Question

如图所示，质量m=1Kg的小球穿在长L=0.8m的斜杆上，斜杆与水平方向成α=37°角，斜杆固定不动，小球与斜杆间的动摩擦因数μ=0.75．小球受水平向左的拉力F=2N，从斜杆的顶端由静止开始下滑，求（sin37°=0.6，cos37°=0.8，g=10m/s²）                                                                                                                            

（1）小球运动的加速度大小；                                                                                

（2）小球运动到斜杆底端时的速度大小．                                                              

                                                                                              

                                                                                                                               

Answer 1

（1）沿杆子方向：Fcosα+mgsinα﹣μN=ma

垂直杆子方向：Fsinα+N=mgcosα

∴

代入数据解得a=2.5  m/s²．

（2）根据v²=2aL得，

v==2 m/s．

答：（1）小球运动的加速度大小为2.5m/s²．

（2）小球运动到斜杆底端时的速度大小为2m/s．