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Question

A block of mass m=2 kg is resting on a rough inclined plane of inclination 30 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 on the block as shown in figure, so that the block does not slip on the plane?( g=10 ms2)

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

The correct option is C 2.68 N
Since, mg sin 30>μmg cos 30
The block has a tnedency to slip downwards, Let F be the minimum force applied on it, so that it does not slip.
Then,
N=F+mg cos 30
mg sin 30=μN=μ(F+mg cos 30
F=mg sin 30μmg cos 30
F=(2)(10)(12)0.5(2)(10)(32)
F=2017.32
F=2.68 N

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