A body of mathrm{m} mathrm{kg} slides from re | vedclass

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Question

Apply $W.E.T.$ from $A$ to $B$

$\Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}}$

$\Rightarrow \mathrm{mg} 	ime

Vedclass · Accepted Answer

$21.9 \mathrm{~m} / \mathrm{s}$

Vedclass · Answer

Apply $W.E.T.$ from $A$ to $B$

$\Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}}$

$\Rightarrow \mathrm{mg} 	imes\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{R}ight)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2-0\left\{\mathrm{v}_{\mathrm{A}}=0 	ext { rest }ight\}$

$\Rightarrow \mathrm{mgR} \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2$

$\Rightarrow \sqrt{\mathrm{gR} \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=\mathrm{v}_{\mathrm{B}}$

$\Rightarrow \sqrt{\frac{10 	imes 14 	imes 2(2.4)}{1.4}}=\mathrm{v}_{\mathrm{B}}$

$\Rightarrow 21.9=\mathrm{v}_{\mathrm{B}}$

Hence option $(2)$ is correct

A body of $\mathrm{m} \mathrm{kg}$ slides from rest along the curve of vertical circle from point $A$ to $B$ in friction less path. The velocity of the body at $B$ is:

(given, $\mathrm{R}=14 \mathrm{~m}, \mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^2$ and $\sqrt{2}=1.4$ )

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