## College Physics (4th Edition)

The tension in the rope at the bottom of the swing is $438~N$
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 $438~N$