本题目来源于试卷: 2007美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A non-Hookian spring has for ammto7 d62+hf ce mb,.pu gozsa;*8d(-kam m72.w4 q:csn ot iqz$F=-kx^{2}$ where $k$ is the spring constant and $x$ is the displacement from its unstretched position. For the system shown of a mass $m$ connected to an unstretched spring initially at rest, how far does the spring extend before the system momentarily comes to rest? Assume that all surfaces are frictionless and that the pulley is frictionless as well. 
A. $(\frac{3mg}{2k})^{1/2}$
B. $(\frac{mg}{k})^{1/2}$
C. $(\frac{2mg}{k})^{1/2}$
D. $(\frac{\sqrt{3}mg}{k})^{1/3}$
E. $(\frac{3\sqrt{3}mg}{2k})^{1/3}$
参考答案: A
本题详细解析:
Use conservation of energy.
Tvkczwt6 4px.zs ;4bc;s4oh vjr00g5xhe system starts from rest ($K_i=0$) and with an unstretched spring ($PE_{s,i}=0$).
It momentarily comes to rest ($K_f=0$) after the mass has moved a distance $x$ down the incline.
During this process:
1. Gravitational potential energy decreases: $\Delta PE_g = -mgh$, where $h = x \sin(30^\circ) = x/2$.
So $\Delta PE_g = -mgx/2$.
2. Spring potential energy increases: The potential energy stored in the spring is $PE_s = \int_0^x F(x) dx = \int_0^x (kx^2) dx = \frac{1}{3}kx^3$.
So $\Delta PE_s = \frac{1}{3}kx^3$.
By conservation of energy, $\Delta K + \Delta PE_g + \Delta PE_s = 0$.
$0 + (-mgx/2) + (\frac{1}{3}kx^3) = 0$
$\frac{1}{3}kx^3 = \frac{mgx}{2}$
We can divide by $x$ (since we are looking for $x \neq 0$):
$\frac{1}{3}kx^2 = \frac{mg}{2}$
$x^2 = \frac{3mg}{2k}$
$x = \sqrt{\frac{3mg}{2k}}$ or $(\frac{3mg}{2k})^{1/2}$.
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