Question

如圖所示，放在傾角為30°的固定斜面上的正方形導(dǎo)線框abcd與重物之間用足夠長(zhǎng)的細(xì)線跨過(guò)光滑的輕質(zhì)定滑輪連接，線框的邊長(zhǎng)為L、質(zhì)量為m、電阻為R，重物的質(zhì)量為m�，F(xiàn)將線框沿bc方向拋出，穿過(guò)寬度為D、磁感應(yīng)強(qiáng)度為B的勻強(qiáng)磁場(chǎng)，磁場(chǎng)的方向垂直斜面向上．線框向下離開(kāi)磁場(chǎng)時(shí)的速度剛好是進(jìn)人磁場(chǎng)時(shí)速度的（n大于1），線框離開(kāi)磁場(chǎng)后繼續(xù)下滑一段距離，然后上滑并勻速進(jìn)人磁場(chǎng)．線框與斜面間的動(dòng)摩擦因數(shù)為，不計(jì)空氣阻力，整個(gè)運(yùn)動(dòng)過(guò)程中線框始終與斜面平行且不發(fā)生轉(zhuǎn)動(dòng)，斜細(xì)線與bc邊平行，cd邊磁場(chǎng)邊界平行．求：
⑴線框在上滑階段勻速進(jìn)人磁場(chǎng)時(shí)的速度υ₂大��；
⑵線框在下滑階段剛離開(kāi)磁場(chǎng)時(shí)的速度υ₁大��；
⑶線框在下滑階段通過(guò)磁場(chǎng)過(guò)程中產(chǎn)生的焦耳熱Q．

Answer 1

【標(biāo)準(zhǔn)解答】（1）線框勻速上滑進(jìn)入磁場(chǎng)，則有

mgsin30° + μmgcos30° + BIL = Mg······································································ ①（2分）

而I = ··········································································································· ②（1分）

E = BLυ₂ ··········································································································· ③（1分）

解得υ₂ =  ···························································································· ④（2分）

（2）對(duì)系統(tǒng)，由動(dòng)能定理知，                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
離開(kāi)磁場(chǎng)后的下滑階段：
 mgsin30° · h – mgh– μmgcos30° · h = 0 – (m + m)υ₁²  ······························· ⑤（2分）

進(jìn)人磁場(chǎng)前的上滑階段：
 – mgsin30° · h + mgh– μmgcos30° · h = (m + m)υ₂² – 0 ······························ ⑥（2分）

由以上三式得：υ₁ =  ·········································································· ⑦（2分）

（3）設(shè)剛進(jìn)入磁場(chǎng)時(shí)速度為υ₀。線框在下滑穿越磁場(chǎng)的過(guò)程，由系統(tǒng)能量守恒有：
(m + m)υ₀² – (m + m)υ₁² = mg(L + D) – mgsin30° ·(L + D) + μmgcos30° ·(L + D) + Q
 ························································································································ ⑧（4分）                                                                                                                                                     
由題設(shè)知υ₀ = nυ₁

解得：Q =  – mg(L + D) ·················································· ⑨（2分）

【思維點(diǎn)拔】本題的關(guān)鍵在于理解多過(guò)程和多對(duì)象設(shè)置情景，先找出解題的突破口即先從受力平衡的線框勻速上滑進(jìn)入磁場(chǎng)的過(guò)程來(lái)分析，列出平衡方程，由于線框勻速上滑進(jìn)入磁場(chǎng)，所以對(duì)應(yīng)的重物也是勻速運(yùn)動(dòng)，細(xì)線的拉力與重物的重力大小相等。而其他過(guò)程，系統(tǒng)受力不平衡，所以細(xì)線的拉力不再與重物的重力大小相等，由于線框切割磁感線的速度在不斷變化，導(dǎo)致安培力和細(xì)線中的拉力大小在不斷變化，無(wú)法用動(dòng)力學(xué)觀點(diǎn)來(lái)求解，但可以用能量觀點(diǎn)來(lái)解決變力做功問(wèn)題，這時(shí)宜以線框和重物整體為研究對(duì)象。