(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