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

From Eq. (6.8a)
`K_(f)= K_(t)+int_(0.1)^(2.01)((-k))/(x)dx`
`=(1)/(2)mv_(t)^(2)-kln(x) |{:(2.01),(0.1):}`
`=(1)/(2) mv_(t)^(2)-k ln (2.01"/"0.1)`
`=2-0.5 ln (20.1)`
`= 2-1.5= 0.5 J`
`v_(f)= sqrt(2K_(f)"/"m)= 1ms^(-1)`
Here, note that ln is a symbol for the natural logarithm to the base e and not the lagarithm to the base `10[ln X= log_(e ) X= 2.303 log_(10)X]`.