Power To Accelerate 1000kg Elevator Cab Upward

[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!

 

[Edel Crine]

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]

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)?

 

[PeroK]

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?

 

[Edel Crine]

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

 

[PeroK]

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.

 

[Edel Crine]

Okay, that's looks right.

Then, for a, I need to divide the work by time, so would it be 11077.5 Watt??

 

[PeroK]

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!

 

[Edel Crine]

That's less than the cruising power!

Oh, it should be bigger... but I think that I might need more help,,,,,,

 

[PeroK]

Oh, it should be bigger... but I think that I might need more help,,,,,,

You didn't answer my question:

For a) is the power requirement constant?

 

[Edel Crine]

You didn't answer my question:

Gravitational acceleration and mass are constant...?

 

[PeroK]

Gravitational acceleration and mass are constant...?

They are. But what about power?

 

[Edel Crine]

They are. But what about power?

It would not be constant for single time?

 

[PeroK]

It would not be constant for single time?

If in doubt, calculate! You used this equation:

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

 

[Cutter Ketch]

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?

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.

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

Yes.

 

[Cutter Ketch]

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

 

[jbriggs444]

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

Yes.

Wrong. @PeroK is leading you down the correct path.

 

[Edel Crine]

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]

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

Yes, do you understand why?

 

[Cutter Ketch]

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.

 

[Edel Crine]

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?

 

[PeroK]

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?

 

[Edel Crine]

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...

 

[PeroK]

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.

 

[Edel Crine]

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?

 

[Edel Crine]

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

 

[PeroK]

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?

 

[Edel Crine]

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...?

 

[jbriggs444]

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$$P=\vec{F}\cdot\vec{v}$$

 

[PeroK]

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.