A block of mass `m=1kg` moving on a horizontal surface with speed `v_(i)=2ms^(-1)` enters a rough patch ranging from `x0.10m to x=2.01m`. The retarding force `F_(r)` on the block in this range ins inversely proportional to x over this range
`F_(r)=-(k)/(x) fo r 0.1 lt xlt 2.01m`
`=0` for `lt 0.1m` and `x gt 2.01m` where `k=0.5J`. What is the final K.E. and speed `v_(f)` of the block as it crosses the patch?

Options

A. ` 2m^(-1)`
B. ` 1m^(-1)`
C. ` 3m^(-1)`
D. ` 0.5m^(-1)`

Answer

`(2)`

Explanation

`K_(f) = K_(i) int _(0.1)^(2.01)(-k)/(x dx) = 1/2 mv_(i)^(2) - kln(x) |_(0.1)^(2.01) `
`v_(f) = sqrt(2K_(f))`/m