1.

A bead of mass kg starts from rest from"A" to move in a vertical plane alongasmooth fixed quarter ring of radius 5m, underthe action of a constant horizontal force F-5N as shown. The speed of bead as it reachespoint "B" isR-5m1) 14.14 m/s 2) 7.07 m/s 3)5 m/s 4) 25 m/sTDYATION OF

Answer»

Work done by the force = force * displacement horizontally W = F * Radius = 5 N * 5 m = 25 J

K E = Work done by Force 5 N + change in PE, due to conservation of energy

1/2 m v² = 25 N + 1/2 kg * 10 m /s² * 5 m = 50 N

v² = 200 => v = 10 √2 m/sec. = 14.14 m/s



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