A metre stick, of mass $400 g$ , is pivoted at one end displaced through an angle $60^{o}$ . The increase in its potential energy is:

1 J

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10 J

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100 J

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1000 J

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Solution

The correct option is A $1 J$
$P E = \frac{m g l}{2} - \frac{m g l}{2} (1 - cos θ)$

$= \frac{0.4 \times 10 \times 1}{2} - \frac{0.4 \times 10 \times 1}{2} (1 - cos 60)$

$= 2 - 1 = 1 J$

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Similar questions

A metre stick , of mass 400 g, is pivoted at one end displaced through an angle $60^{\circ}$ . The increase in its potential energy is

60^{0}

600 g m

60^{\circ}

(g = 10 m / s^{2})

600 g

60^{\circ}

(g = 10 m s^{- 2})

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