Answer
$v = 6.6~m/s$
Work Step by Step
We can find the work done on the block by the force:
$W = \int^{4.0}_{3.0}-6x~dx$
$W = -3x^2~\Big\vert^{4.0}_{3.0}$
$W = (-3)(4.0)^2-[(-3)(3.0)^2]$
$W = -48+27$
$W = -21~J$
We can find the kinetic energy at $x=3.0~m$:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(2.0~kg)(8.0~m/s)^2$
$K = 64~J$
Then the kinetic energy at $x=4.0~m$ is:
$64~J-21~J = 43~J$
We can find the velocity at $x=4.0~m$:
$\frac{1}{2}mv^2 = K$
$v^2 = \frac{2K}{m}$
$v = \sqrt{\frac{2K}{m}}$
$v = \sqrt{\frac{(2)(43~J)}{2.0~kg}}$
$v = 6.6~m/s$
