Consider two incline rough surfaces A and B with angles of inclination 45° and 60°, respectively. Height of the top edge of each surface from ground is same. A particle placed on top of the incline 'A' reaches the bottom with a speed v_A. When the same particle is placed on top of the incline B, it reaches the bottom with speed v_B. Assuming the same coefficient of friction in the two cases :

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v_A=v_B

v_A > v_B

it cannot be decided

v_A = v_B

answer is A.

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Detailed Solution

$\begin{array}{l} W = ΔKE \Rightarrow \frac{1}{2} {mv}_{A}^{2} = mgh - (μmgcos 45^{\circ}) \frac{h}{\sin 45^{\circ}} \\ = mgh - μmghcot 45^{\circ} = mgh - μmgh \\ and \frac{1}{2} {inv}_{B}^{2} = mgh - μmgcot 60^{\circ} = mgh - \frac{μmgh}{\sqrt{3}} \end{array}$
Therefore v_B > v_A

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