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Question

A block of mass 15 kg is resting on a rough inclined plane as shown in figure. The block is tied up by a horizontal string which has a tension of 50 N. The coefficient of friction between the surfaces of contacts is (g=10 m/s2)

A
12
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B
23
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C
34
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D
14
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Solution

The correct option is A 12

T=50 N
Component of force (in y direction)
N=T2+1502N=200/2
f=μ2002 (As we know, f = μN)
Component of force in x direction,
1502=T2+f,
1502=502+μ×2002μ=12
Why this question?
Tip: For questions based on motion along inclined plane, it is better to define the axises along the incline and perpendicular to the incline.

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