A block of mass $$m$$, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant $$k$$. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in its equilibrium position. If now the block is pulled with a constant force $$F$$, the maximum speed of the block is:

At maximum speed, net force on the block is zero (mean position of new SHM): $$F_{\text{net}} = 0 \implies F = kx_0 \implies x_0 = \frac{F}{k}$$

Applying the Work-Energy Theorem from the initial position to this point of maximum speed: $$W_{\text{all}} = \Delta K$$

$$W_F + W_{\text{spring}} = \frac{1}{2}mv_{\text{max}}^2 - 0$$

$$F \cdot x_0 - \frac{1}{2}kx_0^2 = \frac{1}{2}mv_{\text{max}}^2$$

$$F\left(\frac{F}{k}\right) - \frac{1}{2}k\left(\frac{F}{k}\right)^2 = \frac{1}{2}mv_{\text{max}}^2$$

$$\frac{F^2}{k} - \frac{F^2}{2k} = \frac{1}{2}mv_{\text{max}}^2 \implies \frac{F^2}{2k} = \frac{1}{2}mv_{\text{max}}^2$$

$$v_{\text{max}} = \frac{F}{\sqrt{mk}}$$