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]
The chemical potential ene9b,nwy/;ng pn+:yfua9 /ruc)yjhd-k rgy stqv 7vnzpt+j vuk;j2:28e kg/ lxf1 ad7ored in a battery is converted into kinetic energy in a toy car that increqkvv;e87 x2tzgpka2vj : lfu7 +n1d/jases its speed first from $0\,\mathrm{mph}$ to $2\,\mathrm{mph}$ and then from $2\,\mathrm{mph}$ up to $4\,\mathrm{mph}$. Ignore the energy transferred to thermal energy due to friction and air resistance. Compared to the energy required to go from $0$ to $2\,\mathrm{mph}$, the energy required to go from $2$ to $4\,\mathrm{mph}$ is
A. half the amount.
B. the same amount.
C. twice the amount.
D. three times the amount.
E. four times the amount.
参考答案: D
本题详细解析:
By the Work-Energy Theorem, the (z 0pmi7p08pn2ztmrs energy required is equal to the change in kinetic energytm 0p0p8p zz7r snmi2(, $\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
Let $v_0 = 0$, $v_1 = 2\,\mathrm{mph}$, and $v_2 = 4\,\mathrm{mph}$.
Energy for the first interval (0 to 2): $\Delta K_1 = \frac{1}{2}m(v_1)^2 - \frac{1}{2}m(v_0)^2 = \frac{1}{2}m(2^2) - 0 = \frac{1}{2}m(4)$.
Energy for the second interval (2 to 4): $\Delta K_2 = \frac{1}{2}m(v_2)^2 - \frac{1}{2}m(v_1)^2 = \frac{1}{2}m(4^2) - \frac{1}{2}m(2^2) = \frac{1}{2}m(16 - 4) = \frac{1}{2}m(12)$.
The ratio of the energies is $\Delta K_2 / \Delta K_1 = (\frac{1}{2}m(12)) / (\frac{1}{2}m(4)) = 12 / 4 = 3$.
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