题目内容
用如图所示轮组提升重物,绳子能承受的最大拉力为120牛,动滑轮重60牛,绳重与摩擦均不计.(1)用此滑轮组最多能将多重的物体提升?
(2)用此滑轮组提起重240牛的物体,使它以0.5米/秒的速度匀速上升,人加在绳子自由端的拉力多大?此时滑轮组的机械效率多大?拉力的功率多大?
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【答案】分析:(1)因绳子所能承受的最大拉力为120N,故能提起多重物体,即通过F=
(G+G动)来求重物的重.
(2)通过F=
(G+G动)来求拉力的大小;根据公式η=
=
=
=
求出机械效率;根据公式P=
=
=Fv求出拉力的功率.
解答:解:(1)因绳子最多能承受的拉力F=120N,且F=
(G+G动),故最多能提起物体的重量:G=3F-G动=3×120N-60N=300N;
(2)人加在绳子自由端的拉力:F′=
(G′+G动)=
×(240N+60N)=100N;
滑轮组的机械效率:
η=
×100%=
×100%=
×100%=
×100%=
×100%=80%;
拉力移动速度:v=3×0.5m/s=1.5m/s,
拉力的功率:P=
=
=Fv=100N×1.5m/s=150W.
答:(1)此滑轮组最多能将300N的物体提升;
(2)人加在绳子自由端的拉力为100N;滑轮组的机械效率为80%;拉力的功率为150W.
点评:本题是比较基本的滑轮组拉力、机械效率、功率的计算题,熟练运用公式的变形公式F=
(G+G动)、η=
=
=
=
、P=
=
=Fv是解决此题的关键.
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(2)通过F=
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解答:解:(1)因绳子最多能承受的拉力F=120N,且F=
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(2)人加在绳子自由端的拉力:F′=
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滑轮组的机械效率:
η=
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拉力移动速度:v=3×0.5m/s=1.5m/s,
拉力的功率:P=
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答:(1)此滑轮组最多能将300N的物体提升;
(2)人加在绳子自由端的拉力为100N;滑轮组的机械效率为80%;拉力的功率为150W.
点评:本题是比较基本的滑轮组拉力、机械效率、功率的计算题,熟练运用公式的变形公式F=
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