Answer
(a) $v = \sqrt{\frac{F\cdot x}{m}}$
(b)$v = \sqrt{\frac{3~F\cdot x}{4~m}}$
Work Step by Step
The force $F$ is equal to $kx$. The total energy $E$ in the system is $\frac{1}{2}kx^2$. These two equations can be written as:
$k = \frac{F}{x}$
$E = \frac{1}{2}kx^2 = \frac{1}{2}F\cdot x$
(a) When $x=0$, all the of the energy in the system is in the form of kinetic energy. Therefore,
$KE = E$
$\frac{1}{2}mv^2 = \frac{1}{2}F\cdot x$
$v^2 = \frac{F\cdot x}{m}$
$v = \sqrt{\frac{F\cdot x}{m}}$
(b) When the spring is stretched $\frac{x}{2}$, then 25% of the energy $E$ is elastic potential energy stored in the spring and 75% of the energy $E$ is kinetic energy. So,
$KE = 0.75~E$
$\frac{1}{2}mv^2 = \frac{3}{4}\times \frac{1}{2}F\cdot x$
$v^2 = \frac{3~F\cdot x}{4~m}$
$v = \sqrt{\frac{3~F\cdot x}{4~m}}$
