Answer
$v = -3.46~m/s$
Work Step by Step
We can find $v$ when the object passes through $x=0$ in the positive direction:
$\frac{1}{2}mv^2 = 30.0~J$
$v^2 = \frac{(2)(30.0~J)}{m}$
$v = \sqrt{\frac{(2)(30.0~J)}{m}}$
$v = \sqrt{\frac{(2)(30.0~J)}{8.0~kg}}$
$v = 2.74~m/s$
At $x = 5.0~m$, the velocity is zero.
Let $v_0 = 2.74~m/s$ and let $v = 0$
We can find the object's acceleration:
$v^2 = v_0^2+2ax$
$a = \frac{v^2 - v_0^2}{2x}$
$a = \frac{0 - (2.74~m/s)^2}{(2)(5.0~m)}$
$a = -0.75~m/s^2$
Consider the motion from $x=5.0~m$ to $x = -3.0~m$
Let $v_0 = 0$
Let $x = -8.0~m$
We can find $v$ at $x = -3.0~m$:
$v^2 = v_0^2+2ax$
$v^2 = 0+2ax$
$v = -\sqrt{2ax}$
$v = -\sqrt{(2)(-0.75~m/s^2)(-8.0~m)}$
$v = -3.46~m/s$
