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Question

# A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.50. If g=10ms−2, then the magnitude of a force acting upwards at an angle of 60o from the horizontal that will just start the block moving is:

A
5 N
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B
5.36 N
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C
74.6 N
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D
10 N
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Solution

## The correct option is D 74.6 NEquating all the forces. In vertical direction, N=mg−Fsin60o=1×10−F√32……(1) In horizontal direction, Fcos60o≥μN=0.5×(10−F√32) ⇒F2≥5−F√34 ⇒F≥74.64N Hence, minimum required force is 74.6N

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