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?

Explanation

Using work - energy theorem ,
Change in kinetic energy = Work done against retarding force
` rArr K_(f) - K_(i) = W_(r) = int_(x_(i))^(x_(f)) F_(r) . Dx`
` rArr " " = int_(0.1)^(2.01) ((-k)/x) . dx `
` = - k [1n x] _(0.1)^(2.01)`
` = 0.5 [1n (2.01)/(0.1)] ` ` = - 0.5 xx 2.303 xx 1.3032 `
` = - 1.5 J `
` :. " " K_(f)=K_(i) - 1.5 J = 1/2 mv_(i)^(2) - 1.5 J `
` = 1/2 xx 1 xx 2^(2) - 1.5 = 2 - 1.5 = 0.5 J `
`K_(f)= 1/2 mv_(f)^(2)`
`:. " " v_(f) = sqrt((2K_(f))/m) = sqrt((2xx0.5)/1 ) = 1` m/s