Question

如圖9所示，豎直平面內(nèi)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(zhì)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(dòng)摩擦因數(shù)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(dòng)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(jīng)過圖示位置上升0.5 m的過程中，拉力F₂做了多少功？

Answer 1

解析：(1)A、B小球和細(xì)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(guān)系，s_B=0.5 m                                                          ③

由能量關(guān)系，拉力F₁做功為：

W₁=fs_A+m_Bgs_B                                                                                                 ④

得：W₁=0.8 J                                                                  ⑤

(2)設(shè)細(xì)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(shè)拉力F₂做功為W₂，對系統(tǒng)由能量關(guān)系得W₂-fs_A-m_Bgs_B=m_Av_A²₂-m_Av_A²₁          ⑨

代入數(shù)據(jù)解得W₂=6.8 J                                                         ⑩

答案：(1)0.8 J  (2)6.8 J