As shown in the figure, a block of mass m is sliding with a...

By the principle of conservation of mechanical energy, the kinetic energy of the block is completely converted into the elastic potential energy of the spring at the point of maximum compression.

$\dfrac{1}{2} m v^2 = \dfrac{1}{2} k x_{\text{max}}^2$

$x_{\text{max}}^2 = \dfrac{m v^2}{k}$

$x_{\text{max}} = v\sqrt{\dfrac{m}{k}}$

Since the surface is smooth, there is no loss of energy. When the block returns to its initial position, its speed will be the same as its initial speed $v$. However, velocity is a vector quantity. The block will be moving in the opposite direction, so its velocity will be $-\vec{v}$, which is not the same as the initial velocity $\vec{v}$.

Therefore, the maximum compression is $v\sqrt{\dfrac{m}{k}}$ and the velocity will not be the same.