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Byju's Answer

Standard X

Physics

Potential Energy of an Object at a Height

A long rod AB...

Question

A long rod ABC of mass $m$ and length $L$ has two particles of masses $m$ and $2 m$ attached to it as shown in the figure. The system is initially in the horizontal position. The work to be done to keep it vertical with A at the bottom is ( $g =$ acceleration due to gravity)

2 m g L

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\frac{3 m g L}{2}

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\frac{5 m g L}{2}

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3 m g L

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Solution

The correct option is D $3 m g L$
Initially, the rod lies on the ground, thus initial potential energy of the system is zero i.e. $U_{i} = 0$
Mass $m$ of the rod lies at its geometric center B, thus total mass at point B is $2 m$ .
Potential energy of the system finally $U_{f} = (2 m) g (A B) + (2 m) g (A C)$

Or $U_{f} = (2 m) g (\frac{L}{2}) + (2 m) g (L) = 3 m g L$

Work done $W = U_{f} - U_{i}$

$⟹$ $W = 3 m g L - 0 = 3 m g L$

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