mgsinθ=F安
又 F安=ILB,I=
| E |
| R总 |
R总=R1+
| R2RL |
| R2+RL |
| 12R•4R |
| 12R+4R |
联立解得最大速度:vm=
| 3mgR |
| B2L2 |
(2)由能量守恒知,mg•2S0sin30°=Q+
| 1 |
| 2 |
| v | 2m |
解得,Q=mgS0-
| 1 |
| 2 |
| v | 2m |
| 9m3g2R2 |
| 2B4L4 |
答:(1)金属棒下滑的最大速度为
| 3mgR |
| B2L2 |
(2)金属棒由静止开始下滑2S0的过程中,整个电路产生的焦耳热Q为mgS0-
| 9m3g2R2 |
| 2B4L4 |

| E |
| R总 |
| R2RL |
| R2+RL |
| 12R•4R |
| 12R+4R |
| 3mgR |
| B2L2 |
| 1 |
| 2 |
| v | 2m |
| 1 |
| 2 |
| v | 2m |
| 9m3g2R2 |
| 2B4L4 |
| 3mgR |
| B2L2 |
| 9m3g2R2 |
| 2B4L4 |