一物體放在斜面的頂端,受到平行于斜面向下的力F的作用了,已知力F隨時(shí)間變化的圖象及物體運(yùn)動(dòng)的v﹣t圖象所示,6s末剛好到達(dá)斜面的底端,求(g取10m/s2).                                                               

                                                  

(1)物體的質(zhì)量m;                                                                                                      

(2)斜面的長(zhǎng)度L.                                                                                                      

                                                                                                                                       


(1)0﹣2s物體做勻加速直線運(yùn)動(dòng),

則有:mgsina+F1﹣f=ma

加速度a===1m/s2

由甲圖知此時(shí)的拉力F1=3N

2﹣6s做勻速直線運(yùn)動(dòng),沿斜面方向滿(mǎn)足:

mgsina+F﹣f=0

聯(lián)立得:m=1kg

(2)加速階段:

勻速階段:S2=vt2

其中t1=2s           t2=4s

斜面長(zhǎng)度L=S1+S2

聯(lián)立解得:L=10m

答:(1)物體的質(zhì)量1kg;

(2)斜面的長(zhǎng)度10m.


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