Here, `m=1kg, v_(i)=2ms^(-1),k=0.5J`
Initial K.E., `K_(i)=(1)/(2)mv_(i)^(2)=(1)/(2)xx1(2)^(2)=2J`
Work done against friction
`W=int_(x=0.1m)^(x=2.01m)F_(r), dx=int_(x=0.1m)^(x=2.01m)-(k)/(x)dx`
`=-0.5[log_(e)x]_(x=01m)^(x=2.01m)`
`=-0.5"log"_(e)(2.01)/(0.1)`
`W=-0.5xx2.303log_(10)20.10`
`=-0.5xx2.303xx1.303=-1.5J`
`:.` Final K.E., `K_(f)=K_(i)+W=2.0-1.5J`
`=0.5J`
`v_(f)=sqrt((2k_(f))/(m))=sqrt((2xx0.5)/(1))=1ms^(-1)`