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5.Work, Energy, Power and Collision
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The mass of the $bob$ of a simple pendulum of length $L$ is $m$. If the $bob$ is left from its horizontal position then the speed of the $bob$ and tension in the thread in the lowest position of the $bob$ will be respectively
A
$\sqrt {2gL} $ and $3\,mg$
B
$3\,mg$ and $\sqrt {2gL} $
C
$2\,mg$ and $\sqrt {2gL} $
D
${2gL}$ and $3\, mg$
Solution
By the conservation of energy
Potential energy at point $\mathrm{A}=$ Kinetic energy at point $\mathrm{B}$
$m g l=\frac{1}{2} m v^{2} \Rightarrow v=\sqrt{2 g l}$
and tension $=m g+\frac{m v^{2}}{l}$
$\Rightarrow T=m g+\frac{m}{l}(2 g l)$
$\Rightarrow T=3 m g$
Standard 11
Physics
