Question

如图所示，两根不计电阻的金属导线MN与PQ放在水平面内，MN是直导线，PQ的PQ₁段是直导线，Q₁Q₂段是弧形导线，Q₂Q₃段是直导线，MN、PQ₁、Q₂Q₃相互平行，M、P间接入一个阻值R=0.25Ω的电阻。一根质量为1.0 kg不计电阻的金属棒AB能在MN、PQ上无摩擦地滑动，金属棒始终垂直于MN，整个装置处于磁感应强度B=0.5T的匀强磁场中，磁场方向竖直向下。金属棒处于位置（I）时，给金属棒一个向右的速度v₁=4 m/s，同时方向水平向右的外力F₁ ="3" N作用在金属棒上使金属棒向右做匀减速直线运动；当金属棒运动到位置(Ⅱ)时，外力方向不变，大小变为F₂，金属棒向右做匀速直线运动，经过时间t ="2" s到达位置(Ⅲ)。金属棒在位置(I)时，与MN、Q₁Q₂相接触于a、b两点，a、b的间距L₁=1 m，金属棒在位置(Ⅱ)时，棒与MN、Q₁Q₂相接触于c、d两点。已知s₁="7.5" m。求：（1）金属棒向右匀减速运动时的加速度大小?
（2）c、d两点间的距离L₂=?
（3）外力F₂的大小？
（4）金属棒从位置(I)运动到位置(Ⅲ)的过程中，电阻R上放出的热量Q=？

Answer 1

（1）金属棒从位置(I)到位置(Ⅱ)的过程中，加速度不变，方向向左，设大小为a，在位置I时，a、b间的感应电动势为E₁，感应电流为I₁，受到的安培力为F_安1，则
E₁=BL₁ v₁，，F_安1···································①
F_安1="4" N·································································②
根据牛顿第二定律得
F_安1－F₁ =ma·····························································③
a=" 1" m / s²·································································④
（2）设金属棒在位置(Ⅱ)时速度为v₂，由运动学规律得
=－2a s₁···························································⑤
v₂=" 1" m / s·································································⑥
由于在(I)和(II)之间做匀减速直线运动，即加速度大小保持不变，外力F₁恒定，所以AB棒受到的安培力不变即F_安1=F_安2
···························································⑦
m······················································⑧
（3）金属棒从位置(Ⅱ)到位置(Ⅲ)的过程中，做匀速直线运动，感应电动势大小与位置(Ⅱ)时的感应电动势大小相等，安培力与位置(Ⅱ)时的安培力大小相等，所以
F₂= F_安2="4" N······························································⑨
(4) 设位置（II）和(Ⅲ)之间的距离为s₂，则
s₂= v₂t="2" m ································································⑩
设从位置(I)到位置(Ⅱ)的过程中，外力做功为W₁，从位置(Ⅱ)到位置(Ⅲ)的过程中，外力做功为W₂，则
W₁= F₁ s₁="22.5" J ···························································11
W₂= F₂ s₂="8" J······························································12
根据能量守恒得W₁+ W₂·······························13·
解得Q =" 38" J ·····························································14

解析