Question

如圖所示，兩根不計電阻的金屬導線MN與PQ 放在水平面內(nèi)，MN是直導線，PQ的PQ₁段是直導線，Q₁Q₂段是弧形導線，Q₂Q₃段是直導線，MN、PQ₁、Q₂Q₃相互平行，M、P間接入一個阻值R=0.25Ω的電阻。一根質(zhì)量為1.0 kg不計電阻的金屬棒AB能在MN、PQ上無摩擦地滑動，金屬棒始終垂直于MN，整個裝置處于磁感應強度B=0.5T的勻強磁場中，磁場方向豎直向下。金屬棒處于位置（I）時，給金屬棒一個向右的速度v₁=4 m/s，同時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動；當金屬棒運動到位置(Ⅱ)時，外力方向不變，大小變?yōu)镕₂，金屬棒向右做勻速直線運動，經(jīng)過時間t =2 s到達位置(Ⅲ)。金屬棒在位置(I)時，與MN、Q₁Q₂相接觸于a、b兩點，a、b的間距L₁=1 m，金屬棒在位置(Ⅱ)時，棒與MN、Q₁Q₂相接觸于c、d兩點。已知s₁=7.5 m。求：（1）金屬棒向右勻減速運動時的加速度大小?

（2）c、d兩點間的距離L₂=?

（3）外力F₂的大小？

（4）金屬棒從位置(I)運動到位置(Ⅲ)的過程中，電阻R上放出的熱量Q=？

 

Answer 1

（1）金屬棒從位置(I)到位置(Ⅱ)的過程中，加速度不變，方向向左，設大小為a，在位置I時，a、b間的感應電動勢為E₁，感應電流為I₁，受到的安培力為F_安1，則

E₁=BL₁ v₁，，F_安1···································①

F_安1=4N·································································②

根據(jù)牛頓第二定律得

F_安1－F₁ =ma·····························································③

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

（2）設金屬棒在位置(Ⅱ)時速度為v₂，由運動學規(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動，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F_安1=F_安2

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

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

（3）金屬棒從位置(Ⅱ)到位置(Ⅲ)的過程中，做勻速直線運動，感應電動勢大小與位置(Ⅱ)時的感應電動勢大小相等，安培力與位置(Ⅱ)時的安培力大小相等，所以

F₂=F_安2=4N······························································⑨

(4) 設位置（II）和(Ⅲ)之間的距離為s₂，則

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

設從位置(I)到位置(Ⅱ)的過程中，外力做功為W₁，從位置(Ⅱ)到位置(Ⅲ)的過程中，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