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5.Work, Energy, Power and Collision
medium
A bead of mass $\frac {1}{2}\,kg$ 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 $F = 5\,N$ as shown. The speed of bead as it reaches the point $B$ is ................ $\mathrm{m}/ \mathrm{s}$ [Take $g = 10\,m/s^2$ ]
A
$14.14$
B
$7.07$
C
$5$
D
$25$
Solution
From work$-$energy theorem
$\mathrm{W}_{\mathrm{F}}+\mathrm{W}_{\mathrm{mg}}+\mathrm{W}_{\mathrm{N}}=\frac{1}{2} \mathrm{mv}^{2}$
$\mathrm{F} \cdot \mathrm{R}+\mathrm{mgR}+0=\frac{1}{2} \mathrm{mv}^{2}$
$5 \times 5+\frac{1}{2} \times 10 \times 5=\frac{1}{2} \times \frac{1}{2} \times v^{2}$
$\mathrm{v}=\sqrt{200}=14.14 \mathrm{m} / \mathrm{s}$
Standard 11
Physics
