Question

One meter stick having mass $600gm$, pivoted at one end. And is displaced through an angle of ${60}^{\circ }$, the increase in its potential energy is $(g=10m/s^2)$

• A. $1.5J$
• B. $15J$
• C. $150J$
• D. $0.15J$

Answer 1

The correct option is A $1.5J$

The center of Mass will be at the center (mid-way) of the meter stick.

$△u=u_f-u_i$

$=(-m.g.h\cos {60}^o)-(-mgh)$

$mgh-mgh.\cos {60}^o$

$=mg(h-\frac{h}{2})=mg\frac{h}{2}$

$\Rightarrow △u=0.6\times 10\times \frac{1}{4}$

$\Rightarrow △u=+1.5J$

$A$