Question 161 of 194

A pendulum bob has a speed of $3m{s}^{-1}$ at its lowest position. The pendulum is $50cm$ long. The speed of bob, when the length makes an angle of $60^{\circ}$ to the vertical will be $\left(g=10m{s}^{-2}\right)$ ______ $m{s}^{-1}$.

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Answer: 2

Applying work energy theorem:
${w}_{g}+{w}_{T}=\Delta K$
$-mgl\left(1-\cos 60^{\circ}\right)=\dfrac{1}{2}m{v}^{2}-\dfrac{1}{2}m{u}^{2}$
${v}^{2}={u}^{2}-2gl\left(1-\cos 60^{\circ}\right)$
${v}^{2}=9-2\times 10\times 0.5\left(\dfrac{1}{2}\right)$
${v}^{2}=4$
$v=2m{s}^{-1}$