(C) The potential energy lost by the bob as it moves from the $60^\circ$ position to the lowest point is converted into kinetic energy.Using the princ

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(C) The potential energy lost by the bob as it moves from the $60^\circ$ position to the lowest point is converted into kinetic energy.Using the principle of conservation of energy: $mgh = \frac{1}{2}mv^2$,where $h$ is the vertical height dropped.The vertical height $h$ is given by $h = l(1 - \cos \theta)$,where $l = 2\, m$ and $\theta = 60^\circ$.$h = 2(1 - \cos 60^\circ) = 2(1 - 0.5) = 1\, m$.Equating energy: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1} = \sqrt{19.6} \approx 4.43\, m/s$.

Class 11 Physics Work, Energy, Power and Collision Work Energy Theorem and Conservation of Mechanical Energy

Medium

$A$ pendulum bob on a $2\, m$ string is displaced $60^\circ$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path? (Take $g = 9.8\, m/s^2$)

A
$\sqrt{2}$
B
$\sqrt{9.8}$
C
$4.43$
D
$1/\sqrt{2}$

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Work, Energy, Power and Collision

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Work Energy Theorem and Conservation of Mechanical Energy

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$A$ pendulum bob on a $2\, m$ string is displaced $60^\circ$ from the vertical and then released. What is the speed of the bob as it passes through the lowest point in its path? (Take $g = 9.8\, m/s^2$)

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$A$ body of mass $5 \,kg$ is thrown vertically up with a kinetic energy of $490 \,J$. The height at which the kinetic energy of the body becomes half of the original value is (acceleration due to gravity $=9.8 \,ms^{-2}$) (in $\,m$)

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