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Question

A block of mass 20 kg is acted upon by a force F=30N at an angle 530 with the horizontal in downward direction as shown. The coefficient of friction between the block and the horizontal surface is 0.2. The friction force acting on the block by the ground is (g=10m/s2)

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A
40.0 N
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B
30.0 N
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C
18.0 N
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D
44.8 N
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Solution

The correct option is C 18.0 N
Max. frictional force
fmax=μN
=μ(mg+Fsin530)
=0.2(20×10+3045)=44.8N
As Fcosθ=30cos530=30×35=18N<fmax friction force will also be 18 N.

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