作业帮有图 求真相.Suppose the roller coaster in Fig.6-38 (h1 = 37 m,h2 = 12 m,h3 = 20) passes point A with a speed of 1.20 m/s.If the average force of friction is equal to one third of its weight,with what speed will it reach point The distance travel
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有图 求真相.Suppose the roller coaster in Fig.6-38 (h1 = 37 m,h2 = 12 m,h3 = 20) passes point A with a speed of 1.20 m/s.If the average force of friction is equal to one third of its weight,with what speed will it reach point The distance travel
有图 求真相.
Suppose the roller coaster in Fig.6-38 (h1 = 37 m,h2 = 12 m,h3 = 20) passes point A with a speed of 1.20 m/s.If the average force of friction is equal to one third of its weight,with what speed will it reach point The distance traveled is 40.0 m.
中文：(h1 = 37 m,h2 = 12 m,h3 = 20) 一个物体.过了点A的时候的速度是1.2m/s.如果说平均破擦力等于3分之1的这个物体的重量.这个物体到达B的时候是什么速度?2点之间的距离为40m.
有图 求真相.Suppose the roller coaster in Fig.6-38 (h1 = 37 m,h2 = 12 m,h3 = 20) passes point A with a speed of 1.20 m/s.If the average force of friction is equal to one third of its weight,with what speed will it reach point The distance travel
翻译有点问题吧!
distance表示的是在AB之间经过的路程,而跟距离应该用straight line 或者 straight distance,This sentence “The distance traveled is 40.0 m.” should be translated definitely in “滑轮经过了40M的路程” not “AB 亮点之间的距离是40M” There are lots of differences between them.
直接用动能定理就可以解决啊.这题是高一的物理.
利用动能定理，合外力做功等于动能改变量，
先分析都有什么力做功.重力做正功mgh(h为两点高度差)，摩擦力阻力做负功f*s（摩擦力乘以走过的路程）。
然后确定末动能（要求），初动能（可算）。两者做差。
最后列等式解方程就可以了！所有动能定理的题都这思路，掌握方法比告诉你结果强！...
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利用动能定理，合外力做功等于动能改变量，
先分析都有什么力做功.重力做正功mgh(h为两点高度差)，摩擦力阻力做负功f*s（摩擦力乘以走过的路程）。
然后确定末动能（要求），初动能（可算）。两者做差。
最后列等式解方程就可以了！所有动能定理的题都这思路，掌握方法比告诉你结果强！
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根据动能定理，0.5m*(v2^2-v1^2)=mg(h1-h2)+f*l
其中f=mg/3，l=40,v1=1.2
代入就可以解出v2