First slide
Conservation of mechanical energy
Question

A bead of mass 12 kg starts from rest from A to move in vertical plane along a smooth fixed quarter ring of radius 5m , under the action of a constant horizontal force F = 5N as 
shown.The speed of bead as it reaches the point B is [Take g = 10ms-2]

 

 

Moderate
Solution

 

WAll=12mv2, work done due to force is WF=F.S, work done due to weight Wmg=mgh, work done due to normal force=WN=Normal force is perpendicular to displacementWF+Wmg+Wn=12mv2(5×5)+12×10×5+0=12×12×v2v=14.14m/s

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