[單元]
牛頓定律
[來源]
101輔大物理考古題
[題目]
A 1600-kg elevator is carrying passengers having a combined mass of 200kg.
A constant friction force of 4000 N retards its upward.
a) What is the minmum power delivered by the motor to lift the elevator at
a constant speed of 3.00 m/s?
b) What power must the motor deliver at the instant speed of the elevator
is v if it designed to provide an upward acceleration of 1.00 m/s^2 ?
[想法]
第一題：當電梯以3(m/s)的速度上升，所做最小的功。
這應該是可以使用P=Fv求解得P = 4000*3 = 12000(J)
(不曉得正不正確)
第二題：當電梯以速度v上升，同時帶有1(m/s^2)的向上加速度，問必須做多少功。
這題就搞不清楚了....
一樣使用P=Fv嗎？
P=mav？？ 不.... 還有一個向下加速度的9.8啊....Orz
有請解惑。