A block starts moving up an inclined plane of inclination 300 with an initial velocity of v0. It comes back to its initial position with velocity v02. The value of the coefficient of kinetic friction between the block and the inclined plane is close to I1000 . The nearest integer to I is ___ .

Work done by all fores =ΔK.EIn upward journey, -mgh-μmg(h)3=-12mv02In downward journey,mgh-μmg(3h)=12 mv022 Dividing equations: 1+3μ1-3μ=4⇒μ=35 ⇒μ≃3461000

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A block starts moving up an inclined plane of inclination $30^{0}$ with an initial velocity of $v_{0}$ . It comes back to its initial position with velocity $\frac{v_{0}}{2}$ . The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{I}{1000}$ . The nearest integer to I is ___ .

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answer is 346.

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Detailed Solution

$\begin{array}{l} Work done by all fores = Δ K . E \\ I n u p w a r d j o u r n e y, \\ - mgh - μ mg (h) \sqrt{3} = - \frac{1}{2} {mv}_{0}^{2} \\ I n d o w n w a r d j o u r n e y, \\ mgh - μ mg (\sqrt{3} h) = \frac{1}{2} m {(\frac{v_{0}}{2})}^{2} \\ Dividing equations: \frac{1 + \sqrt{3} μ}{1 - \sqrt{3} μ} = 4 \\ \Rightarrow μ = \frac{\sqrt{3}}{5} \\ \Rightarrow μ ≃ \frac{346}{1000} \end{array}$