Question

求解一道物理题。英文的。急急急。。

A skier starts from rest at the top of a hill. The skier coasts down the hill and up a
second hill, as the drawing illustrates. The crest of the second hill is circular, with a
radius of r=36 m. Neglect friction and air resistance. What must be the height h of
the rst hill so that the skier just loses contact with the snow at the crest of the second
hill?(Hint: Here you may want to use both energy conservation and force methods as
part of your solution. In any case you should identify what forces would be directed
radially inwards when the skier is at the crest of the second hill and remember that
these are the forces that would generate the skier's centripetal acceleration. Of these forces, which would vanish if contact between the skier and the hill was lost? This force might then be set to zero in the solution.)

Answer 1

这道题目主要是能量守恒，因为摩擦力和空气阻力忽略不计。以虚线为势能零点。势能Ep＝mgh，动能为Ek=0.5*mv*v，
Ep＝Ek
所以2gh=v*v
要求不接触第二个山顶，那么离心力F＝mv*v/r F大于等于mg的话就不接触第二个山顶
所以临界速度为mv*v/r=mg
所以v*v=36g
又因为2gh=v*v
所以36*g=2gh
h=18（米）
所以第一座山的高度为18米

Answer 2

解：设滑雪者质量为m，到达第二个山坡顶的速度为v。圆心o处为零势能点
由skier just loses contact with the snow at the crest of the second
hill这句话可知，在第二个山坡上滑雪者只受到重力作用，支持力为0
则向心加速度a=G/m=g=v^2/r (1)

Neglect friction and air resistance
忽略摩擦力和空气阻力，故滑雪者只受重力作用，机械能守恒，于是有：
mg（h+r）=mgr+1/2mv^2 (2)
把（1）式代入（2）式有：
mg（h+r）=mgr+(1/2)mgr
化简得：h+r=r+1/2r
代入数据得h=1/2r=18米

Answer 3

设这个人的质量为M，到达第二个顶点是的速度为V
求高度差h
根据能量守恒 Mgh=M(V平方)
根据离心力公式 F =2M(V平方）/r
F=Mg=2M(V平方）/r
(V平方）=gr/2
代换到第一个公式得出 h=r/2=18米