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A body of mass 1.5 kg is allowed to slide down along a quadrant of a circle from the horizontal position. In reaching to the bottom, Its velocity is 8m/s. The work done in overcoming the friction is 12J. The radius of circle is $(g = 10 {ms}^{- 2})$

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Detailed Solution

$mgh = w_{f} + \frac{1}{2} {mu}^{2}$
$1.5 \times 10 \times R = 12 + \frac{1}{2} \times 1.5 \times 64$
$15 R = 12 + 48 \Rightarrow R = 4 m$

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