Question

A metre stick weighing $600g$, is displaced through an angle of ${60}^{\circ }$ in vertical plane as shown. The increase in its potential energy is $(g=10m{s}^{-2})$.

• A. $1.5J$
• B. $15J$
• C. $30J$
• D. $45J$

Answer 1

The correct option is A $1.5J$

Given,

$m=500gm=0.6kg$

$\theta ={60}^0$

Change in potential energy at the center of the meter stick,

$\Delta U=U_f-U_i$

$\Delta U=(-mghcos\theta )-(-mgh)=mgh-mghcos{60}^0$

$\Delta U=mgh(1-\frac{1}{2})=\frac{mgh}{2}$

$\Delta U=\frac{0.6\times 10\times \frac{1}{2}}{2}$

$\Delta U=1.5J$

The correct option is A.