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Q.

In the diagram shown coefficient of static friction (μs) and kinetic friction (μk) between the block and the surface are 0.5 and 0.4 respectively. Then the force of friction acting on the block at t = 5 s is 

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a

20 N

b

16 N

c

12 N

d

10 N

answer is A.

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Detailed Solution

(fmax⁡)S=μS.mg=0.5×4×10N=20N

At  t=5s, F=5×5 N=25N

since F>(Fmax⁡)S, the block will slide on the surface.

∴ Frictional force acting on the block =μKmg=0.4×4×10N=16N

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