First slide

Concept of potential energy

Question

Two ends A and B of a smooth chain of mass ‘m’ and length ‘l’ are situated as shown in figure. If an external agent pulls A till it comes to same level of B, the change in potential energy is $\frac{m g l}{x} J$ . Where x is _______

Moderate

Solution

$Mass of raised part = \frac{m}{6}$
$\begin{array}{l} = Δ h = \frac{l}{3} - \frac{l}{6} = \frac{l}{6} \\ Δ v = Δmg h \\ = \frac{m}{6} \times g \times \frac{l}{6} = \frac{m g l}{36} \end{array}$

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