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

A block of mass m=1kg moving on a horizontal surface with speed v_(i)=

1.	A block of mass m=1kg moving on a horizontal surface with speed v_(i)= 2ms^(-1) enters a rough patch ranging from x= 0.10 m to x= 2.01m. The retarding force F_(r ) on the block in this range in inversely proportioanl to x over this range. F_(r )= (-k)/(x) for 0. 1 lt x lt 2.01m= 0 " for " x lt 0.1m and x gt 2.01m where k= 0.5J. 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_(r )-K_(i)= int Fdx` `rArr K_(F) = K_(i) + underset(0.1)overset(2.01)int ((-k))/(x) dx = (1)/(20 mv_(1)^(2)- kln (x) \|(2.01),(0.1)\|` `=(1)/(2) mv_(1)^(2) - k LN (2.01//0.1) = 2- 0.5 log_(e) (20.1)` `=2 - 1.5= 0.5J` `v_(f) = sqrt(2K_(f)//m)= 1ms^(-1)`