Question

如圖5所示，水平面上，質(zhì)量為m的物塊受到與水平方向夾角為θ的推力F的作用(θ≠0)，物塊做勻速直線運(yùn)動(dòng).現(xiàn)將F撤去，設(shè)此后物塊運(yùn)動(dòng)過(guò)程中加速度的大小為a，則

A.a=0                                                                    B.a=

C.a＞                                                       D.a＜

Answer 1

解析：對(duì)物體受力分析如圖所示：

物體做勻速運(yùn)動(dòng)，可知，滑動(dòng)摩擦力F₁為：

F₁=Fcosθ=μ(mg+Fsinθ)                                                                                      ①

撤掉力F，物體在滑動(dòng)摩擦力F₂作用下減速運(yùn)動(dòng)，F₂=μmg=ma                            ②

F₂＜F₁                                                                                                                   ③

由①②③知：a＜.

答案：D