Question

一个小物体从斜面底端冲上足够长的斜面后,返回到斜面的底端,已知小物体的初动能为E,它返回斜面底端的速度大小为v,克服摩擦力做功为.若小物体冲上斜面的动能变为2E,则有(    )

A.返回斜面底端时的动能为E                  B.返回斜面底端时的动能为E

C.返回斜面底端时的速度大小为2v          D.返回斜面底端时的速度大小为v

Answer 1

AD

解析：上滑过程由动能定理

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

返回到出发点-2fs=-E                                                         ②

=mv²                                                                    ③

小物体以2E初动能上滑,由动能定理

上滑过程：-(f+mgsinθ)s′=0-2E                                                 ④

返回到出发点：-2fs′=E′-2E                                                   ⑤

E′=mv′²                                                                  ⑥

由①②③④⑤⑥得E′=E,v′=v.