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(zhì)量為m₁＝0.5kg的木板A，一質(zhì)量為m₂＝1kg的小物體B以初速度v₀滑上A的上表面的同時(shí)對A施加一個(gè)水平向右的力F，A與B之間的動摩擦因數(shù)為μ＝0.2，g＝10m/s²；小物體B在A上運(yùn)動的路程S與F力的關(guān)系如圖（b）所示。求：v₀、F₁、F₂ 。

Answer 1

（1）4m/s；（2）3N；（3）9N

解析:

（1）由圖象可看出當(dāng)F≤1N時(shí)，B物體在A板上的路程始終等于板長L，當(dāng)F＝1N時(shí)，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(yùn)動。

設(shè)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(shè)B運(yùn)動的位移為S₂，A運(yùn)動的位移為S₁，經(jīng)過t時(shí)間兩者速度均為v，根據(jù)運(yùn)動學(xué)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(yùn)動的路程S＝S_B－S_A ······································································ ⑥

聯(lián)立①②③④⑤⑥解得：S＝··············································································· ⑦

將F＝1N，S＝1m代入，解得：v₀＝4m/s

（2）根據(jù)⑦式分析可知，當(dāng)1N≤F≤F₁時(shí)，隨著F力增大，S減小，當(dāng)F＝F₁時(shí)，出現(xiàn)S突變，說明此時(shí)A、B在達(dá)到共同速度后，恰好再次發(fā)生相對運(yùn)動，B將會從A板左端掉下。

對A、B恰好發(fā)生相對運(yùn)動時(shí)，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(lián)立①⑧解得解得F₁＝3N

（3）此時(shí)B在A上運(yùn)動的路程為S₁＝＝m

當(dāng)F≥F₁時(shí)，物體B在A板上的路程為B相對A向右運(yùn)動的路程的兩倍。

故當(dāng)F＝F₂時(shí)，將S＝0.5S₁代入⑦式解得：F₂＝9N

【評析】本題考查牛頓運(yùn)動定律。滑塊問題是物理模型中非常重要的模型，是學(xué)生物理建模能力培養(yǎng)的典型模型�；瑝K問題的解決非常靈活，針對受力分析、運(yùn)動分析以及牛頓第二定律的掌握，還有相對運(yùn)動的分析，特別是摩擦力的變化與轉(zhuǎn)型，都是難點(diǎn)所在。本題通過非常規(guī)的圖象來分析滑塊的運(yùn)動，能從圖中讀懂物體的運(yùn)動。