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Question

A block of mass m=2kg is resting on a rough inclined plane of inclination 30o as shown in figure. The coefficient of friction between the block and the plane is μ=0.5. What minimum force F should be applied perpendicular to the plane on the block, so that block does not slip on the plane.? (g=10m/s2)
1188246_696131540f214158b4c0774fba4fbc7d.png

A
zero
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B
6.24N
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C
2.68N
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D
4.34N
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Solution

The correct option is C 2.68N
mgsinθ=μ(mgsinθ+f)
mg2=12(mg32+F)
mg(132)=F
20(0.13)=2.68N
1050259_1188246_ans_6a12fc5c2919460499381bb05bb37499.png

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