A block of mass m = 1 kg, moving on a horizontal surface with speed `v_(1) = 2 ms^(-1)` enters a rough patch ranging from x = 0.10 m to x = 2.01 m . The retarding force `F_(r)` on the block is this range in inversely proportional to x over this range .
`F_(r) = (-K)/(x) " for " 0.1 lt x lt 2.01 m and " for " x lt 0.1 m and x gt 2.01 m` F =0
Where k = 0.5 J. What is the final kinetic energy and speed `v_(f)` of the block as it corsses this patch ?

Explanation

From work energy theorem `K_(f) - K_(i) = intFdx`
`rArrK_(f)=K_(1)+underset(0.1)overset(2.01)int((-k)/(x))dx=(1)/(2)mv_(i)^(2)-kln(x)|(2.01),(0.1)|`
`=(1)/(2)mv_(1)^(2)-kln(2.01//0.1)=2-0.5 log_(e) (20.1)`
`2 - 1.5 = 0.5j`
`v_(f)=sqrt(2K_(f)lm)=1ms^(-1)`