One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th e cyclist and the cycle, what increase in speed in Km/h can a rider attain by the end of the
1 answer:
Answer:
1/2 M V^2 = .1 M g H where 10% of PE goes into KE
V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2
V = 31.1 m/s increase in speed during descent
1 km / hr = 1000 m / 3600 sec = .278 m/s
V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr
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