Answer
The block reaches a point $0.10~m$ above the unstretched point.
The block reaches a point $0.20~m$ below the unstretched point.
Work Step by Step
At the point where the spring is unstretched, let $U_g = 0$
We can find the height $y$ above and below the unstretched point that the block reaches:
$K+U_e+U_g = 2.00~J$
$0+\frac{1}{2}ky^2+mgy = 2.00~J$
$\frac{1}{2}ky^2+mgy - 2.00~J = 0$
$\frac{1}{2}(200.0~N/m)y^2+(10.0~N)~y - 2.00~J = 0$
$(100.0~N/m)y^2+(10.0~N)~y - 2.00~J = 0$
We can use the quadratic formula to find the values of $y$:
$y = \frac{-10.0 \pm \sqrt{(10.0)^2-(4)(100)(2.00)}}{(2)(100.0)}$
$y = \frac{-10.0 \pm \sqrt{900}}{200.0}$
$y = -0.20~m, 0.10~m$
The block reaches a point $0.10~m$ above the unstretched point.
The block reaches a point $0.20~m$ below the unstretched point.
