A block of mass  10 kg , moving in  x  direction with a constant  speed of 10 $m{s}^{-1}$, is subjected to a retarding force  F = 0.1x J/m during its travel from  x = 20 m to  30 m. Its final KE will be 

$\text{ Here, }m=10 \mathrm{kg},v_i=10 \mathrm{m}{ \mathrm{s}}^{-1}$

Initial kinetic energy of the block is 

$K_i=\frac{1}{2}m{v}_i^2=\frac{1}{2}\times (10 \mathrm{kg})\times {\left(10 \mathrm{m}{ \mathrm{s}}^{-1}\right)}^2=500 \mathrm{J}$

work done by retarding force

$W={\int }_{x_1}^{x_2}{F}_{r}dx={\int }_{20}^{30}-0.1xdx=-0.1{\left[\frac{x^2}{2}\right]}_{20}^{30}$

$=-0.1\left[\frac{900-400}{2}\right]=-25 \mathrm{J}$

According to work energy theorem,

$\begin{array}{l}W=K_f-K_i\\ K_f=W+K_i=-25 \mathrm{J}+500 \mathrm{J}=475 \mathrm{J}\end{array}$