题目内容


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.                                                              

                                                                                              

                                                                                                                               


(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.


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在探究物体的加速度a与物体所受外力F、物体质量M间的关系时,采用如图所示的实验装置,小车及车中的砝码质量用M表示,盘及盘中的砝码质量用m表示.                                                                     

                                                     

(1)当M与m的大小关系满足                                                           时,才可以认为绳子对小车的拉力大小等于盘和砝码的重力.                                                                                                        

(2)某一组同学先保持盘及盘中的砝码质量m一定来做实验,其具体操作步骤如下,以下做法正确的是        .          

A.平衡摩擦力时,应将盘及盘中的砝码用细绳通过定滑轮系在小车上                         

B.每次改变小车的质量时,不需要重新平衡摩擦力                                                  

C.实验时,先放开小车,再接通打点计时器的电源                                                  

D.用天平测出m以及小车质量M,小车运动的加速度可直接用公式求出                  

(3)另两组同学保持小车及车中的砝码质量M一定,探                  究加速度a与所受外力F的关系,由于他们操作不当,这两组同学得到的a﹣F关系图象分别如图1和图2所示,其原因分别是:                

图1:                                                                                                                     

图2:                                                                                                                     

                                                                                                                                 

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