[力學] 101輔大電梯上升碰阻力做的功率

看板TransPhys作者Trewf (Trewf)時間11年前 (2013/07/03 21:43)推噓0(0推 0噓 9→)

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[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。 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 1.163.19.22

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