[单元]
牛顿定律
[来源]
101辅大物理考古题
[题目]
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]
第一题：当电梯以3(m/s)的速度上升，所做最小的功。
这应该是可以使用P=Fv求解得P = 4000*3 = 12000(J)
(不晓得正不正确)
第二题：当电梯以速度v上升，同时带有1(m/s^2)的向上加速度，问必须做多少功。
这题就搞不清楚了....
一样使用P=Fv吗？
P=mav？？ 不.... 还有一个向下加速度的9.8啊....Orz
有请解惑。