Power of Elevator Motor: 500kg Acceleration & Cruising Speed

  • Thread starter Thread starter Geminiforce
  • Start date Start date
  • Tags Tags
    Elevator Power
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 5K views
Geminiforce
Messages
3
Reaction score
0

Homework Statement


A 500 kg elevator starts from rest. It moves upward for 3.50 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising = ?

Homework Equations


F=ma
P=W/deltaT
W=Fdelta(r)cos(theta)

The Attempt at a Solution


This problem, I first found the acceleration using Vf= Vi +at
a= 0.5m/s^2

Next I plugged this into delta x = volt +1/2a(t^2) and found delta x to be 3.0625
Next, I solved for tension using: T-Mg = Ma
T= M(a+g) = 10,578.75 W
Is this correct? If it isn't, can someone explain to me how to approach this problem?
 
on Phys.org
Geminiforce said:

Homework Statement


A 500 kg elevator starts from rest. It moves upward for 3.50 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising = ?

Homework Equations


F=ma
P=W/deltaT
W=Fdelta(r)cos(theta)

The Attempt at a Solution


This problem, I first found the acceleration using Vf= Vi +at
a= 0.5m/s^2

Next I plugged this into delta x = volt +1/2a(t^2) and found delta x to be 3.0625
Next, I solved for tension using: T-Mg = Ma
Your method looks good up until this point:
Geminiforce said:
T= M(a+g) = 10,578.75 W
Perhaps this is a typo but, tension is a force and not power.
 
Geminiforce said:

Homework Statement


A 500 kg elevator starts from rest. It moves upward for 3.50 s with constant acceleration until it reaches its cruising speed, 1.75 m/s.
(a) What is the average power of the elevator motor during this period?
(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
Pcruising = ?

Homework Equations


F=ma
P=W/deltaT
W=Fdelta(r)cos(theta)

The Attempt at a Solution


This problem, I first found the acceleration using Vf= Vi +at
a= 0.5m/s^2

Next I plugged this into delta x = volt +1/2a(t^2) and found delta x to be 3.0625
Why? The problem did not ask for distance moved.

Next, I solved for tension using: T-Mg = Ma
T= M(a+g) = 10,578.75 W
No, the units are wrong. kg (m/s2 is Newtons, not Watts. This is a force not power. Do you know how to convert from force to power, i.e. from Newtons to Watts?

Is this correct? If it isn't, can someone explain to me how to approach this problem?
 
HallsofIvy said:
Why? The problem did not ask for distance moved.


No, the units are wrong. kg (m/s2 is Newtons, not Watts. This is a force not power. Do you know how to convert from force to power, i.e. from Newtons to Watts?

You're right i accidently put the force in the wrong units...
I meant to put Power=M(a+g)(v)... However the question asks for the average power and I think the power calculated there is the power at 3 second. Would this power i just calculated be the cruising speed?