一道物理题
\u4e00\u9053\u7269\u7406\u9898\uff0c\u6025\uff01\u82e5\u7269\u4f53\u6574\u4e2a\u5e95\u9762\u90fd\u5728\u684c\u5b50\u4e0a\uff0c\u5219P=F/s=200/0.1=2000Pa\uff0c\u82e5\u7269\u4f53\u534a\u4e2a\u5e95\u9762\u5728\u684c\u5b50\u4e0a\uff0c\u5219P=F/(s/2)=200/0.05=4000Pa\uff0c\u538b\u5f3a\u503c\u5e94\u8be5\u4ecb\u4e8e\u4e24\u8005\u4e4b\u95f4\u53732000Pa~4000Pa\u3002\u6240\u4ee5\u8fd9\u9053\u9898\u9009BC\u3002
S1\uff0cS2\u90fd\u90fd\u65ad\u5f00\u65f6\uff0cR1R2R3\u4e32\u8054\uff08\u60c5\u51b5\u4e00\uff09\uff0c\u53ea\u95ed\u5408S2\u65f6\uff0cR1R2\u4e32\u8054\uff08\u60c5\u51b5\u4e8c\uff09\uff0c\u53ea\u95ed\u5408S1\u65f6\uff0c\u7535\u8def\u4e2d\u53ea\u5269\u4e0bR1\uff08\u60c5\u51b5\u4e09\uff09\u3002\u5bf9\u6bd4\u60c5\u51b5\u4e00\u548c\u60c5\u51b5\u4e09\uff0c\u6839\u636ep=I^2R\uff0c\u53ef\u77e5I1/I3=1/6\uff0c\u6545(R1+R2+R3)/R1=6/1(\u4e00\u5f0f)\u3002\u5355\u770b\u60c5\u51b5\u4e00\uff0c\u6839\u636ep=I^2R\uff0cp1/p3=R1/R3\uff0c\u53c8\u56e0\u4e3ap1=1w\uff0c\u6240\u4ee5p3=R3/R1\u3002\u5355\u770b\u60c5\u51b5\u4e8c\uff0c\u6839\u636ep=I^2R\uff0cI=U/R\uff0c\u53ef\u5f97(U/(R1+R2))^2R2=8(\u4e8c\u5f0f)\u3002\u5355\u770b\u60c5\u6982\u51b5\u4e09\uff0c\u6839\u636ep=U^2/R\uff0c\u53ef\u5f97U^2/R1=1(\u4e09\u5f0f)\u3002.\u7531\u4e00\u5f0f\u53ef\u5f97\uff0cR2=5R1-R3(\u56db\u5f0f)\uff0c\u5c06\u56db\u5f0f\u4ee3\u5165\u4e8c\u5f0f\u5e76\u4e0e\u4e09\u5f0f\u505a\u6bd4\uff0c\u53ef\u5f9727(R1)^2-15R1R3+2(R3)^2=0(\u4e94\u5f0f)\uff0c\u4e94\u5f0f\u4e24\u8fb9\u540c\u9664\u4ee5(R1)^2\uff0c\u53ef\u6c42\u53ef\u5f97R3/R1=3\u62164.5\u3002.\u518d\u6839\u636eR1>R2\uff0c\u4ee5\u53ca\u4e00\u5f0f\uff0c\u53ef\u820d\u53bbR3/R1=3\u3002\u6545R3/R1=4.5\uff0c\u5373P3=4.5w\u3002
选C先取A、B为一整体，由牛顿第二定律得：（mA+mB）gsinθ-FfB=（mA+mB）aFfB=μ2FNFN=（mA+mB）gcosθ联立可得a=gsinθ-μ2gcosθ再隔离A物体，设A受到的静摩擦力为FfA，方向沿斜面向上，对A再用牛二律得：mAgsinθ-FfA=mAa可得出FfA=μ2mAgcosθ．
.μ1mAgcosθ
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