Physics energy question

Question

An electric motor, rated at 45.0 kW with an efficiency of 0.79, provides power while lifting a 1200 kg elevator. The motor can lift eight passengers (at constant speed) to a height of 35.0 m in 20.0 s. As a safety mechanism (if the cable snaps), the designers of the elevator placed a mechanical spring (k = 1.25 x 104 N/m) at the bottom of the elevator.

a) The energy transformations involved in this question.

b) The mass of the passengers.

c) Is the stiffness of the spring appropriate for this problem? Please explain. Note: human tolerance is less than 9 g’s.

Answer 1

Let

P = 45 kW = 45000 W be the power rating of the motor,

f = 0.79 be the motor's efficiency,

m = 1200 kg be the mass of the elevator,

M be the mass of the elevator plus all the passengers (to be calculated),

n = 8 be the number of passengers,

h = 35 m be the height,

t = 20 s be the time it takes to lift the full elevator to height h,

k = 12500 N/m be the stiffness of the spring,

g = 9.81 m/s² be gravitational acceleration,

a be the deceleration experienced in a fall (to be calculated)

a)

Electrical energy is transformed partially into heat and partially into

potential energy of the elevator:

79% is transformed into the potential energy and 21% into heat.

To some unknown degree the motor also temporarily gives it some kinetic energy to start with.

It is unknown, because we not know the elevator's speed.

But at the end of the ride all that kinetic energy gets converted to

the elevator's potential energy.

When the cable snaps, the potential energy is transformed into kinetic energy

of the falling elevator.

When the falling elevator depresses the spring, the kinetic energy is

transformed into spring energy.

Thereafter the elevator starts oscillating, during which process

there is a transfer of kinetic, potential, and spring energy.

In addition, eventually all that energy gets transformed into heat

due to some friction, at which point the elevator stops oscillating.

b)

The weight of the full elevator is

Mg

The potential energy obtained by the elevator is

Mgh

That potential energy equals the work performed by the motor.

The power needed to perform that work in the given amount of time is

Mgh/t

That power is provided by the motor after any losses, so

Mgh/t = Pf

Therefore

M = Pft/gh

Substituting actual numbers

M = 45000×0.78×20/(9.81×35) = 2044.56 kg

The mass of the passengers is 2044.56 - 1200 = 844.56 kg.

That amounts to 105.57 kg per passenger.

c)

When the falling elevator hits the spring its starts depressing it.

In that process its total energy Mgh is being transformed into

energy of the spring.

That transfer is complete when the spring reaches maximum depression A.

At that point the elevator experiences maximum deceleration

a = Ak/M

We will be able to calculate that deceleration once we determine A.

We can calculate A from conservation of energy -- the spring energy at

maximum depression, kA²/2, must equal the elevator's maximum energy, Mgh.

Thus

kA²/2 = Mgh

So

A = √(2Mgh/k)

Therefore

a = Ak/M = √(2ghk/M)

The above formula will give us the deceleration in m/s².

But we want it in the number of g's, that is, we want the ratio

a/g = √(2hk/(gM))

Substituting actual numbers

a/g = √(2×35×12500/(9.81×2044.56)) = 6.6

That does not exceed the maximum tolerance of 9 g's.