NEETPhysicsLaws of MotionMCQ+4 / −1
A block of mass 10 kg placed on rough horizontal surface having coefficient of friction m = 0.5, if a horizontal force of 100 N acting on it then acceleration of the block will be
- A10 m/s2
- B5 m/s2
- C15 m/s2
- D0.5 m/s2.
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Correct answer: B

$ \therefore $ Frictional force = fk
= $${\mu _k}R = {\mu _k}mg$$
= 0.5 × 10 × 10
= 50 N [g = 10 m/sec2]
$ \therefore $ Net force acting on the body = F = P – fk
= 100 – 50 = 50 N.
$ \therefore $ Acceleration of the block = a = F/m
= 50/10 = 5 m/sec2 .
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