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]
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d0m8zg)mraa; ent of force versus position for a $4.0\,\mathrm{kg}$ cart constrained to move in one dimension on the $x$ axis. At $x=0$ the cart has a velocity of $-3.0\,\mathrm{m/s}$ (in the negative direction).
Which of the following is closest to the maximum speed of the cart?
A. $1.6\,\mathrm{m/s}$
B. $2.5\,\mathrm{m/s}$
C. $3.0\,\mathrm{m/s}$
D. $4.0\,\mathrm{m/s}$
E. $4.2\,\mathrm{m/s}$
参考答案: E
本题详细解析:
Use the Work-Energy Thenx g8s64,txh7ef ) nguorem: $W_{net} = \Delta KE$.
Initial $KE_i$ at $x=0$ is $KE_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(4.0\,\mathrm{kg})(-3.0\,\mathrm{m/s})^2 = 18\,\mathrm{J}$.
The cart first moves left ($x<0$), where $F=+4\,\mathrm{N}$. This is negative work, $KE$ decreases. The cart stops and reverses.
By conservation of energy (as $F$ is a conservative force in this region), when the cart returns to $x=0$, its $KE$ will again be $18\,\mathrm{J}$, but with $v = +3.0\,\mathrm{m/s}$.
The cart then moves right ($x>0$). The speed will be maximum where the *total* work done (starting from $v=0$ at the leftmost point) is maximum. This occurs where the force $F$ drops to zero, at $x=5\,\mathrm{m}$.
We need $KE_{max} = KE(x=0) + W_{0 \to 5}$.
$W_{0 \to 5}$ is the area under the graph from $x=0$ to $x=5$.
$W_{0 \to 1} = \text{Area(rectangle)} = (1\,\mathrm{m})(4\,\mathrm{N}) = 4\,\mathrm{J}$.
$W_{1 \to 5} = \text{Area(trapezoid)} = \frac{1}{2}(b_1+b_2)h$. $F(1)=4$, $F(5)=0$, $h=4$.
$W_{1 \to 5} = \frac{1}{2}(4+0)(4) = 8\,\mathrm{J}$.
$W_{0 \to 5} = 4\,\mathrm{J} + 8\,\mathrm{J} = 12\,\mathrm{J}$.
$KE_{max} = 18\,\mathrm{J} + 12\,\mathrm{J} = 30\,\mathrm{J}$.
$KE_{max} = \frac{1}{2}mv_{max}^2 \implies 30 = \frac{1}{2}(4.0)v_{max}^2 \implies v_{max}^2 = 15$.
$v_{max} = \sqrt{15} \approx 3.87\,\mathrm{m/s}$.
This value is closest to $4.0\,\mathrm{m/s}$ (D). However, the provided solution is (E). There is a discrepancy between the calculated answer and the provided key.
