# Power To Accelerate 1000kg Elevator Cab Upward

• Edel Crine
In summary: I don't know?Oh, wait it should be 22155 Watt not joule.In summary, the cab must deliver energy at a rate of 11077.5 watts in order to travel at a constant speed.
Edel Crine
Homework Statement
A motor must lift a 1000-kg elevator cab. The cab’s maximum occupant capacity is 400 kg, and its constant “cruising” speed is 1.5 m>s. The design criterion is that the cab must achieve this speed within 2.0 s at constant acceleration, beginning from rest.
(a) When the cab is carrying its maximum capacity, at what rate must the motor deliver energy to get the cab up to cruising speed?
(b) At what constant rate must the motor supply energy as the fully loaded cab rises after attaining cruising speed?
Relevant Equations
a=Δv/t
F=ma
w=Δk=F*Δx
(a) When the cab is carrying its maximum capacity, at what rate must the motor deliver energy to get the cab up to cruising speed?

My shot was finding the acceleration,
a=1.5/2 = 0.75m/s2
Displacement of elevator
Vf2 = vi2 + 2aΔx
Δx = 1.5m
Force of elevator (Since it accelerates upward, the weight of everything will be heavier than before.
w = (1400kg)(9.8+0.75 m/s2)(1.5m) = 22155 J.
Then, should I divide this work done by the time in order to get a rate of energy delivered to the elevator?

And I am not sure for part b...

I appreciate for every help from all of you!

Last edited by a moderator:
If it does, I got 11077.5 J/s = watt as my rate!

for b, isn't the force will be 0 or should I use 9.8m/s2 as acceleration?

Edel Crine said:
If it does, I got 11077.5 J/s = watt as my rate!

for b, isn't the force will be 0 or should I use 9.8m/s2 as acceleration?
Oh, since constant speed means net force exerts on the elevator is zero, the force at this moment should be same with the gravitational force exerts on the elevator (and its contents)?

Edel Crine said:
Oh, since constant speed means net force exerts on the elevator is zero, the force at this moment should be same with the gravitational force exerts on the elevator (and its contents)?
Try doing b) first. Then go back to a). For a) is the power requirement constant?

PeroK said:
Try doing b) first. Then go back to a). For a) is the power requirement constant?
I did b) like this:
F = (1400kg)(9.8m/s2) = 13720 N
P=Fv = (13720N)(1.5m/s) = 20580 Watt

Edel Crine said:
I did b) like this:
F = (1400kg)(9.8m/s2) = 13720 N
P=Fv = (13720N)(1.5m/s) = 20580 Watt
Okay, that's looks right.

PeroK said:
Okay, that's looks right.
Then, for a, I need to divide the work by time, so would it be 11077.5 Watt??

Edel Crine said:
Then, for a, I need to divide the work by time, so would it be 11077.5 Watt??
That's less than the cruising power!

PeroK said:
That's less than the cruising power!
Oh, it should be bigger... but I think that I might need more help,,,,,,

Edel Crine said:
Oh, it should be bigger... but I think that I might need more help,,,,,,

PeroK said:
For a) is the power requirement constant?

PeroK said:
Gravitational acceleration and mass are constant...?

Edel Crine said:
Gravitational acceleration and mass are constant...?
They are. But what about power?

PeroK said:
They are. But what about power?
It would not be constant for single time?

Edel Crine said:
It would not be constant for single time?

If in doubt, calculate! You used this equation:

Edel Crine said:
I did b) like this:
P=Fv = (13720N)(1.5m/s) = 20580 Watt

Edel Crine said:
My shot was finding the acceleration,
a=1.5/2 = 0.75m/s2
Displacement of elevator
Vf2 = vi2 + 2aΔx
Δx = 1.5m

Excellent approach, and so far so good. Now what?

Edel Crine said:
Force of elevator (Since it accelerates upward, the weight of everything will be heavier than before.
w = (1400kg)(9.8+0.75 m/s2)(1.5m) = 22155 J.
Then,

I’m not sure I would say everything is heavier. Things inside the elevator like the riders appear to feel heavier because they are in an accelerated reference frame, but from outside there is no confusion. We see the acceleration and understand that they are not heavier. In any case your equation is correct. a is constant. ma is force. Force times distance is work. Well done.

Edel Crine said:
Then, should I divide this work done by the time in order to get a rate of energy delivered to the elevator?

Yes.

And for part B it is still force times distance. What is the force? What is the distance moved in 1 second?

Edel Crine said:
Then, should I divide this work done by the time in order to get a rate of energy delivered to the elevator?
Cutter Ketch said:
Yes.
Wrong. @PeroK is leading you down the correct path.

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PeroK said:
If in doubt, calculate! You used this equation:
Then would it be:
a) P = (1400kg)(9.8m/s2 + 0.75m/s2)(1.5m/s) = 22155 J...?

PeroK
Edel Crine said:
Then would it be:
a) P = (1400kg)(9.8m/s2 + 0.75m/s2)(1.5m/s) = 22155 J...?

Yes, do you understand why?

jbriggs444 said:
Wrong. @PeroK is leading you down the correct path. @Cutter Ketch is mistaken in this case.

Ahh ... I am mistaken. Breaking the acceleration period into smaller intervals of time the distance traveled during each interval is changing so the rate of work is not constant. My bad.

PeroK said:
Yes, do you understand why?
At this point the elevator is accelerating, so it needs more energy than keeping it as constant because it needs to overcome the gravitational acceleration?

Edel Crine said:
At this point the elevator is accelerating, so it needs more energy than keeping it as constant because it needs to overcome the gravitational acceleration?

I meant:

Why is power not constant for constant acceleration?

jbriggs444
PeroK said:
I meant:

Why is power not constant for constant acceleration?
Oh, wait it should be 22155 Watt not joule.
Ummmmmmm, some energy has been dissipated...? I apologize...

Edel Crine said:
Oh, wait it should be 22155 Watt not joule.
Ummmmmmm, some energy has been dissipated...? I apologize...
There's not a lot of point in doing this question unless you take away from it an understanding of this point. Eventually, with some help, you put the right numbers into the right equations. But, if you don't understand the physics, you'll make the same mistake in the future.

Hint think about how kinetic energy increases under constant acceleration.

PeroK said:
There's not a lot of point in doing this question unless you take away from it an understanding of this point. Eventually, with some help, you put the right numbers into the right equations. But, if you don't understand the physics, you'll make the same mistake in the future.

Hint think about how kinetic energy increases under constant acceleration.
It increases proportionally to the displacement of force?

And the displacement will be different (getting bigger and bigger) in each time interval?!

jbriggs444
Edel Crine said:
And the displacement will be different (getting bigger and bigger) in each time interval?!

Why don't you calculate the KE each second under constant acceleration of say ##2m/s^2## for an object of ##1kg##:

##t = 0, v = 0, KE = 0##

## t = 1s, v = 2 m/s, KE = 2J##

##t = 2s, v = 4m/s, KE = 8J##

What do you notice?

PeroK said:
Why don't you calculate the KE each second under constant acceleration of say ##2m/s^2## for an object of ##1kg##:

##t = 0, v = 0, KE = 0##

## t = 1s, v = 2 m/s, KE = 2J##

##t = 2s, v = 4m/s, KE = 8J##

What do you notice?
The rate of kinetic energy increasing is not constant...?

Edel Crine said:
And the displacement will be different (getting bigger and bigger) in each time interval?!
Yes. If you concentrate on tiny time intervals, the work done in each will be getting bigger and bigger as the displacement gets bigger and bigger because the velocity is getting bigger and bigger.

There is a way to put this observation into a formula. Velocity is the incremental displacement per unit time. If you multiply force by incremental displacement, you get work per incremental time -- i.e. power. So$$P=\vec{F}\cdot\vec{v}$$

Edel Crine said:
The rate of kinetic energy increasing is not constant...?
Exactly. You can work out the average power input required for each second. This power requirement goes up as the speed increases. This is the main reason your car has a maximum speed, with power as the limiting factor. Under constant power, the acceleration soon tails off.

Gosh, now I got it...!
I really appreciate for continuous help from all of you...
With my harsh English, this is the only thing that I can say, all of you are the awesome

Lnewqban and PeroK
Edel Crine said:
Gosh, now I got it...!
I really appreciate for continuous help from all of you...
With my harsh English, this is the only thing that I can say, all of you are the awesome
So what did you answer for a?

## 1. How does the power to accelerate a 1000kg elevator cab upward work?

The power to accelerate a 1000kg elevator cab upward works by using a motor and a system of pulleys and cables. The motor provides the necessary force to lift the cab, while the pulleys and cables help to distribute the weight evenly and prevent the cab from falling.

## 2. What is the maximum speed that can be achieved when accelerating a 1000kg elevator cab upward?

The maximum speed that can be achieved when accelerating a 1000kg elevator cab upward depends on various factors such as the power of the motor, the efficiency of the pulley system, and the weight of the cab itself. However, in general, elevators can reach speeds of up to 20 feet per second.

## 3. How much power is needed to accelerate a 1000kg elevator cab upward?

The amount of power needed to accelerate a 1000kg elevator cab upward depends on the height that the cab needs to travel and the speed at which it needs to reach its destination. On average, an elevator requires about 10 horsepower to lift a 1000kg cab.

## 4. Is it safe to accelerate a 1000kg elevator cab upward at high speeds?

Yes, it is safe to accelerate a 1000kg elevator cab upward at high speeds. Elevators are designed and built to withstand high speeds and can safely transport passengers and cargo without any issues. However, regular maintenance and safety inspections are essential to ensure the safe operation of elevators.

## 5. Can the power to accelerate a 1000kg elevator cab upward be controlled?

Yes, the power to accelerate a 1000kg elevator cab upward can be controlled through the use of a motor controller. This device regulates the amount of power being supplied to the motor, allowing for smooth and controlled acceleration and deceleration of the elevator cab.

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