如圖所示,光滑圓弧的半徑為0.2m,有一質(zhì)量為1.0kg的物體自A點(diǎn)由靜止開(kāi)始下滑到達(dá)B點(diǎn),然后物體沿粗糙水平面繼續(xù)向前最遠(yuǎn)能夠到達(dá)C處,已知B到C的距離為1m.求:(g=10m/s2)                                                   

(1)物體到達(dá)B點(diǎn)時(shí)的速率;                                                                                        

(2)物體與水平面間的動(dòng)摩擦因數(shù).                                                                              

                                                                                          

                                                                                                                                       


(1)設(shè)物體到B點(diǎn)的速度為v,由A到B的過(guò)程,只有重力做功,機(jī)械能守恒,則得:

 mgR=mv2

解之得:v==m/s=2m/s

(2)設(shè)物體在水平面上運(yùn)動(dòng)摩擦力做功W,由A到C為研究過(guò)程,由動(dòng)能定理得:

 mgR﹣μmgs=0

解得:μ===0.2

答:(1)物體到達(dá)B點(diǎn)時(shí)的速率是2m/s;

(2)物體與水平面間的動(dòng)摩擦因數(shù)是0.2.


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