   Chapter 5, Problem 58P

Chapter
Section
Textbook Problem

A 6.50 × 102-kg elevator starts from rest and moves upward for 3.00 s with constant acceleration until it reaches its cruising speed, 1.75 m/s. (a) What is the average power of the elevator motor during this period? (b) How does this amount of power compare with its power during an upward trip with constant speed?

(a)

To determine
the average power of the elevator motor during this period.

Explanation

Section1:

To determine: The motor force.

Answer: The motor force is 6.75×103N .

Explanation:

Given Info:

The mass of the elevator is 6.50×102kg .

The acceleration of the elevator is 1.75m/s .

The time of acceleration of the elevator is 3.0s .

The acceleration of the elevator in the first 3.0s is,

a=vv0t

• v0 is initial velocity
• v is the final velocity
• t is time of acceleration

Formula to calculate the motor force is,

Fmotor=m(a+g)

• a is the acceleration
• m is the mass of the elevator
• g is the acceleration due to gravity

On re-arranging,

Fmotor=m(vv0t+g)

Substitute 6.50×102kg for m, 1.75m/s for v, zero for v0 and 3.0s for t to find the motor force,

Fmotor=(650kg)(1.75m/s03.0s+9.8m/s2)=6

(b)

To determine
The instantaneous power input from the motor.

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