A crane lifts a 650-kg beam vertically upward 23.0 m and then swings it horizontally a distance of 18.0 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed.
A crane lifts a 650-kg beam vertically upward 23.0 m and then swings it horizontally a distance of 18.0 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed.
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:**
A crane lifts a 650-kg beam vertically upward 23.0 m and then swings it horizontally a distance of 18.0 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed.
**Analysis:**
To calculate the work done by the crane, we need to consider two parts of the movement:
1. **Vertical Lift:**
- Mass of the beam (m) = 650 kg
- Height (h) = 23.0 m
- Gravitational acceleration (g) = 9.8 m/s²
- Work done (W_vertical) = m * g * h
2. **Horizontal Movement:**
- Since the problem assumes constant speed and neglects friction, the work done horizontally is zero because there is no force component in the direction of movement.
**Conclusion:**
The total work done by the crane is the sum of the work done in lifting the beam vertically and moving it horizontally:
\[ W_{\text{total}} = W_{\text{vertical}} + W_{\text{horizontal}} = m \cdot g \cdot h + 0 \]
Plug in the values to compute the final result.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7237bf46-4025-4c4f-b170-df7369a74fb6%2F08fe6fe3-c247-4c5f-8ffe-f99902e848d0%2Fpzk7m6r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A crane lifts a 650-kg beam vertically upward 23.0 m and then swings it horizontally a distance of 18.0 m. How much work does the crane do? Neglect friction, and assume the beam moves with constant speed.
**Analysis:**
To calculate the work done by the crane, we need to consider two parts of the movement:
1. **Vertical Lift:**
- Mass of the beam (m) = 650 kg
- Height (h) = 23.0 m
- Gravitational acceleration (g) = 9.8 m/s²
- Work done (W_vertical) = m * g * h
2. **Horizontal Movement:**
- Since the problem assumes constant speed and neglects friction, the work done horizontally is zero because there is no force component in the direction of movement.
**Conclusion:**
The total work done by the crane is the sum of the work done in lifting the beam vertically and moving it horizontally:
\[ W_{\text{total}} = W_{\text{vertical}} + W_{\text{horizontal}} = m \cdot g \cdot h + 0 \]
Plug in the values to compute the final result.
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