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?

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_(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?
Answer» Solution :From Eq. (6.8a) `K_(f)= K_(t)+int_(0.1)^(2.01)((-k))/(X)dx` `=(1)/(2)mv_(t)^(2)-kln(x) \|{:(2.01),(0.1):}` `=(1)/(2) mv_(t)^(2)-k ln (2.01"/"0.1)` `=2-0.5 ln (20.1)` `= 2-1.5= 0.5 J` `v_(f)= SQRT(2K_(f)"/"m)= 1ms^(-1)` Here, note that ln is a symbol for the natural logarithm to the base e and not the lagarithm to the base `10[ln X= log_(e ) X= 2.303 log_(10)X]`.