Laws of Motion

JEE Main level, numerical answer, medium

159 of 210

Question 159 of 210

A body of mass mm is launched up on a rough inclined plane making an angle of 30∘30^{\circ} with the horizontal. The coefficient of friction between the body and plane is x5\dfrac{\sqrt{x}}{5} if the time of ascent is half of the time of descent. The value of xx is

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Answer: 3

ta=12td{t}_{a}=\dfrac{1}{2}{t}_{d}
2saa=122sad…(i)\sqrt{\dfrac{2s}{{a}_{a}}}=\dfrac{1}{2}\sqrt{\dfrac{2s}{{a}_{d}}}\ldots \left(i\right)
aa=gsin⁡θ+μgcos⁡θ{a}_{a}=g\sin \theta +\mu g\cos \theta
=g2+32μg=\dfrac{g}{2}+\dfrac{\sqrt{3}}{2}\mu g
ad=gsin⁡θ−μgcos⁡θ{a}_{d}=g\sin \theta -\mu g\cos \theta
=g2−32μg=\dfrac{g}{2}-\dfrac{\sqrt{3}}{2}\mu g
Using the above values of aa{a}_{a} and ad{a}_{d} and putting in
equation (i)\left(i\right), we will get, μ=35\mu =\dfrac{\sqrt{3}}{5}

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