Question

一物体质量为m=2kg.在倾角为θ=37°的斜面上的A点以初速度v.=3m/s下滑,A点距弹簧上端B的距离AB=4m.

Answer 1

1.fs=mgh.(f=umgcos37.s=2*(4+0.2)-3=5.4m.h=3*sin37=1.8)...解得u=1/(3cos37)

2.E=E动+E势-fx=m(v^2)/2+(4+0.2)*sin37*mg-umgcos37*(4+0.2)=31.4J