An elevator cab and its load have a combined mass of 1600 kg. Find the tension in the supporting cable when the cab, originally moving downward at 12 m/s, is brought to rest with constant acceleration in a distance of 42m.(adsbygoogle = window.adsbygoogle || []).push({});

Taking downwards as positive

My solution:

F net = mg - T = ma

T = m(g - a) (1)

V^2 - Vi^2 = 2a (y - yi)

a = (-12)^2 / 2(42) = -1.7 m/s^2

(1) -----> T = m (g - a) = 1600 (9.8 + 1.7) = 18400 N

Did I do it right ? cuz my friend got the different answer from me, he basically got T = m (g + a).......

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# Homework Help: Force and motion in Newton's laws for elevator

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