题目内容


一物体放在斜面的顶端,受到平行于斜面向下的力F的作用了,已知力F随时间变化的图象及物体运动的v﹣t图象所示,6s末刚好到达斜面的底端,求(g取10m/s2).                                                               

                                                  

(1)物体的质量m;                                                                                                      

(2)斜面的长度L.                                                                                                      

                                                                                                                                       


(1)0﹣2s物体做匀加速直线运动,

则有:mgsina+F1﹣f=ma

加速度a===1m/s2

由甲图知此时的拉力F1=3N

2﹣6s做匀速直线运动,沿斜面方向满足:

mgsina+F﹣f=0

联立得:m=1kg

(2)加速阶段:

匀速阶段:S2=vt2

其中t1=2s           t2=4s

斜面长度L=S1+S2

联立解得:L=10m

答:(1)物体的质量1kg;

(2)斜面的长度10m.


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