Question

A block of mass 1 kg moving on a horizontal surface with speed, u = 2 m/s enters a rough patch from x= 0.1 m to x = 2.01 m. The retarding force f on the block in the range is inversely proportional to x over this range where constant k is 0.5 m. What is the final kinetic energy and velocity of the block as it crosses this patch. Given log 20.1 = 3.001.

Answer 1

hey mate..,.,,.....

here is ur answer...........

F = ma..mv dv/dx = -k/x = -0.5/x

vdv = -0.5 dx/x

v2 / 2 - (2) 2/ 2 = 0.5 ( -1n . 2.01/0.1) = -1.5

v2 = 2 ( 2- 1.5) = 1

v = 1 m/s .

hope it helps♥♥♥

Answer 2

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