(a)Find the useful power output (in W) of an elevator motor that lifts a 2200 kg load a height of 30.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2200 kg is raised in height, but the full 10,000 kg is accelerated. (b)What does it cost (in cents), if electricity is $0.0900 per kW · h? ### Problem Description(a) **Finding Useful Power Output:**Calculate the useful power output in watts (W) of an elevator motor. The motor lifts a 2200 kg load to a height of 30.0 meters in 12.0 seconds, while also increasing the speed from rest to 4.00 meters per second. The total mass of the counterbalanced system is 10,000 kg, meaning only 2200 kg is raised in height, but the full 10,000 kg is accelerated. Enter the answer as a number of watts.- **Attempted Answer:** 130673.33 W- **Status:** Incorrect (indicated by a red cross)(b) **Calculating Cost of Electricity:**Determine the cost in cents, if electricity costs $0.0900 per kilowatt-hour (kW·h). Enter the answer as a number of cents.- **Attempted Answer:** 3.92 cents- **Status:** Incorrect (indicated by a red cross)

Name: (a)Find the useful power output (in W) of an elevator motor that lifts
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Description: (a)Find the useful power output (in W) of an elevator motor that lifts a 2200 kg load a height of 30.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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(a)

Find the useful power output (in W) of an elevator motor that lifts a 2200 kg load a height of 30.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2200 kg is raised in height, but the full 10,000 kg is accelerated. (b)

What does it cost (in cents), if electricity is $0.0900 per kW · h?

### Problem Description

(a) **Finding Useful Power Output:**
Calculate the useful power output in watts (W) of an elevator motor. The motor lifts a 2200 kg load to a height of 30.0 meters in 12.0 seconds, while also increasing the speed from rest to 4.00 meters per second. The total mass of the counterbalanced system is 10,000 kg, meaning only 2200 kg is raised in height, but the full 10,000 kg is accelerated. Enter the answer as a number of watts.

- **Attempted Answer:** 130673.33 W
- **Status:** Incorrect (indicated by a red cross)

(b) **Calculating Cost of Electricity:**
Determine the cost in cents, if electricity costs $0.0900 per kilowatt-hour (kW·h). Enter the answer as a number of cents.

- **Attempted Answer:** 3.92 cents
- **Status:** Incorrect (indicated by a red cross)

Expert Solution

Step 1: Mention the values given in the question

Given,

mass, m = 2200 kg

M = 10

height, h = 30 m

time, t = 12 s

Force acting on the mass M = Ma

Where a is the acceleration

a = (v-u)/t = 4/12 = 0.33 m/s²

Power output of total system due to acceleration is Fv = Mav

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