# Power on an elevator

1. 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?

2. Homework Equations
P = Fv

3. The Attempt at a Solution
F = 2100 N
V = 8 m/s
P = (2100) * (8) = 16800

in significant figures thats 1.7 x 10^4 w
in kilowatts thats 17
why is this wrong?

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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 doesnt work either
the answer is asked for in 2 s.ds and in KW, so i assumed that would be 14, but thats 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 im missing something obvious here. sorry i couldn't be of more help.

alphysicist
Homework Helper