An elevator filled with passengers has a mass of 1.70\times 10^3~\text{kg}1.70×10 3 kg. after starting to move up the elevator accelerates at a rate of 0.600~\text{m/s}^20.600 m/s 2 for 3.00 s. what is the tension in the cable during deceleration?

15.64 KN

Explanation:

Given

Mass of the elevator = 1.70 × 10³ kg = 1700 kg

acceleration of the elevator = - 0.6 m/s²

Tension in the cable = ?

The free body diagram is given in the attached image

The net force acting on a body accelerates the body in the same direction as that in which the resultant is applied.

Force balance,

ma = T - mg

1700 (-0.6) = T - (1700×9.8)

T = 16660 - 1020 = 15640 N = 15.64 KN

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.