Question 61 of 183

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10m$. If it dissipates $10$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is: [Use, $g=10m{s}^{-2}$ ]

A$6\sqrt{5}m{s}^{-1}$B$5\sqrt{6}m{s}^{-1}$C$5\sqrt{5}m{s}^{-1}$D$2\sqrt{5}m{s}^{-1}$

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Answer: (A) $6\sqrt{5}m{s}^{-1}$

Given the length of the pendulum is $\ell =10m$.
The initial potential energy of the pendulum at the horizontal position is given by
$U=mg\ell ...\left(1\right)$
The final kinetic energy of the pendulum at the lowermost position is given by
$K=\dfrac{1}{2}m{v}^{2}...\left(2\right)$
So, from equations (1) and (2), it follows that
$\dfrac{9}{10}mg\ell =\dfrac{1}{2}m{v}^{2} \Rightarrow \dfrac{9}{10}\times 10\times 10=\dfrac{1}{2}{v}^{2} \Rightarrow {v}^{2}=180 \Rightarrow v=\sqrt{180} =6\sqrt{5}m{s}^{-1}$