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 small chunk of ice fal x00r,+ beyg39 krcal a7 s ln,c2h*ubols from rest down a frictionless /b6e9if ;yfbz parabolic ice sheet shown in the figure. At the point labeled A in the diagram, the ice sheet becomes a steady, roug ez 6;yb9ffb/i h incline of angle $30^{\circ}$ with respect to the horizontal and friction coefficient $\mu_{k}$. This incline is of length $\frac{3}{2}h$ and ends at a cliff. The chunk of ice comes to rest precisely at the end of the incline. What is the coefficient of friction $\mu_{k}$? 
A. $0.866$
B. $0.770$
C. $0.667$
D. $0.385$
E. $0.333$
参考答案: B
本题详细解析:
Use the Work-Energy Theorem for the entire path, from the start (top) to th-58qzg5 hcuufe end (cliff uc 85f5-hq gzuedge).
Let the initial height be $H$. Point A is at a height $H-h$.
The end of the ramp is at a lower height.
Let's set the potential energy $PE=0$ at point A.
Initial state (top): $K_i = 0$, $PE_i = mgh$.
State at A: $K_A$, $PE_A = 0$. By conservation of energy (frictionless part): $K_A = PE_i = mgh$.
Final state (end of incline): $K_f = 0$. The ice travels a distance $L = \frac{3}{2}h$.
The vertical drop from A to the end is $\Delta y = L \sin(30^\circ) = (\frac{3}{2}h)(\frac{1}{2}) = \frac{3}{4}h$.
So, $PE_f = -mg(\frac{3}{4}h)$.
Work done by non-conservative forces (friction) $W_{nc} = \Delta K + \Delta PE = (K_f - K_A) + (PE_f - PE_A)$.
The work by friction is $W_f = -f_k L = -(\mu_k N) L$. $N = mg \cos(30^\circ)$.
$W_f = -\mu_k (mg \cos(30^\circ)) (\frac{3}{2}h)$.
Substitute everything in:
$-\mu_k mg (\cos 30^\circ) (\frac{3}{2}h) = (0 - mgh) + (-mg \frac{3}{4}h - 0)$
$-\mu_k mg (\frac{\sqrt{3}}{2}) (\frac{3}{2}h) = -mgh - \frac{3}{4}mgh = -\frac{7}{4}mgh$
Divide by $-mgh$:
$\mu_k (\frac{3\sqrt{3}}{4}) = \frac{7}{4}$
$\mu_k (3\sqrt{3}) = 7$
$\mu_k = 7 / (3\sqrt{3}) = 7\sqrt{3} / 9 \approx 0.7698$.
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