A block of mass 1 kg moving on a horizontal surface with a speed of 2 ms. It then enters a rough patch extending from x = 0.1 m to x = 2.01 m such that the retarding force F, on the block is inversely K. proportional to x over this range. i.e F, =-— for 0.1 < x < 2.01 m where K = 0.5 J. The speed with which the block crosses this patch is (Given In20.1 = 3) 1 ms + 2)2 ms 3) 4 ms- 4)0.5 ms?

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?

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?

A block of mass m= 1 kg , moving on a horizontal surface with speed v_(t)= 2 ms^(-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 "0.1 lt x lt 2.01m = 0 " for "x lt 0.1m" 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?