A block of mass 50kg 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
600√3+0.6
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D
600√3−0.6
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Solution
The correct option is C600√3+0.6
Net normal force acting will be (mg−Fsin30∘)
Limiting friction F=μN=0.6(mg−Fsin30∘)
Horizontal component of the applied force is Fcos30∘
To just start sliding, applied force= friction force, Fcos30∘=0.6(mg−Fsin30∘) F√32=(0.6×50×10)−0.6F2 F(√32+0.62)=300 F=600√3+0.6