Questions

A bead of mass 1/2 kg starts from rest from A to move in a vertical plane along a smooth fixed quarter ring of radius 5 m, under the action of a constant horizontal force F=5N as shown in figure. The speed of bead as it reaches point B is

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a

14.14ms−1

b

7.07ms−1

c

5ms−1

d

25ms−1

detailed solution

Correct option is A

Applying the work-energy theorem, we get 12×mv2−0=F×R+mg×R12×12×v2=5×5+12×10×5=50v=200=14.14ms−1

Similar Questions

A block is attached to a horizontal spring of stiffness k. The other end of the spring is attached to a fixed wall. The entire system lie on a horizontal surface and the spring is in natural state. The natural length of the spring is ${\mathrm{\ell}}_{0}$If the block is slowly lifted up vertically to a height $\frac{5}{12}{\mathrm{\ell}}_{0}$ from its initial position, which of the following is not correct:

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