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 sled loaded with children starts from rest andle32-9zz :hd9-wpzmybt( fq qef u+s/3e8r qsge55l yzf 3xp; slides down a snowy $25^{\circ}$ (with respect to the horizontal) incline traveling 85 meters in 17 seconds. Ignore air resistance. What is the coefficient of kinetic friction between the sled and the slope?
A. $0.36$
B. $0.40$
C. $0.43$
D. $1.00$
E. $2.01$
参考答案: B
本题详细解析:
First, find the accelerai-nk llbda)2z8 z3 8qotion $a$ from kinematics.
Since the sled starts from rest, we use $\Delta x = \frac{1}{2}at^2$.
$85\,\mathrm{m} = \frac{1}{2} a (17\,\mathrm{s})^2 = \frac{1}{2} a (289\,\mathrm{s}^2)$
$a = (85 \times 2) / 289 = 170 / 289 \approx 0.588\,\mathrm{m/s}^2$.
Next, apply Newton's second law. Perpendicular to the incline, the normal force is $N = mg \cos\theta$.
Parallel to the incline, the net force is $F_{net} = mg \sin\theta - f_k = ma$, where $f_k$ is the kinetic friction.
We know $f_k = \mu_k N = \mu_k mg \cos\theta$.
Substitute $f_k$ into the parallel equation: $mg \sin\theta - \mu_k mg \cos\theta = ma$.
Divide by $mg$: $\sin\theta - \mu_k \cos\theta = a/g$.
Solve for $\mu_k$: $\mu_k \cos\theta = \sin\theta - a/g \implies \mu_k = \tan\theta - \frac{a}{g \cos\theta}$.
Using $g \approx 10\,\mathrm{m/s}^2$:
$\mu_k = \tan(25^\circ) - \frac{0.588}{(10) \cos(25^\circ)} \approx 0.466 - \frac{0.588}{10(0.906)} \approx 0.466 - 0.0649 \approx 0.40$.
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