Question

一個小物體從斜面底端沖上足夠長的斜面后,返回到斜面的底端,已知小物體的初動能為E,它返回斜面底端的速度大小為v,克服摩擦力做功為.若小物體沖上斜面的動能變?yōu)?E,則有(    )

A.返回斜面底端時的動能為E                  B.返回斜面底端時的動能為E

C.返回斜面底端時的速度大小為2v          D.返回斜面底端時的速度大小為v

Answer 1

AD

解析：上滑過程由動能定理

-(f+mgsinθ)s=0-E                                                              ①

返回到出發(fā)點-2fs=-E                                                         ②

=mv²                                                                    ③

小物體以2E初動能上滑,由動能定理

上滑過程：-(f+mgsinθ)s′=0-2E                                                 ④

返回到出發(fā)點：-2fs′=E′-2E                                                   ⑤

E′=mv′²                                                                  ⑥

由①②③④⑤⑥得E′=E,v′=v.