A block of massm = 1 kg moving on a horizontal surface with speed v_(1) = 2ms^(-1) enters a rough patch ranging from x = 010 m to x = 2.01. m. 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 0.1 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 ?

A block of massm = 1 kg moving on a horizontal surface with speed v_(1

1.	A block of massm = 1 kg moving on a horizontal surface with speed v_(1) = 2ms^(-1) enters a rough patch ranging from x = 010 m to x = 2.01. m. 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 0.1 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 ?
Answer» Solution :From work energy theorem `K_(f) - K_(i) = intFdx` `rArrK_(f)=K_(1)+underset(0.1)overset(2.01)INT((-k)/(X))DX=(1)/(2)mv_(i)^(2)-kln(x)\|(2.01),(0.1)\|` `=(1)/(2)mv_(1)^(2)-kln(2.01//0.1)=2-0.5 log_(e) (20.1)` `2 - 1.5 = 0.5j` `v_(f)=sqrt(2K_(f)lm)=1ms^(-1)`