动滑轮重60牛用100牛匀速拉物体G上升速度为0.1不记绳重摩擦求G、自由端移动速度、拉力功率、拉力对G的功率
n=3、、、、请答全、、谢谢了、、
人气:436 ℃ 时间:2019-08-20 04:24:22
解答
n=3拉力F=100N,动滑轮中G2=60N,物体上升速度v1=0.1m/snF=G1+G2物体重G1=nF-G2=3*100-60=240N自由端速度v2=nv1=3*0.1=0.3m/s拉力功率P2 = Fv2 = 100*0.3 = 30W拉力对G的功率P1 = GV1 = 240 * 0.1 = 24W...
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