A 72.0 kg man is standing on a spring scale in an elevator at rest. The elevator ascends, attaining it's max velocity of 1.20 m/s in .800s. It travels with this constant velocity for 5.00s. The elevator then undergoes a uniform acceleration in the negative y direction for 1.5s and comes to rest. What does the spring scale register during the time interval it is slowing down?
There are other parts to this but I have those answers and I used T-mg=ma. I calculated the upward acceleration to be 1.5m/s^2. Other parts of the question asked what the scale read at rest, 705.6N, during the first .800s, 813.6N and at constant speed, 7056N.
The Attempt at a Solution
Gotta say I have no clue how to solve for the last part, what does the spring scale register as it slows. I tried t=72.0kg(-1.5m/s^2)+72.00kg(9.8g) but that's wrong. Can somebody lead me in the right direction.