Question

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 is 11,000 N. The speed v_1 of the elevator at the beginning of the 8 m descent is most nearly (A)4 m/s (B) 10 m/s (C) 13 m/s (D) 16 m/s E) 21 m/s

Answer 1

The speed  $V_{i}$ 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.

$V_{f}$ ² = $V_{i}$² + 2ay

$V_{i}$ = √ ( 2 x 1.2m/s² x 8 m )

$V_{i}$ = √19.2 m²/s²

$V_{i}$ = 4.38 m/s   ≈ 4 m/s

Hence, the initial velocity of the elevator is 4m/s.

Read more about the Equation of kinematics:

brainly.com/question/12351668

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