# Thread: Elevator work, power [Physics]

1. ## Elevator work, power [Physics]

I threw this one in the physics help forum, but nobody seems to go there so i figured i would try the other topics forum ^^

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s.

(a). What is the average power of the elevator motor during this time interval?

Couldn't i use the total work thru kinetic energy? W = 1/2 mVf^2 - 1/2 mVi^2 ?
I tried that the first time for this problem, teacher said that i'm wrong. I'm not sure what to do other than ths.

(b) How does this power compare with the motor power when the elevator moves at its crusing speed?

As always, any help is Greatly appreciated....

2. Originally Posted by 92stealth
I threw this one in the physics help forum, but nobody seems to go there so i figured i would try the other topics forum ^^

A 650-kg elevator starts from rest. It moves upward for 3.00 s with constant acceleration until it reaches its cruising speed of 1.75 m/s.

(a). What is the average power of the elevator motor during this time interval?

Couldn't i use the total work thru kinetic energy? W = 1/2 mVf^2 - 1/2 mVi^2 ?
I tried that the first time for this problem, teacher said that i'm wrong. I'm not sure what to do other than ths.

(b) How does this power compare with the motor power when the elevator moves at its crusing speed?

As always, any help is Greatly appreciated....
a) Assumption: The motor is 100% efficient.

You can use the equations of motion to find both distance and acceleration:

a = a, u= 0, v= 1.75, t= 3

v = u + at

$a = \frac{v-u}{t} = \frac{1.75}{3} = \frac{7}{12}$

u=0, v= 1.75, t=3, s=s

$s = \frac{1}{2}(u+v)t = \frac{1}{2}(1.75+0)(3) = \frac{21}{8}$

Since a and s are known you can use

$F = ma$ and $W = Fs$ so $W = mas = 650 \times \frac{7}{12} \times \frac{21}{8} = 995J$ (3sf)

3. Exactly the answer i got, just got there a different way, he said that i was wrong. Wonder if my teacher is the one that looked at it wrong...thanks man

4. Originally Posted by 92stealth
Exactly the answer i got, just got there a different way, he said that i was wrong. Wonder if my teacher is the one that looked at it wrong...thanks man
I'm afraid your thinking is wrong ... and your teacher is correct.

$W_{motor} = \frac{1}{2}m(v_f^2 - v_0^2) + mg\Delta y$

$W_{gravity} = -mg \Delta y$

$W_{net} = \frac{1}{2}m(v_f^2 - v_0^2)$

(a). What is the average power of the elevator motor during this time interval?
$P_{avg} = \frac{W_{motor}}{\Delta t}$

(b) How does this power compare with the motor power when the elevator moves at its cruising speed?
it's greater ... at cruising speed $P = Fv = mgv$