♠ 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 SHOULD BE APPROPRIATE AND WITH DETAILED SOLUTION...
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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♥♥♥
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♥♥♥
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