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

Explanation

Work energy theorem ,
`DeltaK = W `
` :. K_(f) - K_(i) = int_(x_(i))^(x_(f)) F dx`
` :. K_(f) =K_(i) + int_(0.1)^(2.01) (-k)/x dx `
` =1/2 mv_(i)^(2) - k [ " In x"]_(0.1)^(2.01)`
` = 1/2 x1xx(2)^(2) - 0.5 " in" (2.01)/(0.1)`
` = 2- 0.5 xx 2.303 xxlog_(10)20.1 `
` = 2- 0.5 xx 2.303 xx1.3032 `
` :. 1/2 mv_(f)^(2) =0.5 f `
` :. v_(f)^(2) =(2xx0.5)/m`
` :. v_(f) = sqrt(1/1)`
` :. v_(f) =1 ms^(-1)` .