（1）动滑轮重G动=nF-G=3×40N-100N=20N；机械效率η= G nF = 100N 3×40N =83.3%．（2）绳子自由端移动的距离S= W F = 180J 40N =4.5m；物体升高的高度h= S 3 = 4.5m 3 =1.5m．答：可使物体上升1.5m．
