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A bead of mass `1/2 kg` starts from rest from A to move in a vertical place along a smooth fixed quarter ring of radius `5 m`, under the action of a constant horizontal force `f=5 N` as shown. The speed of bead as it reaches the point (B) is [Take `g=10 ms^(-2)`] A. (a) `14.14ms^-1`B. (b) `7.07ms^-1`C. (c) `5ms^-1`D. (d) `25ms^-1` |
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Answer» Correct Answer - A Applying the work-energy theorem, we get `1/2xxmv^2-0=FxxR+mgxxR` `1/2xx1/2xxv^2=5xx5+1/2xx10xx5=50` `v=sqrt(200)=14.14ms^-1` |
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