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 therorem `K_(f)-K_(i)=int Fdx`
`rArr K_(f)=K_(i)+ int_(0.1)^(2.01)((-k))/(x)dx =(1)/(2)mv_(i)^(2)-k1n(x):|_(0.1)^(2.01)`
`=(1)/(2)mv_(i)^(2)-k1n(2.01//0.1)=2-0.5 log_(e )(20.1)`
`=2-1.5=0.5 J " " v_(f)=sqrt(2K_(f )//m)=1ms^(-1)`