an elevator and its load weight a total of 800 kg. find the tension T in the supporting cable when the elevator, originally moving downward at 20 ms is brought to rest with constant retardation in a distance of 50m.
Answers
Answered by
66
Here,
m = 800 kg
u = 20 ms
v = 0 ms
s = 50 m
g = 9.8 ms-2
u2 = -2as
=> 202 = -2a×50
=> a = - 4 ms-2
Now the acceleration acting upwards by the cable = 4 ms-2
So, the elevator experiences an downward acceleration = 4
Net acceleration downwards = 9.8 + 4 = 13.8 ms-2
Tension in the cable will be 800×13.8 = 11040 N
m = 800 kg
u = 20 ms
v = 0 ms
s = 50 m
g = 9.8 ms-2
u2 = -2as
=> 202 = -2a×50
=> a = - 4 ms-2
Now the acceleration acting upwards by the cable = 4 ms-2
So, the elevator experiences an downward acceleration = 4
Net acceleration downwards = 9.8 + 4 = 13.8 ms-2
Tension in the cable will be 800×13.8 = 11040 N
Answered by
9
Answer:
11040 N
Explanation:
Here,
m = 800 kg
u = 20 ms
v = 0 ms
s = 50 m
g = 9.8 ms-2
u2 = -2as
=> 202 = -2a×50
=> a = - 4 ms-2
Now the acceleration acting upwards by the cable = 4 ms-2
So, the elevator experiences an downward acceleration = 4
Net acceleration downwards = 9.8 + 4 = 13.8 ms-2
Tension in the cable will be 800×13.8 = 11040 N
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