化學應用物理的問題
2008-02-11 10:30 pm
an elevator weighing 459kg carries 3 people weighing 253kg from the ground to the top floor of the building. a height of 641feet.a. how much energy(J)is required to lift the elevator and people to the top floor.b(i)if the elevator falls from the top(641ft),what is theV ife the elevator and people to the top?what is Vwhen it is 200ft.how long will it take the elevator and three people to reach ground?
回答 (1)
2008-02-11 11:32 pm
✔ 最佳答案
Energy required to life the elevator and 3 people equals to the potential energy at 641 ft.Potential energy = mgh = (459 + 253) kg x 9.81 ms^-2 x (641 x 0.3048) m = 1364652 J
When the elevator start drop, the velocity V at top is 0 ms^-1.
When the elevator drop to 200 ft,
gain of kinetic energy = lost of potential energy
1/2 m v^2 = mgh
1/2 v^2 = gh
1/2 v^2 = 9.81 x (641 - 200) x 0.3048
v^2 = 2637.26
v = 51.35 ms^-1
Total energy at ground equals total potential energy at top. Then, the final velocity at ground,
1/2 mv^2 = 1364652 J
v^2 = 1364652 x 2 / (459+253)
v = 61.91 ms^-1
The time T for the elevator and 2 people accelerate from 0 ms^-1 to 61.91 ms^-1 under gravity,
9.81 ms^-2 x T = 61.91 ms^-1
T = 6.31 s
收錄日期: 2021-04-23 23:37:47
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https://hk.answers.yahoo.com/question/index?qid=20080211000051KK01754