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(C) Given: Mass $m = 0.5\ g = 0.5 	imes 10^{-3}\ kg$, Spring constant $k = 1.25\ N/cm = 125\ N/m$, Compression $x = 2.0\ cm = 0.02\ m$, Distance $d =

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$9$

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(C) Given: Mass $m = 0.5\ g = 0.5 	imes 10^{-3}\ kg$, Spring constant $k = 1.25\ N/cm = 125\ N/m$, Compression $x = 2.0\ cm = 0.02\ m$, Distance $d = 10\ cm = 0.1\ m$, Friction force $f = 0.0475\ N$.According to the Work-Energy Theorem, the work done by the spring minus the work done against friction equals the change in kinetic energy:$W_{spring} - W_{friction} = \Delta K$$\frac{1}{2} k x^2 - f d = \frac{1}{2} m v^2$Substituting the values:$\frac{1}{2} 	imes 125 	imes (0.02)^2 - 0.0475 	imes 0.1 = \frac{1}{2} 	imes 0.5 	imes 10^{-3} 	imes v^2$$0.5 	imes 125 	imes 4 	imes 10^{-4} - 0.00475 = 0.25 	imes 10^{-3} 	imes v^2$$0.025 - 0.00475 = 0.25 	imes 10^{-3} 	imes v^2$$0.02025 = 0.25 	imes 10^{-3} 	imes v^2$$v^2 = \frac{0.02025}{0.25 	imes 10^{-3}} = \frac{20.25}{0.25} = 81$$v = 9\ m/s$.

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