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?

Answer

`0.5`

Explanation

By work -energy theorem, `Delta K= W_(r) or K_(f)- K_(i)= underset(x_(i))overset(x_(f))int F_(r)dx`
`therefore K_(f)= K_(i) + underset(0.1)overset(2.01)int ((-k))/(x) dx`
`= (1)/(2) mv_(i)^(2)- k underset(0.1)overset(2.01)int (1)/(x) dx`
`= (1)/(2) xx 1 xx 2^(2)- k [ln x]_(0.1)^(2.01)`
`= 2- 0.5 ["ln " (2.01)/(0.1)]=2- 0.5 ln 20.1`
`= 2- 0.5 xx 2.303 log 20.1`
`=2 - 0.5 xx 2.303 xx 1.3032= 2- 1.5` = 0.5J
Final speed, `v_(f)= sqrt((2K_(f))/(m))= sqrt((2 xx 0.5)/(1))=1 ms^(-1)`