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 point object of ma gzjo*y 63.hm3.e98hmquobw / v-lshg)byn. e(v kp1i r)w*nrlm): rgzwy 6g x2 ,/yefzhk1n5ss $m$ is connected to a cylinder of radius $R$ via a massless rope. At time $t=0$ the object is moving with an initial velocity $v_{0}$ perpendicular to the rope, the rope has a length $L_{0},$ and the rope has a non-zero tension. All motion occurs on a horizontal frictionless surface. The cylinder remains stationary on the surface and does not rotate. The object moves in such a way that the rope slowly winds up around the cylinder. The rope will break when the tension exceeds $T_{max}$ Express your answers in terms of $T_{max}$ m, $L_{0},$ R, and $v_{0}$. What is the kinetic energy of the object at the instant that the rope breaks?
A. $\frac{{mv_{0}}^{2}}{2}$
B. $\frac{{mv_{0}}^{2}R}{2L_{0}}$
C. $\frac{m{v_0}^2R^2}{2{L_0}^2}$
D. $\frac{m{v_{0}}^{2}{L_{0}}^{2}}{2R^{2}}$
E. none of the above
参考答案: A
本题详细解析:
The tension force from the rope is always perpendicular to the objecty+5akn : gen5u's velocity vector.
Therefore, the tension force does no work (n :a5k5+eguny$W_T = \int \vec{T} \cdot d\vecs = 0$).
The motion is on a horizontal frictionless surface, so gravity and the normal force also do no work.
By the Work-Energy Theorem, $W_{net} = \Delta K = 0$.
The kinetic energy of the object is conserved throughout the motion.
Its final kinetic energy is the same as its initial kinetic energy, $K_{final} = K_{initial} = \frac{1}{2}mv_0^2$.
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