Physics 11

A 62kg cyclist changes the speed of a 12kg bicycle from 8.2m/s to 12.7m/s. Determine the work done.

I've already answered & gotten this question correct, however it made me confused about the concept.

I found Ek. & Ek to find change in Ek which was 3479.85. He did work against both friction & inertia & yet this one formula covered both & I didn't have to add anything together. Why not? As far as I understood, to get Work total you need to do the formula for each work (friction & inertia) then add.

Thank you!! ☺

asked by Kayla
  1. No, the work done was the difference in kinetic energies, assuming the work to overcome friction at each speed was constant.

    posted by bobpursley
  2. So does that mean if I have work against friction & acceleration I can use that one formula to cover both? :)

    posted by Kayla
  3. Yes, if the friction involved in attaining different speeds was the same....in real life, it is not. Friction of air is dependent on velocity squared as a rule of thumb.

    posted by bobpursley
  4. Ok, that's super helpful thank you very much!! :)

    posted by Kayla

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