- #1
theunloved
- 43
- 1
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.
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 ? because my friend got the different answer from me, he basically got T = m (g + a)...
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 ? because my friend got the different answer from me, he basically got T = m (g + a)...