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 energy stored in ah+aveo; nl9e - battery is converted into i vht1muqz*t-wi xu8kh*y *5)kinetic energy in a toy car 1*hht5y*u )ziiqkxw8m*u v-t that increases 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 energy required i p/ 6g vqfv;k(/ fvnc7(c.esmds equal to the change in kinetic energyc.ffcv e(s/q pmn67 ;v(k/vdg, $\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
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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