Power of Elevator: Calculating Motor Power Needed

[blayman5]

Homework Statement

An elevator has a mass of 956 kg and car-
ries a maximum load of 764 kg. A constant
frictional force of 4150 N retards its motion
upward.
The acceleration of gravity is 9.8 m/s2 .
What power must the motor deliver at a
instantaneous speed of 4.69m/s if the elevator
is designed to provide an upward acceleration
of 1.07 m/s2? Answer in units of W.

The Attempt at a Solution

F=ma
N-mg=(me+ml)a
N=(me+ml)(a+g)
F=N-frictional force
P=Fv

I was wondering if this progression of the problem would be logical to solve for the power.

 

[blayman5]

I got it:

Fy=ma
N-mg-f=ma
N=(me+ml)(g+a) + friction

P= Nv

=107144.616

 

[thomate1]

Your approach is logical, but there are a few things that need to be clarified in order to solve for power. First, I'm assuming that the elevator is initially at rest and is accelerating upward at 1.07 m/s^2. Also, I'm assuming that the elevator is moving at a constant speed of 4.69 m/s, so there is no change in acceleration.

With that in mind, here is the solution:

First, let's calculate the total force needed to accelerate the elevator upward:
F = (m_elevator + m_load)(a + g)
F = (956 kg + 764 kg)(1.07 m/s^2 + 9.8 m/s^2)
F = 17,091 N

Next, we need to take into account the frictional force that is retarding the motion of the elevator. Since the elevator is moving at a constant speed, the net force must be zero. Therefore, we can set the frictional force equal to the total force:
F_friction = 17,091 N

Now, we can solve for the power:
P = F_friction * v
P = (17,091 N)(4.69 m/s)
P = 80,119 W

So, the motor must deliver 80,119 W of power to maintain a constant speed of 4.69 m/s while also providing an upward acceleration of 1.07 m/s^2.