A 1500kg elevator ascends 25 m in 6 s. The motion consists of a steady acceleration from rest followed by a period of a constant velocity of 5 m/s and finally a constant deceleration to rest. If the tension in the supporting cable during acceleration is 25.2 kN, what is the tension during deceleration?

2. Originally Posted by bobby77
A 1500kg elevator ascends 25 m in 6 s. The motion consists of a steady acceleration from rest followed by a period of a constant velocity of 5 m/s and finally a constant deceleration to rest. If the tension in the supporting cable during acceleration is 25.2 kN, what is the tension during deceleration?
I presume the magnitudes of the acceleration and deceleration are the same?

Consider a free body diagram of the elevator as it accelerates downward. I will consider the positive direction to be upward. (I usually choose it in the direction of the acceleration, but since we've got accelerations in different directions I'm not going to be fussy about it.) There is a tension T acting upward and a weight w = mg downward. The acceleration of the elevator is downward. Thus, Netwon's 2nd says:
$\sum F = T - mg = m(-a)$ (Note the - sign on the acceleration!)
We are going to need a. Solving the equation gives:
$a = g - \frac{T}{m}$.

Now we do the elevator decelerating, which means it's accelerating in the direction opposite to the velocity, which means the acceleration is now upward. The free body diagram looks the same (except now I'm calling the tension T') and I'm keeping the upward direction positive. Newton's 2nd now looks like:
$\sum F = T' - mg = ma$ (Now a is upward so it's positive.)
Solving for T' gives:
$T' = ma + mg = m \left ( g- \frac{T}{m} \right ) + mg$
$T' = 2mg - T$

I'll let you plug the numbers in.

-Dan