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!! ☺
4 answers
No, the work done was the difference in kinetic energies, assuming the work to overcome friction at each speed was constant.
So does that mean if I have work against friction & acceleration I can use that one formula to cover both? :)
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.
Ok, that's super helpful thank you very much!! :)
1 answer