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Question

A small block is freely sliding down from the top of a rough inclined plane whose angle of inclination is 450. The block reaches bottom then it completes a smooth vertical circle. If the coefficient of friction is 0.5, the ratio of minimum vertical height of inclined plane to radius of vertical circle is:

A
3:1
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B
5:1
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C
5:2
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D
7:3
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Solution

The correct option is B 5:1
μ=0.5.
Wo by friction=μNa
=(0.5)(mg2)(2H)
=mgH2
minimum velocity required for completely velocity lively=5gR work energy therefore on work
mgH=mV2o2+mgH2.
mgH2=m2(5gR)2
gH=5gR.
HR=51


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