An electric motor is rated to have a maximum power output of 0.79 hp. If this motor is being used to lift a crate of mass 220 kg, how fast (i.e., at what speed) can it lift the crate? Hint: Assume the only other force on the crate is the force of gravity. m/s
An electric motor is rated to have a maximum power output of 0.79 hp. If this motor is being used to lift a crate of mass 220 kg, how fast (i.e., at what speed) can it lift the crate? Hint: Assume the only other force on the crate is the force of gravity. m/s
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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![**Problem Description:**
An electric motor is rated to have a maximum power output of **0.79 hp**. If this motor is being used to lift a crate of mass **220 kg**, at what speed can it lift the crate?
**Hint:** Assume the only other force on the crate is the force of gravity.
[Input Box] ________ m/s
---
**Details for Educational Context:**
This scenario presents a real-world application of power and mechanics, focusing on how the power output of an electric motor affects its ability to lift an object against gravitational force.
- **Power (P):** The rate at which work is done, in this case, it is given as 0.79 horsepower (hp).
- **Mass (m):** The mass of the object being lifted, which is 220 kg.
- **Gravitational Force (Fg):** The force due to gravity, calculated as mass (m) times the acceleration due to gravity (g ≈ 9.8 m/s²).
- **Speed (v):** The velocity at which the crate is lifted.
By relating power, force, and speed, this problem helps in understanding concepts such as conversion of units (horsepower to watts), and the calculation of the lifting speed using the power formula:
\[
P = F \cdot v
\]
Where:
- \( F \) is the force equivalent to the weight of the crate (\( F = m \cdot g \))
- \( v \) is the velocity or speed of lifting.
By solving this problem, learners can better grasp the application of physics in mechanical and engineering contexts.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F31367789-01e4-4129-b30b-49084de51bdb%2F97b22655-16fb-46b6-afda-907499e228bb%2Fg9c8vyj_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
An electric motor is rated to have a maximum power output of **0.79 hp**. If this motor is being used to lift a crate of mass **220 kg**, at what speed can it lift the crate?
**Hint:** Assume the only other force on the crate is the force of gravity.
[Input Box] ________ m/s
---
**Details for Educational Context:**
This scenario presents a real-world application of power and mechanics, focusing on how the power output of an electric motor affects its ability to lift an object against gravitational force.
- **Power (P):** The rate at which work is done, in this case, it is given as 0.79 horsepower (hp).
- **Mass (m):** The mass of the object being lifted, which is 220 kg.
- **Gravitational Force (Fg):** The force due to gravity, calculated as mass (m) times the acceleration due to gravity (g ≈ 9.8 m/s²).
- **Speed (v):** The velocity at which the crate is lifted.
By relating power, force, and speed, this problem helps in understanding concepts such as conversion of units (horsepower to watts), and the calculation of the lifting speed using the power formula:
\[
P = F \cdot v
\]
Where:
- \( F \) is the force equivalent to the weight of the crate (\( F = m \cdot g \))
- \( v \) is the velocity or speed of lifting.
By solving this problem, learners can better grasp the application of physics in mechanical and engineering contexts.
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