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

A block of mass m = 1 kg , moving on a horizontal surface with speed v

1.	A block of mass m = 1 kg , moving on a horizontal surface with speed v_(i) = 2 ms^(-1) enters a rough patch ranging from v_(i) = 2 ms^(-1) enters a rough patch ranging from x = 0.10 m tox = 2.01m The retarding force F_(r ) on the block in this rangeis inversely proportional to x over this range , F_(r) =(-k)/x "for " 0.1 lt x lt 2.01 m = 0 " for " x lt 0.1 m and x gt 2.01where k 0.5 J .What is the final kinetic energy and speed v_(f) of the block as it crosses this patch ?
Answer» Solution :`K_(f) = 2.8 J , v_(f) = 2.37 ms^(-1)`