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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 ms2)

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a

5.73m/sec2

b

8.0m/sec2

c

3.17m/sec2

d

10.0m/sec2

answer is A.

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Detailed Solution

Kinetic friction μkR=0.2mg−Fsin ⁡30∘=0.25×10−40×12=0.2(50−20)=6N Acceleration of the block = Fcos 30∘− Kinetic friction  Mass =40×32−65=5.73m/s2.
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