# UCM problem

## Homework Statement

A 1.2 m long pendulum reaches a speed of 4.0 m/s at the bottom of its swing.
http://members.shaw.ca/barry-barclay/Self-Tests/test07/q02.gif
What is the tension in the string at this position?

Fnet=Mv^2/r

## The Attempt at a Solution

Im my FBD I had FT acting up, and fg acting down.

Ft-Fg=mv^2/r
Ft-mg=mv^2/r
Ft -(9.8)(3)=(3)(4.0)^2/(1.2)
Ft=40 +(9.8)(3.0)

EDIT Realized my mistake less than a second after posting this. :) Mods can delete.