A pendulum of mass $m kg$ has velocity $v$ at point $A$ . What is the change in potential energy when the pendulum reaches point $B$ as shown in figure ?

\frac{m g l}{2}

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\frac{2 m g l}{5}

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\frac{m g l}{5}

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\frac{4 m g l}{5}

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Solution

The correct option is C $\frac{m g l}{5}$
Taking position $A$ as reference point.
Therefore potential energy at point $A$ is zero.
$\Rightarrow U_{A} = 0$
Potential energy at point $B$ is $U_{B} = m g x$
and $x = l - l cos θ = l (1 - cos θ)$
Therefore,
$U_{B} = m g l (1 - cos θ)$

Change in potential energy between $A$ and $B$ is :
$Δ U = U_{B} - U_{A} = m g l (1 - cos θ) - 0$
$Δ U = m g l (1 - cos θ)$
$θ = 37^{\circ}$ $\Rightarrow Δ U = \frac{m g l}{5}$

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