A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8m by a cable in which the tension is 11,000 N. The speed vi of the elevator at the beginning of the 8 m descent is most nearly?

Question

Jordan Bartlett

Physics

4 October, 00:47

A descending elevator of mass 1,000 kg is uniformly decelerated to rest over a distance of 8m by a cable in which the tension is 11,000 N. The speed vi of the elevator at the beginning of the 8 m descent is most nearly?

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Answers (1)

Know the Answer?

Aliana Nguyen · Answer 1

Vi = 4.38 m/s

Explanation:

Given

m = 1000 kg;

y = 8m;

T = 11000 N;

∑Fy = m*a

W - T = m * a

(1000kg * 9.8 m/s²) - 11000N = 1000 kg * a

a = - 1.2 m/s²

Using the equation from parabolic motion

Vf ² = Vi² + 2*a*y

Solve to Vi

Vi = √ 2 * 1.2m/s² * 8 m

Vi = √19.2 m²/s²

Vi = 4.38 m/s