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
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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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.
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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