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Question

A block of mass 50 kg can slide on a rough horizontal surface. The coefficient of friction between the block and the surface is 0.6. The least force of pull acting at an angle of 30 from the horizontal which causes the block to just slide is

A
600
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B
3+0.6600
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C
6003+0.6
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D
60030.6
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Solution

The correct option is C 6003+0.6

Net normal force acting will be (mgFsin30)
Limiting friction F=μN=0.6(mgFsin30)
Horizontal component of the applied force is Fcos30
To just start sliding, applied force= friction force,
Fcos30=0.6(mgFsin30)
F32=(0.6×50×10)0.6F2
F(32+0.62)=300
F=6003+0.6

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