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Q.

A duster weighs 0.5N.  It is pressed against a vertical board with a horizontal force of 11N. If the co-efficient of friction is 0.5, the minimum force that must be applied on the duster parallel to the board to move it upwards is

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a

0.4N

b

0.7N

c

 7N

d

 6N

answer is C.

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Detailed Solution

Fmin=W+f=W+μN

⇒Fmin=0.5+0.5×11=6 N

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