# Physics question

Discussion in 'Pre-Medical - MD' started by byeh2004, Nov 18, 2005.

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1. ### byeh2004 Senior Member 5+ Year Member

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Hey guys I'm a bit stumped on this one question for this week's physics hw

A 500 kg elevator starts from rest. It moves upward for 5.00 seconds with constant acceleration until it reacfhes its cruising speed, 1.75 m/s.

What is the average power of the elevator motor during this period?

Thanks for the help!

2. ### princessd3 Senior Member 7+ Year Member

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Let me take a try. Express Power in terms of Work. Find Work using change in potential energy (mgh). Find Height using one of the big 5 formulas. Think its Vfinal squared= Vinital squared etc (don't remember but it involves h).

There is an easier way to do this. Think there is an expression for power using velocity. Don't remember.

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3. ### Shredder User 5+ Year Member

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can you just find the potential energy at the top of the ascent and divide by time. so,

E=mgh
h=Vot + .5at^2 = .5a*25
a=1.75/5
E=(500)(9.8)(4.375)
E/t = P = 4290 W

P=FV, that might work if you find avg V.
yes it works--avg V is just Vfinal/2 since its constant acceleration.
P=(500kg)(9.8m/s^2)(1.75/2 m/s) = 4290 W

ah, that was invigorating. good old simple physics is so much better than engineering.

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4. ### willow18 10+ Year Member

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I smell our neighborhood philanthropist on his way to lock this one up.

5. ### Shredder User 5+ Year Member

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oh, that makes sense. i was wondering why questions of this nature dont appear more often. shouldve posted on mcat forum instead

+i doubt even philanthropists patrol 24/7. including friday night. or do they...

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6. ### Dave_D Senior Member 7+ Year Member

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Wait a sec. E=MGH is only potential change but there's an acceleration as well so don't we have to us W=E=FD ? If so doesn't that change it slightly to be
E=(Force total)*Height=(Fnet +Fweight)*height

which is
E=(500*0.35m/s^2 + 500*9.8m.s^2)4.375m=22203J and that's 4440W

7. ### princessd3 Senior Member 7+ Year Member

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Hmmm... Power is work done per unit time. Work is equal to change in potential energy. I don't think the acceleration of the object has anything to do with the change in potential energy in moving the object from point A to point B.

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8. ### Dave_D Senior Member 7+ Year Member

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Actually while it's not part of the potential energy we're supposed to be figuring out the power the motor supplies so yes it's important. Work is force times distance and work and energy are really the same thing. The reason you get a potential energy of mgh is that you apply a force of the weight over a distance of the height, IE (mg)*h. mg is the work, h is the distance. But in this example the motor is applying extra force so there is a net acceleration upward. So simply using mgh to get energy doesn't work because the force applied is greater than the weight. Another way to see this is true is to realize that there's kinetic energy in this set up as well as gravitation potential energy since at the end the elevator is moving upward. So you can theoretically add up the potential and kinetic energies to get the total energy or work the motor added to the system and then divide by time to get power. So if you use 1/2mv^2 and add that to the potential energy you get total energy or my amount of 4440.(Which is actually an easier way of getting it than I originally did.)

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