Question

A rod of length 1m and mass 16kg hinged at one end, is initially hanging vertical. The other end is now raised slowly until it makes an angle of ${30}^o$ with the vertical. The required work is: (use g = 10 $m/s^2$)

Answer 1

Weight, W=$mg$=16$\times$10=160N

Work required =$\frac{mg}{2}\cos \theta$

$\theta ={30}^o;\cos \theta =0.866$

$\therefore W=\frac{160}{2}\times 0.866\approx 69.28J$