一、The figure below shows a block with mass $m = 0.400 \, \mathrm{kg}$ sits on a frictionless air track, connected to a spring with force constant $k = 10.00 \, \mathrm{N/m}$ (equilibrium position, $x = 0$). You pull on the block, stretching the spring $0.100 \, \mathrm{m}$ ($x = 0.100$), and release it from rest. The block moves back toward its equilibrium position. What is its $x$-velocity when $x = 0.080 \, \mathrm{m}$?  【Hint】: 1. Hooke's law: $F_{x} = -kx$ . 2. Work by a spring force: $W_{s} = \frac{1}{2} kx_{i}^{2} - \frac{1}{2} kx_{f}^{2}$ .
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