電纜的壓力為p = fs = 2500n 0.01 m2 = 2.5×105 pa;
(2)重力所做的功為W = GH = 2500n×600m = 1.5×106J;
(3)吊車下降時間t = SV = 3000m2/s = 1500s。
總重力的冪為P = WT = 1.5×106j 1500s = 1000 w .
答:
(1)吊車在水平索道上靜止時,纜繩上的壓力為2.5×105 Pa;
(2)重力做功為1.5×106j;
(3)總重力的功率是1000 W .