Question

A pendulum bob has a speed of $3\text{m/s}$ at its lowest position. The pendulum is $50\text{cm}$ long. The speed of the bob, when the length makes an angle of ${60}^{\circ }$ with the vertical, will be $(g=10{\text{m/s}}^2)$

Answer 1

Applying work energy theorem:
$W_g+W_T=\Delta K$
$mgl(1-\cos {60}^{\circ })=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$
$v^2=u^2-2gl(1-\cos {60}^{\circ })$
$=9-2\times 10\times 0.5\left(\frac{1}{2}\right)=4$
$\therefore v=2\text{m/s}$