Question #118176

a) John is riding a bicycle and comes across a hill of height 7.30 m. At the base of the hill, he is traveling at 6.00 m/s. When he reaches the top of the hill, he is traveling at 1.00 m/s. Jonathan and his bicycle together have a mass of 85.0 kg. Ignore friction in the bicycle mechanism and between the bicycle tires and the road.

i) What is the total external work done on the system of Jonathan and the bicycle between the time he starts up the hill and the time he reaches the top?

ii) What is the change in potential energy stored in Jonathanâ€™s body during this process

i) What is the total external work done on the system of Jonathan and the bicycle between the time he starts up the hill and the time he reaches the top?

ii) What is the change in potential energy stored in Jonathanâ€™s body during this process

Expert's answer

i) the total external work can be calculated as the difference between kinetic energies of the system, so

"A_{ext} = E_{kin,fin}-E_{kin,init} = \\dfrac12 m(v_{fin}^2-v_{init}^2) = \\dfrac12\\cdot85.0\\,\\mathrm{kg}\\cdot(1.0^2\\,\\mathrm{m^2\/s^2} - 6.0^2\\,\\mathrm{m^2\/s^2}) = -1487.5\\,\\mathrm{J}."

ii) The work done by gravity can be calculated as "U = -mgh = -85.0\\,\\mathrm{kg}\\cdot9.8\\mathrm{N\/kg}\\cdot7.30\\,\\mathrm{m} = -6080.9\\,\\mathrm{J}."

We can see that the change of kinetic energy is not sufficient for covering such energy, so Jonathan had to spend his potential energy stored in him. The amount of such energy is

"\\Delta E = -U + A_{ext} = 6080.9\\,\\mathrm{J} - 1487.5\\,\\mathrm{J} = 4593.4\\,\\mathrm{J}."

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