An elevator weighs 4000 kg. When the upward tension in the supporting cable is 48000 N, then what is the upward acceleration? Starting from rest,how far does it rise in 3 seconds?
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9
weight of elevator = mg = 40000 N
Tension = 48000 N
net force = 8000 N
acceleration = 8000/4000 m/s2 = 2m/s2
H in 3s = ut + 1/2at2 = 0 + 1/2 * 2 * 9 = 9 m
Tension = 48000 N
net force = 8000 N
acceleration = 8000/4000 m/s2 = 2m/s2
H in 3s = ut + 1/2at2 = 0 + 1/2 * 2 * 9 = 9 m
Answered by
1
The vertically upward force on the elevator is the tension T in the cable.
T=50,000 N.(given).
The downward force on the elevator is weight of the elevator ,W=mg=5,000 x9.8=49,000 N.
The resultant upward force on the elevator is
ma=T-W=50,000–49,000=1000N
The acceleration of the elevator in upward direction is a=(1000)/m=(1000)/(5000) =(1/5) m/s^2.
Starting from rest, the distance travelled in time t(=10s) is
s=(1/2)at^2=(1/2)(1/5)(100)=10m.
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