Elevator Physics: Power & Mass Analysis

[Newton22]

Homework Statement

The mass of the cab is 2000kg and the mass of the counterweight in an elevator is 2600kg. Assume the mass of the cab remains constant.
The cab must descend a distance of 300m from the top floor to the bottom floor, and must do so at a constant speed of 10m/s.

A)What must the power of the motor be in order to accomplish this? Give your answer in Watts.

B) If passengers enter the cab and the mass of the cab is increased, how does this affect the power required?

Homework Equations

P = Fv

P = (10m/s)F

The Attempt at a Solution

The pully is assumed to have no friction and no mass.

The total force that is opposing the downward motion is :
F(counterweight) - F(cab)
25500N - 19600N = - 5900N

Don't know what to next. I think all we need to know is the force but what does the height have anything to do with it?

Please help and thanks

 

[Simon Bridge]

What is the relationship between work and mechanical energy?

 

[Newton22]

What is the relationship between work and mechanical energy?

W = Kf - Ki
W = .5MVf^2 - .5MVi^ 2

 

[Newton22]

Also
W = Ui - Uf

 

[Simon Bridge]

OK - so how does height relate to the change in mechanical energy for the elevator system?

 

[Newton22]

OK - so how does height relate to the change in mechanical energy for the elevator system?

W = Uf - Ui

Uf of the cab is 0
Uf of the counterweight is 2600*9.81*300 = 7651800J
Ui of the cab is 2000*9.81*300 = 5886000J

Uf - Ui
7651800J - 5886000J = 1765800J
So the work is going to be 1765800J

 

[Simon Bridge]

The power is the rate that the work is done rather than the total work.

 

[Newton22]

The power is the rate that the work is done rather than the total work.

How can we relate potential energy to the power?

Maybe P = (Ui - Uf)/t how can we solve this

 

[Simon Bridge]

Off the problem statement - you are told the cab etc has a constant speed.
How do you find time from speed and distance?

This is the same as putting $F=(m_c-m_{cw})g$ then doing $P=Fv = (m_c-m_{cw})gv$
... but you asked the question :)

If the elevator did not do the constant speed - then this would give an average power off the average speed.

 

[Newton22]

Off the problem statement - you are told the cab etc has a constant speed.
How do you find time from speed and distance?

This is the same as putting $F=(m_c-m_{cw})g$ then doing $P=Fv = (m_c-m_{cw})gv$
... but you asked the question :)

If the elevator did not do the constant speed - then this would give an average power off the average speed.

aha so you solve for the time knowing speed and distance. Would it then be correct to insert this time into the equation P = (Ui-Uf)/t ?

 

[Simon Bridge]

That would be correct :)