A block of mass `m= 1 kg`, moving on a horizontal surface with speed `v_(t)= 2 ms^(-1)` enters a rough patch ranging from `x= 0.10m" to "x= 2.01m`. The retarding force `F_(r )` on the block in this range is inversely proportional to x over this range,
`F_(r )= (-k)/(x)" for "0.1 lt x lt 2.01m = 0 " for "x lt 0.1m" and "x gt 2.01m`
where `k= 0.5 J`. What is the final kinetic energy and speed `v_(f)` of the block as it crosses this 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]`.