Question

一個(gè)小物體從斜面底端沖上足夠長(zhǎng)的斜面后,返回到斜面的底端,已知小物體的初動(dòng)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(dòng)能變?yōu)?E,則有(    )

A.返回斜面底端時(shí)的動(dòng)能為E                  B.返回斜面底端時(shí)的動(dòng)能為E

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

Answer 1

AD

解析：上滑過(guò)程由動(dòng)能定理

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

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

=mv²                                                                    ③

小物體以2E初動(dòng)能上滑,由動(dòng)能定理

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

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

E′=mv′²                                                                  ⑥

由①②③④⑤⑥得E′=E,v′=v.