Answer
The tension in the rope at the bottom of the swing is $438~N$
Work Step by Step
We can find the tension at the bottom of the swing:
$\sum F = \frac{mv^2}{r}$
$T-mg = \frac{mv^2}{r}$
$T = mg + \frac{mv^2}{r}$
$T = (35.0~kg)(9.80~m/s^2) + \frac{(35.0~kg)(4.20~m/s)^2}{6.50~m}$
$T = 438~N$
The tension in the rope at the bottom of the swing is $438~N$
