A) 5 N
B) 6 N
C) 5.88 N
D) 8 N
Correct Answer: A
Solution :
\[{{F}_{l}}=\mu mg=0.6\times 1\times 9.8=5.88\ N\] Pseudo force on the block = \[ma=1\times 5=5\ N\] Pseudo is less then limiting friction hence static force of friction = 5 N.
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