（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 / s
2·································································④
（2）设金属棒在位置(Ⅱ)时速度为
v2，由运动学规律得
=－2
a 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
