物理學(xué)案（新教材粵教版）第六章專題強(qiáng)化九動(dòng)力學(xué)和能量觀點(diǎn)的綜合應(yīng)用

專題強(qiáng)化九動(dòng)力學(xué)和能量觀點(diǎn)的綜合應(yīng)用目標(biāo)要求1.會(huì)用功能關(guān)系解決傳送帶、滑塊—木板模型綜合問題.2.會(huì)利用動(dòng)力學(xué)和能量觀點(diǎn)分析多運(yùn)動(dòng)組合問題．題型一傳送帶模型例1(多選)如圖所示為某建筑工地所用的水平放置的運(yùn)輸帶，在電動(dòng)機(jī)的帶動(dòng)下運(yùn)輸帶始終以恒定的速度v0＝1m/s順時(shí)針傳動(dòng)．建筑工人將質(zhì)量m＝20kg的建筑材料靜止地放到運(yùn)輸帶的最左端，同時(shí)建筑工人以v0＝1m/s的速度向右勻速運(yùn)動(dòng)．已知建筑材料與運(yùn)輸帶之間的動(dòng)摩擦因數(shù)為μ＝，運(yùn)輸帶的長(zhǎng)度為L(zhǎng)＝2m，重力加速度大小為g＝10m/s2.以下說法正確的是()A．建筑工人比建筑材料早s到右端B．建筑材料在運(yùn)輸帶上一直做勻加速直線運(yùn)動(dòng)C．因運(yùn)輸建筑材料電動(dòng)機(jī)多消耗的能量為10JD．運(yùn)輸帶對(duì)建筑材料做的功為10J聽課記錄：_______________________________________________________________________________________________________________________________________例2如圖所示，繃緊的傳送帶與水平面的夾角θ＝30°，傳送帶在電動(dòng)機(jī)的帶動(dòng)下，始終保持v0＝2m/s的速率運(yùn)行，現(xiàn)把一質(zhì)量為m＝10kg的工件(可視為質(zhì)點(diǎn))輕輕放在傳送帶的底端，經(jīng)過時(shí)間t＝s，工件被傳送到h＝m的高處，g取10m/s2，求：(1)工件與傳送帶間的動(dòng)摩擦因數(shù)；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)電動(dòng)機(jī)由于傳送工件多消耗的電能．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________1．傳送帶問題的分析方法(1)動(dòng)力學(xué)角度：首先要正確分析物體的運(yùn)動(dòng)過程，做好受力分析，然后利用運(yùn)動(dòng)學(xué)公式結(jié)合牛頓第二定律求物體及傳送帶在相應(yīng)時(shí)間內(nèi)的位移，找出物體和傳送帶之間的位移關(guān)系．(2)能量角度：求傳送帶對(duì)物體所做的功、物體和傳送帶由于相對(duì)滑動(dòng)而產(chǎn)生的熱量、因放上物體而使電動(dòng)機(jī)多消耗的電能等，常依據(jù)功能關(guān)系或能量守恒定律求解．2．功能關(guān)系分析(1)傳送帶克服摩擦力做的功：W＝fs傳．(2)系統(tǒng)產(chǎn)生的內(nèi)能：Q＝fs相對(duì)，s相對(duì)表示相對(duì)路程．(3)功能關(guān)系分析：W＝ΔEk＋ΔEp＋Q.題型二滑塊—木板模型例3如圖甲所示，長(zhǎng)木板A放在光滑的水平面上，質(zhì)量為m＝2kg的另一物體B(可看成質(zhì)點(diǎn))以水平速度v0＝2m/s滑上原來靜止的長(zhǎng)木板A的上表面．由于A、B間存在摩擦力，之后A、B速度隨時(shí)間變化情況如圖乙所示．下列說法正確的是(g取10m/s2)()A．木板A最終獲得的動(dòng)能為2JB．系統(tǒng)損失的機(jī)械能為4JC．木板A的最小長(zhǎng)度為2mD．A、B間的動(dòng)摩擦因數(shù)為聽課記錄：______________________________________________________________________________________________________________________________________例4(2023·黑龍江省佳木斯一中高三檢測(cè))如圖所示，在光滑的水平面上放置一個(gè)足夠長(zhǎng)的木板B，在B的左端放有一個(gè)可視為質(zhì)點(diǎn)的小滑塊A，A、B間的動(dòng)摩擦因數(shù)μ＝，A的質(zhì)量m＝1kg，B的質(zhì)量M＝2kg,g＝10m/s2.現(xiàn)對(duì)A施加F＝7N的水平向右的拉力，1s后撤去拉力F，求：(結(jié)果可以用分?jǐn)?shù)表示)(1)撤去拉力F前小滑塊A和長(zhǎng)木板B的加速度大小a1、a2；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)A相對(duì)于B靜止時(shí)的速度大小v；(3)整個(gè)過程中由于摩擦生成的熱量Q.________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________1．動(dòng)力學(xué)分析：分別對(duì)滑塊和木板進(jìn)行受力分析，根據(jù)牛頓第二定律求出各自的加速度；從放上滑塊到二者速度相等，所用時(shí)間相等，由t＝eq\f(Δv2,a2)＝eq\f(Δv1,a1)，可求出共同速度v和所用時(shí)間t，然后由位移公式可分別求出二者的位移．2．功和能分析：對(duì)滑塊和木板分別運(yùn)用動(dòng)能定理，或者對(duì)系統(tǒng)運(yùn)用能量守恒定律．如圖所示，要注意區(qū)分三個(gè)位移：(1)求摩擦力對(duì)滑塊做功時(shí)用滑塊對(duì)地的位移s滑；(2)求摩擦力對(duì)木板做功時(shí)用木板對(duì)地的位移s板；(3)求摩擦生熱時(shí)用相對(duì)位移Δs.題型三多運(yùn)動(dòng)組合問題例5(2023·浙江舟山市模擬)某游樂場(chǎng)的游樂裝置可簡(jiǎn)化為如圖所示的豎直面內(nèi)軌道BCDE，左側(cè)為半徑R＝m的光滑圓弧軌道BC，軌道的上端點(diǎn)B和圓心O的連線與水平方向的夾角α＝30°，圓弧軌道與粗糙水平軌道CD相切于點(diǎn)C，DE為傾角θ＝30°的光滑傾斜軌道，一輕質(zhì)彈簧上端固定在E點(diǎn)處的擋板上．現(xiàn)有質(zhì)量為m＝1kg的小滑塊P(可視為質(zhì)點(diǎn))從空中的A點(diǎn)以v0＝eq\r(2)m/s的初速度水平向左拋出，恰好從B點(diǎn)沿軌道切線方向進(jìn)入軌道，沿著圓弧軌道運(yùn)動(dòng)到C點(diǎn)之后繼續(xù)沿水平軌道CD滑動(dòng)，經(jīng)過D點(diǎn)(不計(jì)經(jīng)過D點(diǎn)時(shí)的能量損失)后沿傾斜軌道向上運(yùn)動(dòng)至F點(diǎn)(圖中未標(biāo)出)，彈簧恰好壓縮至最短．已知C、D之間和D、F之間距離都為1m，滑塊與軌道CD間的動(dòng)摩擦因數(shù)為μ＝，重力加速度g＝10m/s2，不計(jì)空氣阻力．求：(1)小滑塊P經(jīng)過圓弧軌道上B點(diǎn)的速度大?。?2)小滑塊P到達(dá)圓弧軌道上的C點(diǎn)時(shí)對(duì)軌道壓力的大??；(3)彈簧的彈性勢(shì)能的最大值；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(4)試判斷滑塊返回時(shí)能否從B點(diǎn)離開，若能，求出飛出B點(diǎn)的速度大?。蝗舨荒?，判斷滑塊最后位于何處．________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________例6如圖所示，AB、FG均為半徑R＝m的四分之一光滑圓弧軌道，半徑O1B、O2F均豎直，C點(diǎn)在B點(diǎn)的正下方，C、D兩點(diǎn)在同一高度上，DE為傾角θ＝53°、長(zhǎng)度L1＝2m的粗糙斜軌道，EF為粗糙水平直軌道．一物塊(視為質(zhì)點(diǎn))從A點(diǎn)由靜止滑下，從B點(diǎn)水平飛出后恰好落到D點(diǎn)，并且物塊落到D點(diǎn)時(shí)的速度方向與DE軌道平行，物塊經(jīng)過EF軌道后恰好能到達(dá)G點(diǎn)．物塊與DE、EF兩軌道間的動(dòng)摩擦因數(shù)均為μ＝eq\f(1,3)，取重力加速度大小g＝10m/s2，不計(jì)物塊經(jīng)過E點(diǎn)的能量損失，不計(jì)空氣阻力．求：(sin53°＝，結(jié)果可保留分?jǐn)?shù))(1)C、D兩點(diǎn)間的距離x；________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(2)物塊從B點(diǎn)運(yùn)動(dòng)到E點(diǎn)的時(shí)間t；(3)EF軌道的長(zhǎng)度L2以及物塊最后停