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/s2 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/s2 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

There are two forces acting on the elevator:

- The tension in the cable, T, upward

- The weight of the elevator+passenger, downward, which is given by

W = mg

where m=1700 kg is the mass and g=9.81 m/s^2 is the acceleration of gravity

According to Newton's second law, the resultant of these forces must be equal to the product between mass and acceleration:

T - mg = ma

where

a = 1.20 m/s^2 is the acceleration, also upward

Solving the equation for T, we find the tension in the cable:

$T=mg+ma=m(g+a)=(1700 kg)(9.81 m/s^2)(1.20 m/s^2)=18717 N$

(b) 16677 N

In this second part of the trip, the elevator continues at constant velocity. This means that the acceleration is zero:

a = 0

So Newton's second law becomes:

T - mg = ma = 0

Therefore, the tension in the cable will be equal to the weight of the elevator+passenger:

T = mg = (1700 kg)(9.81 m/s^2)=16677 N

(c) 15657 N

In this third part of the trip, the elevator has a deceleration of

a = -0.60 m/s^2

and we use a negative sign since the acceleration is now downward.

Therefore, Newton's second law is

T - mg = ma

And substituting all the data, we find the new tension in the cable:

$T=mg+ma=m(g+a)=(1700 kg)(9.81 m/s^2-0.60 m/s^2)=15657 N$

(d) 19.35 m, 0 m/s

The distance covered by the elevator in part a) of the trip is

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

The final velocity reached in this part is

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

In the second part, the elevator moves at constant velocity of

$v_2 = v_1 = 1.8 m/s$

so the distance covered is

$d_2 = v_2 t = (1.8 m/s)(8.50 s)=15.3 m$

The distance covered in the third part will be

$d_3 = v_2 t + \frac{1}{2}at^2 = (1.8 m/s)(3.0 s) + \frac{1}{2}(-0.6 m/s^2)(3.0 s)^2=2.7 m$

While the final velocity is

$v_3 = v_2 + at = 1.8 m/s + (-0.6 m/s^2)(3.0 s)=0$

and the total distance covered (so, the heigth of the elevator above the ground) is

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

Answer 2

The total distance traveled by elevator during the entire time of acceleration, at constant velocity and deceleration is 4.2618 m.

Given to us

Mass inside the lift = 1700 kg

Accleration of the lift in upward direction = 1.50 m/s²

Time for accleratio of the lift = 1.50 s

A.)  

If we look carefully at the lift, the lift is under three forces which are,

1. The gravitational force, mg which is acting downwards,

2. The force due to the acceleration in the upward direction, ma

3. The tension in the string, T

Taking all the vertical forces,

$\sum F_y = 0\\T = ma + mg\\T = m(a+g)$

Substitute the values,

$\rm T = m(a+g)\\\\T = 1700(1.50+9.81)\\\\T = 19,227\ N\\\\T = 19.227\ kN$

Thus, the tension in the cable supporting the elevator is 19.227 kN.

B.)

As given to us, the elevator continues upward at a constant velocity, for 8.50 seconds. therefore, there will be no acceleration in the lift. And only gravitational force will act on the elevator.

Taking all the vertical forces,

$\sum F_y = 0\\T = ma + mg\\T = m(a+g)$

Substitute the values,

$\rm T = m(a+g)\\\\T = 1700(0+9.81)\\\\T = 16,677\ N\\\\T = 16.677 \ kN$

Thus, the tension in the cable supporting the elevator is 16.677 kN.

C.)

As given to us, the elevator decelerates at a rate of 0.600 m/s2 for 3.00 seconds. therefore, there will be a negative acceleration of -0.6 m/s2 in the lift.

Taking all the vertical forces,

$\sum F_y = 0\\T = ma + mg\\T = m(a+g)$

Substitute the values,

$\rm T = m(a+g)\\\\T = 1700(-0.6+9.81)\\\\T = 15,657\ N\\\\T = 15.657 \ kN$

Thus, the tension in the cable supporting the elevator is 15.657 kN.

D.)

As given to us the acceleration of the elevator at a rate of 1.20 m/s2 for 1.50 s.

We know that the initial velocity of the elevator is 0, because it was at rest, therefore, according to the first equation of motion.

$\rm S = ut + \frac{1}{2}at^2\\\\S = (0)t + \frac{1}{2}(1.2)(1.50^2)\\\\S = 1.35\ m$

We will find the final velocity of the lift,

$\rm v^2 - u^2 = 2as\\\\v^2 - 0 = 2(1.2)(1.35)\\\\v = 1.8\ m/s$

Now, the elevator travels with the same velocity for the next 8.50 seconds, therefore,

$\rm S_1 = \dfrac{1.8}{8.50}\\\\S_1 = 0.2118\ m$

Now, the elevator decelerates, therefore,

$S_2 = ut + \frac{1}{2}at^2$

Since the elevator is moving with 1.8 m/s and is going to rest,

$S_2 = 1.8(3) + \frac{1}{2}(-0.6)(3^2)\\\\S_2 = 2.7\ m$

The total distance traveled by elevator,

$\text{Total distance traveled by elevator}=S + S_1 + S_2\\\\\text{Total distance traveled by elevator}=1.35 + 0.2118+ 2.7\\\\\text{Total distance traveled by elevator}=4.2618\ m$

Hence, the total distance traveled by elevator during the entire time of acceleration, at constant velocity and deceleration is 4.2618 m.

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