Question

如圖（a）所示，在足夠長的光滑水平面上，放置一長為L＝1m、質量為m₁＝0.5kg的木板A，一質量為m₂＝1kg的小物體B以初速度v₀滑上A的上表面的同時對A施加一個水平向右的力F，A與B之間的動摩擦因數為μ＝0.2，g＝10m/s²；小物體B在A上運動的路程S與F力的關系如圖（b）所示。求：v₀、F₁、F₂。

Answer 1

（1）4m/s；（2）3N；（3）9N

（1）由圖象可看出當F≤1N時，B物體在A板上的路程始終等于板長L，當F＝1N時，剛好不從A板右端掉下，此后A和B一起相對靜止并加速運動。
設B物體的加速度為a₂，A板的加速度為a₁，分別由牛頓第二定律：
μm₂g＝m₂a₂ ············································································································ ①
F＋μm₂g＝m₁a₁ ······································································································· ②
設B運動的位移為S₂，A運動的位移為S₁，經過t時間兩者速度均為v，根據運動學公式：
S_B＝t ············································································································· ③
S_A＝t ·················································································································· ④
v＝v₀－a₂t＝a₁t········································································································· ⑤
B在A上相對A向右運動的路程S＝S_B－S_A ······································································ ⑥
聯立①②③④⑤⑥解得：S＝··············································································· ⑦
將F＝1N，S＝1m代入，解得：v₀＝4m/s
（2）根據⑦式分析可知，當1N≤F≤F₁時，隨著F力增大，S減小，當F＝F₁時，出現S突變，說明此時A、B在達到共同速度后，恰好再次發生相對運動，B將會從A板左端掉下。
對A、B恰好發生相對運動時，B的加速度為a₂，則整體加速度也為a₂，由牛頓第二定律：
F₁＝(m₁＋m₂)a₂········································································································ ⑧
聯立①⑧解得解得F₁＝3N
（3）此時B在A上運動的路程為S₁＝＝m
當F≥F₁時，物體B在A板上的路程為B相對A向右運動的路程的兩倍。
故當F＝F₂時，將S＝0.5S₁代入⑦式解得：F₂＝9N
【評析】本題考查牛頓運動定律。滑塊問題是物理模型中非常重要的模型，是學生物理建模能力培養的典型模型。滑塊問題的解決非常靈活，針對受力分析、運動分析以及牛頓第二定律的掌握，還有相對運動的分析，特別是摩擦力的變化與轉型，都是難點所在。本題通過非常規的圖象來分析滑塊的運動，能從圖中讀懂物體的運動。