- #1

Edel Crine

- 89

- 12

- Homework Statement
- A motor must lift a 1000-kg elevator cab. The cab’s maximum occupant capacity is 400 kg, and its constant “cruising” speed is 1.5 m>s. The design criterion is that the cab must achieve this speed within 2.0 s at constant acceleration, beginning from rest.

(a) When the cab is carrying its maximum capacity, at what rate must the motor deliver energy to get the cab up to cruising speed?

(b) At what constant rate must the motor supply energy as the fully loaded cab rises after attaining cruising speed?

- Relevant Equations
- a=Δv/t

F=ma

w=Δk=F*Δx

(a) When the cab is carrying its maximum capacity, at what rate must the motor deliver energy to get the cab up to cruising speed?

My shot was finding the acceleration,

a=1.5/2 = 0.75m/s2

Displacement of elevator

Vf2 = vi2 + 2aΔx

Δx = 1.5m

Force of elevator (Since it accelerates upward, the weight of everything will be heavier than before.

w = (1400kg)(9.8+0.75 m/s2)(1.5m) = 22155 J.

Then, should I divide this work done by the time in order to get a rate of energy delivered to the elevator?

And I am not sure for part b...

I appreciate for every help from all of you!

My shot was finding the acceleration,

a=1.5/2 = 0.75m/s2

Displacement of elevator

Vf2 = vi2 + 2aΔx

Δx = 1.5m

Force of elevator (Since it accelerates upward, the weight of everything will be heavier than before.

w = (1400kg)(9.8+0.75 m/s2)(1.5m) = 22155 J.

Then, should I divide this work done by the time in order to get a rate of energy delivered to the elevator?

And I am not sure for part b...

I appreciate for every help from all of you!

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