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 `K_(f) - K_(i) = int_(x_(1))^(x_(f)) Fdx`
`K_(f) = K_(i) int_(0.1)^(2.01) ((-K))/(x) dx`
`= (1)/(2) mv_(i)^(2)- k 1 n (x)_(0.1)^(2.01)`
`(1)/(2) mv_(i)^(2) - k i n ( 2 . 01 // 0.1)`
= 2 - 0.51 n (20.1)
= 2 - 1.5 - 0.5 J
`v_(f) = sqrt(2 K_(f) //m) = 1ms^(-1)`
Here, note that 1 n is a symbol for the natural logarithm to the base e and not the logarithm to the base `10 [ 1 n X = log_(e) X = 2.303 log_(10) X]`