滑雪运动应该在哪些地方开展?

“驼峰效应”告诉我们：如果你的心情正处在幽暗的境地,千万不要放弃,因为爬上山顶就可以迎来愉悦的新生活.
答案选 C

踩在雪板上不会下陷 压力一定时 增大受力面积 减小压强
滑雪时曲腿降低重心不容易摔倒 物体重心越低 支撑面越大 越稳定
需要急转弯时,用雪杖支撑地面 力可以改变物体的运动状态

因为滑雪运动是自高向低进行的,重心前倾.因此向后支撑滑行是为了掌握平衡、易于调整方向.

你可以这么算呀,完全没有错.
但是你这么作求出的是位移x,而题目要求时间,你还是要求加速度.
如果你直接分析受力,得出加速度,然后t=v/a 直接就算出来了.
所以不求时间的题一般动能定理等简便,要求时间的话你可以用牛顿第二定律求加速度再做,当然你仍然用动能定理有时也可以.
有问题追问

解题思路：根据牛顿第二定律分别求出物体在斜面上和水平滑行的加速度大小．根据运动学速度公式得出物体经过P时的速度vp与物体运动时间的关系，求出物体在斜面上运动的时间，再由位移公式求解斜面AB的长度S．

设物体在斜面上滑动时加速度大小为a₁、所用时间为t₁，在水平面上滑动时加速度大小为a₂，
则根据牛顿第二定律有：
mgsinθ-μmgcosθ=ma₁
得：a₁=g（sinθ-μcosθ）
μmg=ma₂
得：a₂=μg
又：v_p=a₁t₁-a₂（t-t₁）
解得：t₁=10s
故斜面的长度为：s＝
1
2a1t12
代入解出：S=220m
答：斜面AB的长度S是220m．

点评：
本题考点： 牛顿第二定律；匀变速直线运动的速度与时间的关系；匀变速直线运动的位移与时间的关系．

考点点评： 本题运用牛顿第二定律和运动学公式结合处理动力学问题，加速度是必求的量，关键是得到物体经过P时的速度vp与物体运动时间的关系．

解题思路：（1）根据匀变速直线运动的位移时间公式求出下滑的加速度，对物体进行受力分析，根据牛顿第二定律求出下滑过程中的摩擦力．
（2）在垂直斜面方向上，受力平衡，求出正压力，根据滑动摩擦力的公式即可求得动摩擦力因素．

（1）以滑雪者为研究对象，其受力情况如图所示，受到重力G=mg，斜坡的支持力N和摩擦力f．

将重力G沿斜面和垂直斜面方向分解为G₁和G₂．
由几何知识得：
G₁=mgsinθ
G₂=mgcosθ
设下滑的加速度为a，由公式
v2t−
v20＝2as可得：
a＝

v2t−
v20
2s＝
202−0
2×50m/s2＝4m/s2
由牛顿第二定律得：F_合=ma=80×4N=320N
由受力分析图可知：F_合=G₁-f=mgsinθ-f
联立解得：f=80N
（2）斜面的支持力 N=G₂=mgcosθ
滑动摩擦力公式 f=μN
两式联立解得：μ=0.12
答：（1）滑雪者在下滑过程中受到的摩擦力f为80N；
（2）滑板与雪地之间的动摩擦因数μ为0.12．

点评：
本题考点： 牛顿第二定律．

考点点评： 解决本题的关键知道加速度是联系力学和运动学的桥梁，通过加速度可以根据力求运动，也可以根据运动求力．

（1）滑雪板做的又宽又长，增大了受力面积，在压力一定时减小了对雪面的压强，避免陷入雪中；
（2）运动员如图所示的姿势，使背上流过的空气流速大于从滑雪板下面流过的空气流速，而气体在流速大的地方压强小、在流速小的地方压强大，从而产生向上的压力差，能使他在空中滑行更远的距离．
故答案为：（1）为什么滑雪板做的又宽又长？；增大了受力面积，在压力一定时减小了对雪面的压强，避免陷入雪中；（2）为什么飞起时，运动员如图所示的姿势？使背上流过的空气流速大于从滑雪板下面流过的空气流速，而气体在流速大的地方压强小、在流速小的地方压强大，从而产生向上的压力差，能使他在空中滑行更远的距离．

解题思路：（1）对于下滑过程，合外力对人和雪橇做的功等于人和雪橇做的动能变化量，根据动能定理求解；
（2）根据合外力做功等于重力与阻力做功之和，求解人和雪橇克服阻力做的功．

（1）以人和雪橇整体为研究对象，对于下滑过程，根据动能定理得：
W合＝
1
2m
v2t−
1
2m
v20
代入数据可得：
W 合＝3000J；
（2）以人和雪橇整体为研究对象：
W 合＝WG+Wf=mgh+W_f；
代入数据得：W_f=-6000J
所以，克服摩擦力做的功为6000J．
答：
（1）下滑过程中，合外力对人和雪橇做的功是3000J；
（2）下滑过程中，人和雪橇克服阻力做的功是6000J．

点评：
本题考点： 动能定理的应用．

考点点评： 本题考查功的计算是借助于动能定理求得，也可以由功的表达式W=FS，但F必须是恒力，且关键是确定在力的方向上移动的距离，计算时注意力和距离要对应．

(1)用公式a=vt^2-v0^2/2s 就可以求出到达山坡底端 时速度v的大小为16m/s 既 2=vt^2-2^2/2x42 ,vt=16
(2)同样用公式a=vt^2-v0^2/2s 可以求出第二段的加速度为-4m/s 既 a=0-16^2/2x32 ,a=-4m/s ,然后用公式a=vt-v0/t 可以求出第二段所用时间,既 -4=0-16/t ,t=4s,然后第一段的时间也是 2=16-2/t ,t=7s ,所以总时间为7+4=11s
刚花几分钟给你做的,我也是高一学生.

当然是阿尔卑斯山啦.驰名中外.
冲浪的话澳大利亚,夏威夷

（1）滑雪运动员利用雪橇从雪山上快速滑下后在雪地上留下雪橇的划痕；说明力可以使物体发生形变；
（2）滑雪运动员在滑行过程中做曲线运动；说明力可以改变物体运动状态；
（3）滑雪运动员滑行到终点停止运动；说明力可以改变物体运动状态．
故答案为：雪地上留下雪橇的划痕；力可以使物体发生形变．

解题思路：（1）根据牛顿第二定律求出动摩擦因数时0.25时的加速度，根据匀变速直线运动的速度时间公式求出滑雪者从静止开始到动摩擦因数发生变化所经历的时间．
（2）根据牛顿第二定律求出动摩擦因数为0.125时的加速度，根据匀变速直线运动的位移公式求出开始匀加速运动的位移，从而得出第二段匀加速直线运动的位移，根据速度位移公式求出滑雪者到达B处的速度．
（3）第（2）问滑雪者到达B处的速度为16m/s，知10m/s可能在斜面上，可能在水平面上，分别根据牛顿第二定律求出运动的加速度．

（1）在AB段：速度小于4m/s时：
a₁=gsin37°-μ₁gcos37°=10×0.6-0.25×10×0.8=4m/s²
t1＝
v
a1＝
4
4s＝1s
（2）s1＝
1
2a1t12＝
1
2×4×12＝2m
速度大于4m/s时：
a₂=gsin37°-μ₂gcos37°=10×0.6-0.125×10×0.8=5m/s²
s₂=L-s₁=26-2m=24m
vB＝
v2+2a2s2＝
42+2×5×24=16m/s
（3）
受力分析如图
当速度等于10m/s时动摩擦因素为0.125，分别在AB段与BC段两处，在AB段时加速度为
a₂=gsin37°-μ₂gcos37°=10×0.6-0.125×10×0.8=5m/s²
在BC段时加速度为a₃=-μ₂g=-1.25m/s²．
答：（1）滑雪者从静止开始到动摩擦因数发生变化所经历的时间为1s．
（2）滑雪者到达B处的速度为16m/s．
（3）滑雪者速度大小为10m/s时的受力如图，加速度大小分别为5m/s²，1.25m/s²．

点评：
本题考点： 牛顿第二定律；匀变速直线运动的速度与时间的关系；匀变速直线运动的位移与时间的关系．

考点点评： 本题综合考查了牛顿第二定律和运动学公式，知道加速度是联系力学和运动学的桥梁，通过加速度可以根据力求运动，也可以根据运动求力．

解题思路：解决本题的关键是掌握力的作用效果：力可以改变物体的形状、力可以改变物体的运动状态．

（l）滑雪运动员利用雪橇从雪山上快速滑下后在雪u上留下雪橇的划痕；说明力可以使物体发生形变；
（l）滑雪运动员在滑行过程中做曲线运动；说明力可以改变物体运动状态；
（如）滑雪运动员滑行到终点停止运动；说明力可以改变物体运动状态．
故答案为：雪u上留下雪橇的划痕；力可以使物体发生形变．

点评：
本题考点： 力的作用效果．

考点点评： 这类题目可以从力的作用效果角度分析，即力可以改变物体的形状、力可以改变物体的运动状态

1,0.784牛
2. 3860帕
3. 7.65克/立方厘米

找来一段History of Alpine Skiing供参考
Before skis were used for fun and leisure,the ski was used for work and transportation.The oldest known version is a wide,short ski found in Sweden that has been shown to be over 4500 years old,and cave and rock drawings suggest that skis were used even long before then.These first skis may have been used by a hunter or a traveler,as they were commonly used during the long winters in Northern Russia and the Scandinavian countries.Early skis were not made for speed,but to designed to keep a traveler on top of the snow as they went about their business.
The people from the Telemark area of Norway have been largely credited with developing skiing into a sport,somewhere in the early 1700's.They invented the Telemark and the Christiana (now known as the Christie) turns as methods of artfully controlling speeds on downhill descents.They were also quite fond of jumping.Thus,disciplines in both alpine and Nordic skiing owe their existence to these early pioneers.
The distinction that now we make between the different disciplines of skiing was not made centuries ago.All of the early skis had a boot mounted to the ski only at the toe,with the heel free to move up and down,what we now call a Nordic ski.
The first organized events in skiing,jumping and a type of cross-country race,started in the early 1800's,and both used the Nordic system.While the Nordic ski has seen many changes over the years,with several different varieties existing today,these two disciplines of skiing still exist with much of the same spirit as they had over 150 years ago.
Skiing grew more popular in the early 1900's,as Europeans learned about all of the fun their Norwegian neighbors were having.The sport still utilized Nordic equipment during much of this time.This is illustrated with events included in the first Winter Olympic Games.The inaugural 1924 Games in Chamonix,France had only 5 sports,and the skiing events were both Nordic:Ski Jumping and Nordic Combined.This continued until Cross-Country Skiing first made the Olympic agenda as a stand alone event during the 1932 Winter Games in Lake Placid.
As skiing was taken to ever more challenging terrain,however,technique and equipment adapted to the challenge.The Telemark turn was adequate for the flatter,rolling terrain of Norway,but lacked the control necessary for the steeper slopes of the Alps and other European mountains.This lead to the birth of Alpine skiing.
Alpine ski equipment used a boot that was mounted to the ski at both the toe and the heel,and gave more control to the skier,allowing him/her to negotiate steeper slopes and ski at faster speeds.The Alpine skiing disciplines of downhill and slalom came about with this new equipment,and the 1936 Winter Games in Innsbruck,Austria,saw the first introduction of an alpine ski event:the combined,which added a skier's results in both events.
It was during the 1930's that alpine skiing became a popular European pastime,as ski lifts were invented that eliminated the labor of climbing a mountain before experiencing an exhilarating descent.The ski area industry began in earnest after the Second World War,when Austria and Switzerland developed the first Alpine Ski Resorts.
During the past 60 years different schools of thought have grew up around skiing,with advocates for different techniques and disciplines vying for the recognition as the best form of the sport.
Giant Slalom combined aspects of both previous disciplines,and first made the Olympic Games during the 1952 Oslo Winter Olympics.
Super G,A hybrid of Giant Slalom and Downhill,added a fourth alpine ski discipline when it was added to World Cup events in 1983,and the Olympics in 1988.
During this time equipment manufacturers developed faster and safer equipment,and athletes combined this new equipment with better training and technique to continuously improve the sport.
While the debate still goes on about which is the premier skiing discipline,one thing is for sure:the growth in popularity of all forms of skiing,and the fact that they are sports that can be enjoyed for a lifetime,has meant that each skiing discipline has grown.Each has been embraced and championed by its aficionados.
The legacy left to us by skiing's long history is that each winter brings us an increasing number of ways to enjoy winter,and to race against each other on snow.
Ever heard this before:"First one down the hill wins!"

解题思路：（1）根据牛顿第二定律求出滑雪者在斜坡上从静止开始加速至速度v₁=4m/s期间的加速度，再根据速度时间公式求出运动的时间．
（2）再根据牛顿第二定律求出速度大于4m/s时的加速度，球心速度为4m/s之前的位移，从而得出加速度变化后的位移，根据匀变速直线运动的速度位移公式求出滑雪者到达B处的速度．
（3）根据动能定理分别求出在水平面上速度减为4m/s之前的位移和速度由4m/s减小到零的位移，两个位移之和为滑行的最大距离．

（1）设滑雪者质量为m，滑雪者在斜坡上从静止开始加速至速度v₁=4m/s期间，由牛顿第二定律有：mgsin37°-μ₁mgcos37°=ma₁
解得：a1＝4m/s2
故由静止开始到动摩擦因数发生变化所经历的时间：t＝
v1
a1＝1s
（2）则根据牛顿定律和运动学公式有：x1＝
1
2a1t2
mgsin37°-μ₂mgcos37°=ma₂
x₂=L-x₁

v2B−
v21＝2a2x2
代入数据解得：v_B=16m/s
（3）设滑雪者速度由v_B=16m/s减速到v₁=4m/s期间运动的位移为x₃，速度由v₁=4m/s减速到零期间运动的位移为x₄，则由动能定理有：−μ2mgx3＝
1
2m
v21−
1
2m
v2B
−μ1mgx4＝0−
1
2m
v21
所以滑雪者在水平雪地上运动的最大距离为：x=x₃+x₄=99.2m．
答：（1）滑雪者从静止开始到动摩擦因数发生变化所经历的时间为1s．
（2）滑雪者到达B处的速度为16m/s．
（3）滑雪者在水平雪地上运动的最大距离为99.2m．

点评：
本题考点： 动能定理的应用；匀变速直线运动的位移与时间的关系；牛顿第二定律．

考点点评： 本题综合运用了牛顿第二定律、动能定理等规律，关键理清滑雪者的运动过程，正确地受力分析，运用牛顿定律或动能定理解题．