A ring of mass m = 0.2kg is attached to one end of a spring of force constant
k=100 N/m and natural length l = 10 cm. The ring is constrained to move on a rough vertical wire in the shape of an ellipse of major axis 24cm and minor axis 16cm with its centre at origin (O). Initially the ring is at top point A with other end of the spring being fixed to the origin. Normal reaction of the wire on the ring is zero at A. The ring is given a horizontal velocity $V = 10 m s^{- 1}$ towards right at A so that it just reaches point B. Find the work done against friction (in J) on the ring as it moves from A to B. $(g = 10 m / s^{2})$

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answer is 10.16.

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Detailed Solution

$W_{f} = (k_{f} + U_{f}) - (k_{i} + U_{i})$
$= 0 + \frac{1}{2} K {(2 c m)}^{2}$
$- [\frac{1}{2} m {(10)}^{2} + \frac{1}{2} K {(2 c m)}^{2} + m g (8 c m)]$
$= - \frac{1}{2} \times 0.2 \times 100 - 0.2 \times 10 \times 0.08$
$= - 10 - 0.16 = - 10.16 J$