Question

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?

Answer 1

Answer:

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

Answer 2

The initial speed of the elevator at the beginning of the 8 m descent is most nearly is 4.38 m/sec and this can be determined by using the equation of kinematics.

Given :

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.

First, determine the acceleration by using Newton's second law motion.

$\rm \sum F = ma$

W - T = ma

where W is the weight, T is the tension, a is the acceleration, and m is the mass.

Now, substitute all the known terms in the above formula.

$\rm 1000\times 9.8 -11000=1000\times a$

Simplify the above expression.

9800 - 11000 = 1000a

$\rm a = -1.2 m/sec^2$

Now, use the equation of kinematics in order to determine the initial velocity.

$\rm v^2=u^2-2as$

Now, substitute all the known terms in the above formula.

$\rm u = \sqrt{2\times 1.2\times 8}$

u = 4.38 m/sec

For more information, refer to the link given below:

https://brainly.com/question/862972