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| *Vultures Knob>>>Street Bike |
Riding Up Hill? |
Suppose a street is 300 feet long, and quite steep (it goes up a hill). If I ride my bicycle up the hill in a zig-zag pattern that is 600 feet long, the average force I must exert is: a) 1/4 b) 1/3 c) 1/2 d) equal to the average force I would exert going straight up. Again, up the same hill along the zig-zag path the energy I must expend is: a) 1/4 b) 1/3 c) 1/2 d) equal to the energy I would spend going straight up. C) 1/2 the force -- for twice the number of pedal revolutions. D) same energy -- same hill, height x your weight (no, i ain't going there - people who live in glass houses ....). i hope my answer is correct as I'm really not sure. working out the second part 1st we get. the answer as d. the the following equations F1 and W1 represent the force and work done in moving the body straight up the hill. F2 and W2 are the force and work in a zig zag path. E=mgh height of the hill remains constant in both cases,and also mass and g. so the total energy will be the same in both cases. now working backwards since the potential energy is the work done to get the body up the hill W1=F1*S W2=F2*S since work is the same in both cases and the overall displacement is 300feet we should get F1*S=F2*S since S is equal it will cancel out leaving us with F1=F2. hence the answer here is also d. |
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