An elevator and its load have a total mass of 800kg .find the tension T in the supporting cable when the elevator originally moving downwards at10m/s is brought to rest with constant acceleretion in distance of25m

Answer 1

25 = (10/2) t
t = 5 seconds to stop

a = 10/5 = 2 m/s^2 up

F = m a

T - m g = m * 2
g = 9.81 so
T = 800 (11.81) Newtons

Answer 2

WTF YOU TWAT!! DAMON YOU ARE FREAKIN WRONG. I FEEL BAD FOR THE PPL YOU HELP. First of all in this case the elevator is accelerating downward, hence the tension in the rope will decrease and not increase. PURE LOGIC what do you feel if you pull someone up by rope? More tension... If you slowly let go of the rope neglect friction in your hand it is easier = less tension. So based upon that logic we can find the answer using NEWTONS 2nd LAW. We also have to use a kinematic equation to find the acceleration.
vf^2 = vi^2 +2a(x) we get an acceleration of 2 m/s^2 in the downward direction. Hence considering if the elevator was in equilibrium the tension of the rope would be 8000N. since it is acceleration downward we have to subtract m(a). So holistically the equation can be derived from logic.
MG - m(a). Now fundamentally mg exemplifies the elevator's weight in equilibrium position which also means tension rope if the elevator was not in motion, so the tension in rope would be 8000N - 800(2) = 6400N. LOL your math is flawed wtf is this ----->T-MG = M*2

Answer 3

Is there a moderator who can get Kenneth banned.
We don't need this.
Not to mention
T - mg = ma is correct. Look at the directions. T and a are up, g is down.

Answer 4

I remember my first time harming a human being. The pure rush of oxytocin was overwhelming.