A block of mass m = 1kg moving on horizontal surface with speed u = 2m/s enters a rough horizontal patch ranging from x = 0.10 m to x = 2.00m. If the retarding force f r on the block in this range is inversely proportional to x over this range i.e.
f r = 
0.10 < x < 2.00
= 0 for x < 0.10 and x > 2.00
If k = 0.5 J then the speed of this block as it crosses the patch is (use 
n 20 = 3)
Text Solution
Verified by ExpertsThe correct answer is:
B
W = 
= – k log e 
= –1.5 J
∴ W = Δ K
× 1 × v 2 – 
× 1× 4 = –1.5
⇒ v = 1 m/s
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