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]
As shown below, Lily i h:o ,3krmala -y 7z5na6uu6i r yqugbp- (u)k:kf7 8hxs using the rope through a fixed pulley to move a box with constant speed h k7(fg)8bu-x:pquy k $v$. The kinetic friction coefficient between the box and the ground is $\mu < 1$; assume that the fixed pulley is massless and there is no friction between the rope and the fixed pulley.  Then, while the box is moving, which of the following statements is correct?
A. The magnitude of the force on the rope is constant.
B. The magnitude of friction between the ground and the box is decreasing.
C. The magnitude of the normal force of the ground on the box is increasing.
D. The pressure of the box on the ground is increasing.
E. The pressure of the box on the ground is constant.
参考答案: B
本题详细解析:
Let $T$ be the tension in the rope and $\theta$ be the angle the rope makes with the horizontal.
Since the box moves at a constant speed $v$, the net horizontal force is zero: $T \cos \theta - f_k = 0 \implies f_k = T \cos \theta$.
The net vertical force is also zero: $N + T \sin \theta - mg = 0 \implies N = mg - T \sin \theta$.
We also know $f_k = \mu N$.
Substitute $N$: $f_k = \mu(mg - T \sin \theta)$.
Now set the two expressions for $f_k$ equal: $T \cos \theta = \mu(mg - T \sin \theta)$.
$T \cos \theta = \mu mg - \mu T \sin \theta \implies T(\cos \theta + \mu \sin \theta) = \mu mg \implies T = \frac{\mu mg}{\cos \theta + \mu \sin \theta}$.
Now find $f_k$: $f_k = T \cos \theta = \frac{\mu mg \cos \theta}{\cos \theta + \mu \sin \theta}$.
Divide numerator and denominator by $\cos \theta$: $f_k = \frac{\mu mg}{1 + \mu \tan \theta}$.
As the box moves to the right, it gets closer to the pulley, so the angle $\theta$ increases. As $\theta$ increases (from $0$ to $90^\circ$), $\tan \theta$ increases.
Since $\tan \theta$ is in the denominator, the friction force $f_k$ must decrease.
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