[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。