(1)对雪橇受力分析如图所示,据牛顿运动定律有:
Fx-
f=
ma ;
N+
Fy=
mg 又:
f=
μN;
Fx=
Fcos37°;
Fy=
Fsin37°
故:
![](http://www.00-edu.com/d/file/2022-10-10/618865a201c522c4256fcfb85ddab98d.jpg)
(2)
v2=
at2=2.6×2 = 5.2(m/s)
撤去
F后,据牛顿第二定律有:-
μmg=
ma′故:
a′= -
μg =-0.20×10 =-2.0 m/s
2 (1分)
由于:
![](http://www.00-edu.com/d/file/2022-10-10/6adf13759eaff75a2c17fc2924a64471.jpg)
(1分)
则撤去
F后,再经3s,即5s末时雪橇速度为:
v5 ="0 " (1分)
(3)前2s内雪橇位移:
![](http://www.00-edu.com/d/file/2022-10-10/d50e319a24f647fd22ff68837e9cd828.jpg)
(1分)
后3s内雪橇的位移:
![](http://www.00-edu.com/d/file/2022-10-10/82468453d7d253e42aa28fcd5f374a35.jpg)
(1分)
雪橇5s内位移: s=
s1+
s2=" (5.2+6.76)" =11.96(m) (2分)