Question

（12分）一水平傳送帶以v₁＝2m/s的速度勻速運(yùn)動(dòng)，將一粉筆頭無(wú)初速度放在傳送帶上，達(dá)到相對(duì)靜止時(shí)產(chǎn)生的劃痕長(zhǎng)L₁＝4m�，F(xiàn)在讓傳送帶以a₂＝1.5m/s²的加速度減速，將該粉筆頭無(wú)初速度放在傳送帶上，求粉筆頭在傳送帶上的劃痕長(zhǎng)度L₂。求粉筆頭相對(duì)傳送帶運(yùn)動(dòng)的位移L₂。（取g＝10m/s²）

Answer 1

(12分)解:設(shè)二者之間的動(dòng)摩擦因數(shù)為,第一次粉筆頭打滑時(shí)間

          為t,則依據(jù)傳送帶比粉筆頭位移大L₁得:

                                                                                  

     粉筆頭的加速度解得            

     第二次粉筆頭先加速到與傳送帶速度相同,由于>,

     故二者不能共同減速,粉筆頭以的加速度減速到靜止.

                                                                                                                  

     設(shè)二者達(dá)到的相同速度為_共,由運(yùn)動(dòng)等時(shí)性得:

      解得=0.5                                                           

     此過(guò)程傳送帶比粉筆頭多走

                                                                                                                 

     粉筆頭減速到零的過(guò)程粉筆頭比傳送帶多走

                                                                               

     劃痕長(zhǎng)度為L(zhǎng)₂=s₁-s₂=0.83m.