Question

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 1

From. Eqs.
`K_(f) = K_(i) + int_(0.1)^(2.01)(-K)/xdx`
`=1/2mv_(i)^(2)-kln(x)\_(0.1)^(2.01)`
`=1/2mv_(i)^(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 in is a symbol for the natural logarithum to the base e and not the logarithim to the base 10 [`ln X=log_(c)X= 2.303 log_(10)X]`.