Question

如圖9所示，豎直平面內放一直角桿，桿的水平部分粗糙，豎直部分光滑，兩部分各套有質量分別為m_A="2.0" kg和m_B="1.0" kg的小球A和B，A球與水平桿間動摩擦因數μ=0.20，A、B間用不可伸長的輕繩相連，圖示位置處OA="1.5" m，OB="2.0" m.g取10 m/s².

圖9
（1）若用水平力F₁沿桿向右拉A，使A由圖示位置向右極緩慢地移動0.5 m，則該過程中拉力F₁做了多少功？
（2）若用水平力F₂沿桿向右拉A，使B以1 m/s的速度勻速上升，則在B經過圖示位置上升0.5 m的過程中，拉力F₂做了多少功？

Answer 1

（1）0.8 J  (2)6.8 J

（1）A、B小球和細繩整體豎直方向處于平衡，A受到水平桿的彈力為
N=（m_A+m_B）g                                                                ①
則A受到的摩擦力為f=μ（m_A+m_B）g                                            ②
由幾何關系，s_B="0.5" m                                                         ③
由能量關系，拉力F₁做功為：
W₁=fs_A+m_Bgs_B                                                                                                 ④
得：W₁="0.8" J                                                                 ⑤
(2)設細繩與豎直方向的夾角為θ，因繩不可伸長，所以v_Bcosθ=v_Asinθ              ⑥
則=v_Bcotθ₁= m/s，                                                      ⑦
=v_Bcotθ₂= m/s                                                          ⑧
設拉力F₂做功為W₂，對系統由能量關系得W₂-fs_A-m_Bgs_B=m_Av_A²₂-m_Av_A²₁         ⑨
代入數據解得W₂="6.8" J                                                         ⑩