Answer
The maximum kinetic energy is $~~13~J$
Work Step by Step
The force on the block from the spring is greater at the point where the spring is stretched the most, and then decreases to zero as the block approaches the equilibrium point. The block attains maximum kinetic energy at the point where the force from the spring is equal in magnitude to the force from the kinetic friction which is $80~N$
We can find that point $x$:
$kx = 80~N$
$x = \frac{80~N}{4000~N/m}$
$x = 0.020~m$
We can find the kinetic energy $K_2$:
$K_2+U_{e2} = K_1+U_{e1}+W$
$K_2 = 0+U_{e1}-U_{e2}+W$
$K_2 = \frac{1}{2}k~(x_1^2-x_2^2)-F~\Delta x$
$K_2 = \frac{1}{2}(4000~N/m)~[(0.10~m)^2-(0.020~m)^2]-(80~N)(0.080~m)$
$K_2 = 13~J$
The maximum kinetic energy is $~~13~J$
