Question

A motor must lift a 1500-kg elevator cab. The cab's maximum occupant capacity is 400 kg, and its constant "cruising" speed is 1.3 m/s . The design criterion is that the cab must achieve this speed within 2.0 s at constant acceleration, beginning from rest.
When the cab is carrying its maximum capacity, at what average rate must the motor deliver energy to get the cab up to cruising speed?

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

Answer 1

Answer:

12900 W

24200 W

Explanation:

Given:

v₀ = 0 m/s

v = 1.3 m/s

t = 2.0 s

Find: a and Δx

v = at + v₀

(1.3 m/s) = a (2.0 s) + (0 m/s)

a = 0.65 m/s²

Δx = ½ (v + v₀) t

Δx = ½ (1.3 m/s + 0 m/s) (2.0 s)

Δx = 1.3 m

While accelerating:

Newton's second law:

∑F = ma

F − mg = ma

F = m (g + a)

F = (1500 kg + 400 kg) (9.8 m/s² + 0.65 m/s²)

F = 19855 N

Power = work / time

P = W / t

P = Fd / t

P = (19855 N) (1.3 m) / (2.0 s)

P ≈ 12900 W

At constant speed:

Newton's second law:

∑F = ma

F − mg = 0

F = mg

F = (1500 kg + 400 kg) (9.8 m/s²)

F = 18620 N

Power = work / time

P = W / t

P = Fd / t

P = Fv

P = (18620 N) (1.3 m/s)

P ≈ 24200 W