A block of mass m = 1 kg moving on a horizontal surface with speed v_(i)=2ms^(-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. 1lt x lt 2.01m = 0 for x lt 0.1m and x gt 2.01 m 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 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)=2ms^(-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. 1lt x lt 2.01m = 0 for x lt 0.1m and x gt 2.01 m 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 therorem `K_(f)-K_(i)=int FDX` `rArr K_(f)=K_(i)+ int_(0.1)^(2.01)((-k))/(x)dx =(1)/(2)mv_(i)^(2)-k1n(x):\|_(0.1)^(2.01)` `=(1)/(2)mv_(i)^(2)-k1n(2.01//0.1)=2-0.5 log_(e )(20.1)` `=2-1.5=0.5 J "" v_(f)=sqrt(2K_(f )//m)=1MS^(-1)`