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(D) According to the law of conservation of energy,the initial velocity is $0$ and the final velocity is $0$. $mgx - \frac{1}{2} kx^2 = 0$ $x = \frac{

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All of the above

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(D) According to the law of conservation of energy,the initial velocity is $0$ and the final velocity is $0$. $mgx - \frac{1}{2} kx^2 = 0$ $x = \frac{2mg}{k}$. When the sphere falls through a distance $x/2$,the extension in the spring is $x/2$. The spring force is $F_s = k(x/2) = k \cdot \frac{2mg}{2k} = mg$. Since the spring force and the gravitational force are equal in magnitude and opposite in direction,the net force is $0$,which means the acceleration $a = 0$ at the position $x/2$. At the lowest point $(x)$,the net force is $F_{net} = kx - mg = k(\frac{2mg}{k}) - mg = mg$. Thus,$ma = mg \Rightarrow a = g$ (upwards). Therefore,all statements are correct.

$A$ sphere of mass $m$ is attached to the lower end of a light vertical spring with force constant $k$. The upper end of the spring is fixed. The sphere is released from rest at the natural (unstretched) length of the spring and comes to rest again after falling through a distance $x$.

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