#### Answer

The speed of the block is 3.3 m/s

#### Work Step by Step

We can find the work done by the force as;
$W = \int_{0}^{4.0}~F_x~dx$
$W = \int_{0}^{4.0}~(20-5x)~dx$
$W = (20x-2.5x^2)~\Big\vert_{0}^{4.0}$
$W = 40~J$
The kinetic energy will be equal to the sum of the work done by the force and the work done by friction;
$KE = W+W_f$
$\frac{1}{2}mv^2 = W-mg~\mu_K~d$
$v^2 = \frac{2W-2mg~\mu_K~d}{m}$
$v = \sqrt{\frac{2W-2mg~\mu_K~d}{m}}$
$v = \sqrt{\frac{(2)(40~J)-(2)(2.6~kg)(9.80~m/s^2)(0.25)(4.0~m)}{2.6~kg}}$
$v = 3.3~m/s$
The speed of the block is 3.3 m/s.