A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8 m by a cable in which the tension i
1 answer:
The speed of the elevator at the beginning of the 8 m descent is nearly 4 m/s. Hence, option A is the correct answer.
We are given that-
the mass of the elevator (m) = 1000 kg ;
the distance the elevator decelerated to be y = 8m ;
the tension is T = 11000 N;
let us determine the acceleration 'a' by using Newton's second law of motion.
∑Fy = ma
W - T = ma
(1000kg x 9.8 m/s² ) - 11000N = 1000 kg x a
9800 - 11000 = 1000
a = - 1.2 m/s²
Using the equation of kinematics to determine the initial velocity.
² =
² + 2ay
= √ ( 2 x 1.2m/s² x 8 m )
= √19.2 m²/s²
= 4.38 m/s ≈ 4 m/s
Hence, the initial velocity of the elevator is 4m/s.
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