Answer
v = 16.8 m/s
Work Step by Step
We can find the tension $T$ in the string.
$T~cos(30.0^{\circ}) = mg$
$T = \frac{mg}{cos(30.0^{\circ})}$
The horizontal component of the tension provides the centripetal force for the lunch box to round the curve.
$T~sin(30.0^{\circ}) = \frac{mv^2}{r}$
$\frac{mg}{cos(30.0^{\circ})}~sin(30.0^{\circ}) = \frac{mv^2}{r}$
$v^2 = gr~tan(30.0^{\circ})$
$v = \sqrt{gr~tan(30.0^{\circ})}$
$v = \sqrt{(9.80~m/s^2)(50.0~m)~tan(30.0^{\circ})}$
$v = 16.8~m/s$
