The bob of a pendulum is released from a horizontal position A as shown in the figure. The length of the pendulum is 2 m. If $10 %$ of the initial energy of the bob is dissipated as heat due to the friction of air, what would be the speed of the bob when it reaches the lowermost point B? Take $g = 10 m s^{- 2}$ .

$3 m s^{- 1}$

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$4 m s^{- 1}$

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$5 m s^{- 1}$

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$6 m s^{- 1}$

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Solution

The correct option is D
$6 m s^{- 1}$

PE at A = mgh. Since 10% of this energy is lost, KE at point B $= m g h \times \frac{90}{100} = 0.9 m g h$ .
Now, $\frac{1}{2} m v^{2} = 0.9 m g h$
$\Rightarrow v^{2} = 1.8 g h = 1.8 \times 10 \times 2 = 36$
$\Rightarrow v = 6 m s^{- 1}$
Hence, the correct choice is (d).

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The correct option is D 6 ms−1 PE at A = mgh. Since 10% of this energy is lost, KE at point B =mgh×90100=0.9 mgh. Now, 12mv2=0.9 mgh ⇒v2=1.8 gh=1.8×10×2=36 ⇒v=6 ms−1 Hence, the correct choice is (d).