# hw7a' - HW#7a Note: numbers used in solution steps can be...

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HW#7a Note: numbers used in solution steps can be different from the question part. Page 1 of 4 Note: numbers used in solution steps can be different from the question part. You can practice the methods in the solution and verify with the numbers and answers given in the question part. Or you should practice the methods in the solution and verify your calculation with numbers in your webassign. Problem 8-10 are the same questions from previous HWs Problem 1. A new record for running the stairs of the Empire State Building was set on February 3, 2003. The 86 flights, with a total of 1576 steps, was run in 9 minutes and 33 seconds. Consider a scenario where a man ran up 1600 steps of the Empire State Building in 10 minutes and 59 seconds. If the height gain of each step was 0.20 m, and the man's mass was 80.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower. (381 W, 0.511 hp) Solution: the man did work to lift himself up to such a known height within known amount of time. Power = Work / Time The total time was 10 minutes and 59 seconds =659 s, What is the total work done by the man? We know that the work done by the man increased its mechanical energy. W man = W nc = Ef - Ei Before he starts and after he stops the velocity was both zero, and KE were both zero. So the work done by the man is W man = W nc = E f - E i = mgh f - mgh i =mg h f = 80*9.8*(0.2*1600) = 250880 J The man spent 659 seconds to do 250880 J of work. This work increased his potential energy by 250880 J. (His height was increased for 320 m) Power = Work / Time =250880/659 (J/s) = 250880/659 (Watt) = 381 W 1hp= 1 horse power =746 W 381W =381/746 hp = 0.511 hp It looks like the fast or strongest man can be as “powerful” as a half horse. A forum (under communication) was created within Webassign. I posted student question and my answers there. I wrote there: notice that since it is not constant acceleration, you will not be able to find out the instantaneous velocity at particular moments. You can only find v average. You should not use v average to calculate KE initial or KE final or KE at one point or one particular moment. In this one, KE at the very initial point and the stopping point are considered to both be zero. Problem 2, A kayaker paddles with a power output of 47.0 W to maintain a steady speed of 1.50 m/s. (a) Calculate the resistive force exerted by the water on the kayak. (31.3 N) (b) If the kayaker increases her power output by a factor of 1.4 , and the resistive force due to the water remains the same, by what factor does the kayaker's speed change? (1.4) Solution: At the beginning you may be surprised to see that the man was paddling hard, constantly do work to the kayak, why is the speed unchanged? Why didn’t KE increase? Don’t forget that Wnc include not only work done by human, but also work done by resistance. There must be resistance force which did negative work, as a result the total Wnc=0 and KE didn’t change. The total net force must also be zero, human paddling force canceled with the resistance force.

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## This note was uploaded on 08/25/2010 for the course PHYS 111 taught by Professor Jamesm.lockhart during the Spring '09 term at S.F. State.

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hw7a' - HW#7a Note: numbers used in solution steps can be...

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