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!! ☺

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

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

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  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.

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

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