An 888.0 kg elevator is moving downward with a velocity of 0.800 m/s. It decelerates uniformly and comes to a stop in a distance of 0.2667 m. What is the tension on the cable while it decelerates

The value of tension on the cable T = 1065.6 N

Explanation:

Mass = 888 kg

Initial velocity ( u )= 0.8

Final velocity ( V ) = 0

Distance traveled before come to rest = 0.2667 m

Now use third law of motion = - 2 a s

Put all the values in above formula we get,

⇒ 0 = - 2 × a ×0.2667

⇒ a = 1.2

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

T = 1065.6 N

This is the value of tension on the cable.

the answer would be   dignified - apex

not too sure thinking all accelerations are pointed down