Answer
The speed of the block is $0.50~v_f$ at a distance of $0.50~m$ from the top of the incline.
Work Step by Step
We can find an expression for the acceleration:
$v_f^2 = v_0^2+2ad$
$a = \frac{v_f^2 - v_0^2}{2d}$
$a = \frac{v_f^2 - 0}{(2)(2.0)}$
$a = \frac{v_f^2}{4.0}$
We can find the distance when the velocity is $0.50~v_f$:
$(0.50~v_f)^2 = v_0^2+2ad$
$d = \frac{(0.50~v_f)^2-v_0^2}{2a}$
$d = \frac{0.25~v_f^2}{(2)(\frac{v_f^2}{4.0})}$
$d = 0.50~m$
The speed of the block is $0.50~v_f$ at a distance of $0.50~m$ from the top of the incline.
