Pls include a diagram in your answer. **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 * h2. **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.

Given: The mass of the beam is 650 kg. The vertical displacement of the beam is 23 m. The horizontal…

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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Question

Pls include a diagram in your answer.

**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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