equal to KEf - KEi, what does that tell you?

    For part (b), think kinematics.
  4. Feb 26, 2008 #3
    O yeah it's joules for part (a)

    For part (b) it would be just velocity/time = -average acceleration then I would just multiply that by the amount of time? correct?
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