如圖所示,兩根不計(jì)電阻的金屬導(dǎo)線MN與PQ 放在水平面內(nèi),MN是直導(dǎo)線,PQ的PQ1段是直導(dǎo)線,Q1Q2段是弧形導(dǎo)線,Q2Q3段是直導(dǎo)線,MN、PQ1、Q2Q3相互平行,M、P間接入一個(gè)阻值R=0.25Ω的電阻。一根質(zhì)量為1.0 kg不計(jì)電阻的金屬棒AB能在MN、PQ上無(wú)摩擦地滑動(dòng),金屬棒始終垂直于MN,整個(gè)裝置處于磁感應(yīng)強(qiáng)度B=0.5T的勻強(qiáng)磁場(chǎng)中,磁場(chǎng)方向豎直向下。金屬棒處于位置(I)時(shí),給金屬棒一個(gè)向右的速度v1=4 m/s,同時(shí)方向水平向右的外力F1 =3 N作用在金屬棒上使金屬棒向右做勻減速直線運(yùn)動(dòng);當(dāng)金屬棒運(yùn)動(dòng)到位置(Ⅱ)時(shí),外力方向不變,大小變?yōu)镕2,金屬棒向右做勻速直線運(yùn)動(dòng),經(jīng)過(guò)時(shí)間t =2 s到達(dá)位置(Ⅲ)。金屬棒在位置(I)時(shí),與MN、Q1Q2相接觸于a、b兩點(diǎn),a、b的間距L1=1 m,金屬棒在位置(Ⅱ)時(shí),棒與MN、Q1Q2相接觸于c、d兩點(diǎn)。已知s1=7.5 m。求:(1)金屬棒向右勻減速運(yùn)動(dòng)時(shí)的加速度大小?

(2)c、d兩點(diǎn)間的距離L2=?

(3)外力F2的大?

(4)金屬棒從位置(I)運(yùn)動(dòng)到位置(Ⅲ)的過(guò)程中,電阻R上放出的熱量Q=?

 

(1)金屬棒從位置(I)到位置(Ⅱ)的過(guò)程中,加速度不變,方向向左,設(shè)大小為a,在位置I時(shí),a、b間的感應(yīng)電動(dòng)勢(shì)為E1,感應(yīng)電流為I1,受到的安培力為F1,則

E1=BL1 v1,,F1···································①

F1=4N·································································②

根據(jù)牛頓第二定律得

F1F1 =ma·····························································③

a= 1 m / s2·································································④

(2)設(shè)金屬棒在位置(Ⅱ)時(shí)速度為v2,由運(yùn)動(dòng)學(xué)規(guī)律得

=-2a s1···························································⑤

v2= 1 m / s·································································⑥

由于在(I)和(II)之間做勻減速直線運(yùn)動(dòng),即加速度大小保持不變,外力F1恒定,所以AB棒受到的安培力不變即F1=F2

···························································⑦

m······················································⑧

(3)金屬棒從位置(Ⅱ)到位置(Ⅲ)的過(guò)程中,做勻速直線運(yùn)動(dòng),感應(yīng)電動(dòng)勢(shì)大小與位置(Ⅱ)時(shí)的感應(yīng)電動(dòng)勢(shì)大小相等,安培力與位置(Ⅱ)時(shí)的安培力大小相等,所以

F2=F2=4N······························································⑨

(4) 設(shè)位置(II)和(Ⅲ)之間的距離為s2,則

s2=v2t=2 m ································································⑩

設(shè)從位置(I)到位置(Ⅱ)的過(guò)程中,外力做功為W1,從位置(Ⅱ)到位置(Ⅲ)的過(guò)程中,外力做功為W2,則

W1= F1s1=22.5 J···························································11

W2= F2s2=8 J······························································12

根據(jù)能量守恒得W1+W2·······························13·

解得Q = 38J ·····························································14

解析:略

 

練習(xí)冊(cè)系列答案
相關(guān)習(xí)題

科目:高中物理 來(lái)源: 題型:

如圖所示,兩根豎直固定的金屬導(dǎo)軌ad和bc相距l(xiāng)=0.2m,另外兩根水平金屬桿MN和EF可沿導(dǎo)軌無(wú)摩擦地滑動(dòng),MN桿和EF桿的電阻分別為0.2Ω(豎直金屬導(dǎo)軌的電阻不計(jì)),EF桿放置在水平絕緣平臺(tái)上,回路NMEF置于勻強(qiáng)磁場(chǎng)內(nèi),磁場(chǎng)方向垂直于導(dǎo)軌平面向里,磁感應(yīng)強(qiáng)度B=1T,試求:
(1)EF桿不動(dòng),MN桿以0.1m/s的速度向上運(yùn)動(dòng)時(shí),桿MN兩端哪端的電勢(shì)高?MN兩端電勢(shì)差為多大?
(2)當(dāng)MN桿和EF桿的質(zhì)量均為m=10-2kg.MN桿須有多大的速度向上運(yùn)動(dòng)時(shí),EF桿將開(kāi)始向上運(yùn)動(dòng)?此時(shí)拉力的功率為多大?

查看答案和解析>>

科目:高中物理 來(lái)源: 題型:

精英家教網(wǎng)如圖所示,兩根電阻不計(jì)的光滑金屬導(dǎo)軌ab、cd豎直放置,導(dǎo)軌間距為L(zhǎng),上端接有兩個(gè)定值電阻R1、R2,已知R1=R2=2r.將質(zhì)量為m、電阻值為r的金屬棒從圖示位置由靜止釋放,下落過(guò)程中金屬棒保持水平且與導(dǎo)軌接觸良好.自由下落一段距離后金屬棒進(jìn)入一個(gè)垂直于導(dǎo)軌平面的勻強(qiáng)磁場(chǎng),磁場(chǎng)寬度為h.金屬棒出磁場(chǎng)前R1、R2的功率均已穩(wěn)定為P.則金屬棒離開(kāi)磁場(chǎng)時(shí)的速度大小為
 
,整個(gè)過(guò)程中通過(guò)電阻R1的電量為
 
.(已知重力加速度為g)

查看答案和解析>>

科目:高中物理 來(lái)源: 題型:

(2013?淮安模擬)如圖所示,兩根等高光滑的
14
圓弧軌道,半徑為r、間距為L(zhǎng),軌道電阻不計(jì).在軌道頂端連有一阻值為R的電阻,整個(gè)裝置處在一豎直向上的勻強(qiáng)磁場(chǎng)中,磁感應(yīng)強(qiáng)度為B.現(xiàn)有一根長(zhǎng)度稍大于L、質(zhì)量為m、電阻不計(jì)的金屬棒從軌道的頂端ab處由靜止開(kāi)始下滑,到達(dá)軌道底端cd時(shí)受到軌道的支持力為2mg.整個(gè)過(guò)程中金屬棒與導(dǎo)軌電接觸良好,求:
(1)棒到達(dá)最低點(diǎn)時(shí)的速度大小和通過(guò)電阻R的電流.
(2)棒從ab下滑到cd過(guò)程中回路中產(chǎn)生的焦耳熱和通過(guò)R的電荷量.
(3)若棒在拉力作用下,從cd開(kāi)始以速度v0向右沿軌道做勻速圓周運(yùn)動(dòng),則在到達(dá)ab的過(guò)程中拉力做的功為多少?

查看答案和解析>>

科目:高中物理 來(lái)源: 題型:

精英家教網(wǎng)如圖所示,兩根光滑的平行金屬導(dǎo)軌處于同一水平面內(nèi),相距0.3m,導(dǎo)軌左端PQ間用電阻R=0.2Ω相連接,導(dǎo)軌電阻不計(jì),導(dǎo)軌上停放著一金屬桿MN,桿的電阻為0.1Ω,質(zhì)量為0.1kg,始終與導(dǎo)軌保持良好接觸,整個(gè)裝置處于豎直向下的勻強(qiáng)磁場(chǎng)中,磁感強(qiáng)度為0.5T.現(xiàn)對(duì)金屬桿施加適當(dāng)?shù)乃搅,使桿由靜止開(kāi)始沿導(dǎo)軌勻加速運(yùn)動(dòng),問(wèn):
(1)要使P點(diǎn)電勢(shì)高于Q點(diǎn)電勢(shì),桿應(yīng)向哪個(gè)方向運(yùn)動(dòng)?
(2)當(dāng)金屬桿的速度v=2m/s時(shí),電阻R上消耗的電功率多大?
(3)要使電阻R上的電壓每秒均勻增加0.05V,桿的加速度a應(yīng)為多大?

查看答案和解析>>

科目:高中物理 來(lái)源:2012-2013學(xué)年江蘇省南通市高三第三次調(diào)研測(cè)試物理試卷(解析版) 題型:計(jì)算題

如圖所示,兩根等高光滑的圓弧軌道,半徑為r、間距為L(zhǎng),軌道電阻不計(jì).在軌道頂端連有一阻值為R的電阻,整個(gè)裝置處在一豎直向上的勻強(qiáng)磁場(chǎng)中,磁感應(yīng)強(qiáng)度為B.現(xiàn)有一根長(zhǎng)度稍大于L、質(zhì)量為m、電阻不計(jì)的金屬棒從軌道的頂端ab處由靜止開(kāi)始下滑,到達(dá)軌道底端cd時(shí)受到軌道的支持力為2mg.整個(gè)過(guò)程中金屬棒與導(dǎo)軌電接觸良好,求:

(1)棒到達(dá)最低點(diǎn)時(shí)的速度大小和通過(guò)電阻R的電流.

(2)棒從ab下滑到cd過(guò)程中回路中產(chǎn)生的焦耳熱和通過(guò)R的電荷量.

(3)若棒在拉力作用下,從cd開(kāi)始以速度v0向右沿軌道做勻速圓周運(yùn)動(dòng),則在到達(dá)ab的過(guò)程中拉力做的功為多少?

 

查看答案和解析>>

同步練習(xí)冊(cè)答案