Q.

A block of mass 5 kg is placed on a horizontal surface with coefficient of friction μ=0.2, then the maximum and minimum value of force F respectively for which the block remains at rest (g=10ms2)

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a

15 N, 10 N

b

10 N, 25 N

c

25 N, 5 N

d

5 N, 10 N

answer is C.

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Detailed Solution

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f=μmg=0.2×5×10=10N

Fmin+10=15

Fmin=5N

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Fmax=15+f=25N

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A block of mass 5 kg is placed on a horizontal surface with coefficient of friction μ=0.2, then the maximum and minimum value of force F respectively for which the block remains at rest (g=10ms−2)