Question

An elevator filled with passengers has a mass of 1700 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s^{2} 2 for 1.50 s. Calculate the tension in the cable supporting the elevator. (b) The elevator continues upward at constant velocity for 8.50 s. What is the tension in the cable during this time? (c) The elevator decelerates at a rate of 0.600 m/s^{2} 2 for 3.00 s. What is the tension in the cable during deceleration? (d) How high has the elevator moved above its original starting point, and what is its final velocity?

Answer 1

(a) 18717 N

Newton's second law in this situation can be written as:

$\sum F = T-W = ma$ (1)

where

T is the tension in the cable, pointing upward

W is the weight of the elevator+passengers, pointing downward

m is the mass of the elevator+passengers (1700 kg)

a is the acceleration of the system (1.20 m/s^2, upward)

The weight is equal to the product between the mass, m, and the gravitational acceleration, g:

$W=mg=(1700 kg)(9.81 m/s^2)=16,677 N$

So now we can solve eq.(1) to find T, the tension in the cable:

$T=W+ma=16,677 N +(1700 kg)(1.20 m/s^2)=18,717 N$

(b) 16677 N

In this situation, the elevator is moving with constant velocity: this means that its acceleration is zero,

a = 0

So Newton's second law becomes

$\sum F = T-W = 0$

and so we find

$T=W=16,677 N$

(c) 15657 N

During the deceleration phase, Newton's second law can be written as:

$\sum F = T-W = ma$ (1)

Where the acceleration here points downward (because the elevator is decelerating), as the weight W, so we can write it as a negative number:

a = -0.600 m/s^2

we can solve the equation to find T, the tension in the cable:

$T=W+ma=16,677 N +(1700 kg)(-0.600 m/s^2)=15,657 N$

(d) 19.35 m, 0 m/s

Distance covered during the first part of the motion; we know that

u = 0 is the initial velocity

a = 1.20 m/s^2 is the acceleration

t = 1.50 s is the time

So the distance covered is given by

$d_1=ut + \frac{1}{2}at^2 = (0)(1.50 s)+\frac{1}{2}(1.20 m/s^2)(1.50 s)^2=1.35 m$

and the final velocity after this phase is

$v_1=u+at=0+(1.20 m/s^2)(1.50 s)=1.8 m/s$

During the 2nd part of the motion, the elevator moves at constant speed of 1.8 m/s for t=8.50 s, so the distance covered here is

$d_2 = v_1 t =(1.8 m/s)(8.50 s)=15.3 m$

Finally, in the third part the elevator decelerates at a = -0.600 m/s^2 for t = 3.00 s. So, the distance covered here is

$d_3 = v_1 t + \frac{1}{2}at^2=(1.8 m/s)(3.00 s) + \frac{1}{2}(-0.600 m/s^2)(3.00 s)^2=2.7 m$

and the final velocity is

$v_3 = v_1 +at = 1.8 m/s +(-0.600 m/s^2)(3.00 s)=0$

and the total distance covered is

$d=d_1 +d_2+d_3=1.35 m+15.30 m+2.70 m=19.35 m$