A stationary elevator and its contents have a combined mass of 2000kg. The elevator is suspended by a single cable. If the elevator is decending at 4.0m/s squared, what are the values of all the forces acting at this point?

You posted this twice. Someone, if they want.to, will get to you. Remember, we don't do YOUR homework.

Remember that \sum \vec F = m \vec a. Here the forces acting in the elevator are the tension in the cable (which is positive value because it operates in the positive y direction) and the force of gravity in the negative y direction. You have a value for \vec a (be careful to use the correct sign), now solve for \vec T.