Q.

A block of mass m = 5 kg is resting on a rough horizontal surface for which the coefficient of friction is 0.2. When a force F = 40 N is applied, the acceleration of the block will be (g = 10 rn/s2)

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a

5.73 m/sec2

b

8.0 m/sec2

c

3.17 m/sec2

d

10.0 m/sec2

answer is A.

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Detailed Solution

Kinetic friction = μkR = 0.2(mg - F sin 300)= 0.2 (5×10-40×12) = 0.2 (50-20) = 6 NAcceleration of the block= F cos 300-Kinetic frictionMass= 40 ×32-65 = 5.73 m/s2
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