Question

如圖所示，質量m=1Kg的小球穿在長L=0.8m的斜桿上，斜桿與水平方向成α=37°角，斜桿固定不動，小球與斜桿間的動摩擦因數(shù)μ=0.75．小球受水平向左的拉力F=2N，從斜桿的頂端由靜止開始下滑，求（sin37°=0.6，cos37°=0.8，g=10m/s²）                                                                                                                            

（1）小球運動的加速度大��；                                                                                

（2）小球運動到斜桿底端時的速度大小．                                                              

                                                                                              

                                                                                                                               

Answer 1

（1）沿桿子方向：Fcosα+mgsinα﹣μN=ma

垂直桿子方向：Fsinα+N=mgcosα

∴

代入數(shù)據(jù)解得a=2.5  m/s²．

（2）根據(jù)v²=2aL得，

v==2 m/s．

答：（1）小球運動的加速度大小為2.5m/s²．

（2）小球運動到斜桿底端時的速度大小為2m/s．