Question

如图所示，将边长为a、质量为m、电阻为R的正方形导线框竖直向上抛出，穿过宽度为b、磁感应强度为B的匀强磁场，磁场的方向垂直纸面向里．线框向上离开磁场时的速度刚好是进人磁场时速度的一半，线框离开磁场后继续上升一段高度，然后落下并匀速进人磁场．整个运动过程中始终存在着大小恒定的空气阻力f且线框不发生转动．求：

（1）线框在下落阶段匀速进人磁场时的速度V₂； 
(2）线框在上升阶段刚离开磁场时的速度V₁； 
（3）线框在上升阶段通过磁场过程中产生的焦耳热Q．

Answer 1

（1）线框在下落阶段匀速进入磁场瞬间
mg = f +                                                                        ①
解得     v₂ =                                                                          ②
（2）线框从离开磁场至上升到最高点的过程
(mg + f ) h = mv₁ ²                                                              ③
线框从最高点回落至磁场瞬间
(mg - f ) h = mv₂ ²                                                              ④
③、④式联立解得
v₁ =                                                                    ⑤
=                                                            ⑥
（3）线框在向上通过通过过程中
mv₀²-mv₁² = Q +（mg + f）(a + b)                                     ⑦
v₀ = 2 v₁                                                                                   
Q = m [ (mg)² – f ² ] -（mg + f）(a + b)                         ⑧

解析