The mass of a simple pendulum bob is $100 g m$ . The length of the pendulum is $1 m$ . The bob is drawn aside from the equilibrium position so that the string makes an angle of $60^{o}$ with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

1 J

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0.49 J

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0.94 J

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1.2 J

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Solution

The correct option is B $0.49 J$
Kinetic energy = Change in potential energy = $m g l - l c o s 60^{0}$ = $(0.1 k g)) (9.8 m s^{- 2}) (1 m) - (1 m) (0.5) = 0.49 J$

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The mass of a simple pendulum bob is 100gm. The length of the pendulum is 1m. The bob is drawn aside from the equilibrium position so that the string makes an angle of 60o with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

The correct option is B 0.49JKinetic energy = Change in potential energy = mgl−lcos600=(0.1kg))(9.8ms−2)(1m)−(1m)(0.5)=0.49J

The mass of a simple pendulum bob is $100 g m$ . The length of the pendulum is $1 m$ . The bob is drawn aside from the equilibrium position so that the string makes an angle of $60^{o}$ with the vertical and let go. The kinetic energy of the bob while crossing its equilibrium position will be:

The correct option is B $0.49 J$
Kinetic energy = Change in potential energy = $m g l - l c o s 60^{0}$ = $(0.1 k g)) (9.8 m s^{- 2}) (1 m) - (1 m) (0.5) = 0.49 J$