本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A block of mass $m=3.0\,\mathrm{kg}$ is moving on a horizontal surface towards a massless spring with spring constant $k=80.0\,\mathrm{N/m}$. The coefficient of kinetic friction between the block and the surface is $\mu_{k}=0.50$. The block has a speed of $2.0\,\mathrm{m/s}$ when it first comes in contact with the spring. How far will the spring be compressed?
A. $0.19\,\mathrm{m}$
B. $0.24\,\mathrm{m}$
C. $0.39\,\mathrm{m}$
D. $0.40\,\mathrm{m}$
E. $0.61\,\mathrm{m}$
参考答案: B
本题详细解析:
Use the Work-Energy Theorem. Th),w 7xb;tlkmo e initial kinetic energy is converted into spring potentbxo7t lwkm ;),ial energy and work done by friction.
$KE_i = W_{friction} + PE_{spring, f}$
$\frac{1}{2}mv^2 = (f_k x) + \frac{1}{2}kx^2$
The friction force is $f_k = \mu_k N = \mu_k mg = (0.50)(3.0\,\mathrm{kg})(10\,\mathrm{m/s^2}) = 15\,\mathrm{N}$.
$\frac{1}{2}(3.0)(2.0)^2 = (15)x + \frac{1}{2}(80.0)x^2$
$6 = 15x + 40x^2$
$40x^2 + 15x - 6 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{-15 + \sqrt{15^2 - 4(40)(-6)}}{2(40)}$ (must be positive displacement)
$x = \frac{-15 + \sqrt{225 + 960}}{80} = \frac{-15 + \sqrt{1185}}{80} \approx \frac{-15 + 34.42}{80} = \frac{19.42}{80} \approx 0.243\,\mathrm{m}$.
| 
