Calculating Power for Elevator Ascent: 2100 N at 8.0 m/s

  • Thread starter Thread starter onestarburns
  • Start date Start date
  • Tags Tags
    Elevator Power
AI Thread Summary
To calculate the power required for a 2100 N elevator ascending at a constant speed of 8.0 m/s, the correct formula is P = Fv, where F is the force (2100 N) and v is the velocity (8 m/s). This results in a power output of 16,800 W or 17 kW. A misunderstanding arose regarding the force calculation, as the elevator's acceleration is zero at constant speed, negating the need for additional force calculations. The confusion about significant figures and units was also noted, but the initial power calculation was deemed correct. The discussion emphasizes the importance of understanding the conditions of motion when applying physics equations.
onestarburns
Messages
12
Reaction score
0

Homework Statement



An elevator weighing 2100 N ascends at a constant speed of 8.0 m/s. How much power must the motor supply to do this?

Homework Equations


P = Fv

The Attempt at a Solution


F = 2100 N
V = 8 m/s
P = (2100) * (8) = 16800

in significant figures that's 1.7 x 10^4 w
in kilowatts that's 17
why is this wrong?
 
Physics news on Phys.org
Hello onestarburns,

2100 N = 214.2 kg

F = m*a = 214.2kg * 8m/s^2 = 1713.6 N

W = F*change in dis. = 1713.6N * 8m = 13,708.8 J

P = W/change in time = 13,708.8N/1s = 13,708.8Watts
or
P = Fv = 1713.6N * 8m/s = 13,708.8 Watts

I believe this is correct.
If the elevator 'weighs' 2100 N than its mass is 214.2 kg.
So the force it takes to move this mass 8meters in one second every second is: 1713.6 N
The work that you accomplish by applying 1713.6N of force over 8meters is: 13,708.8 N*m

I am assuming that the question is asking for the work output per second. However if the question elaborates more than what you have given the answer could be different.
So the power required to do this amount of work in one second is: 13,708.8 J*s
 
that doesn't work either
the answer is asked for in 2 s.ds and in KW, so i assumed that would be 14, but that's not working
 
Well personally i hate significant figures so i can't help you there. And without more information in the question i don't really know what else can be done. Let me know if I am missing something obvious here. sorry i couldn't be of more help.:frown:
 
Your answer of 16800 W seems correct to me.

------------------------------

Hi robertm,

There is an error in your second equation. The elevator is moving at constant speed so a=0.
 
thanks alphysicist that's why i was confused. It looked as if the op's first answer was correct so i assumed the question contained additional info that was missing in the post.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top