Answer
The speed of the toolbox just as it reaches the edge will be 5.23 m/s.
Work Step by Step
$K_2 + U_2 = K_1 + U_1 + W_f$
$\frac{1}{2}mv^2+0 = 0 + mgh-F_f~d$
$v^2 = \frac{2mgd~sin(\theta)-2F_f~d}{m}$
$v = \sqrt{\frac{2mgd~sin(\theta)-2F_f~d}{m}}$
$v = \sqrt{\frac{(2)(85.0~N)(4.25~m)~sin(36^{\circ})-(2)(22.0~N)(4.25~m)}{85.0~N~/~9.80~m/s^2}}$
$v = 5.23~m/s$
The speed of the toolbox just as it reaches the edge will be 5.23 m/s.
