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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What is the correct option?
![### Work Calculation Problem
**You may use a calculator.**
---
**Problem:**
A forklift exerts an upward force of \( 2.00 \times 10^3 \) N on a box as it moves the box 5.00 m higher. How much work does the forklift do?
---
**Answer Choices:**
- \( \circ \ 1.00 \times 10^4 \) J
- \( \circ \ 500 \) J
- \( \circ \ 0.00 \) J
- \( \circ \ 1.00 \times 10^3 \) J
**Selected Answer:**
- \( \bullet \ 0.00 \) J
---
**Explanation:**
In this problem, we are asked to calculate the work done by a forklift. Work can be calculated using the formula:
\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]
Where:
- Force (\(F\)) = \( 2.00 \times 10^3 \) N
- Distance (\(d\)) = 5.00 m
- \( \theta \) is the angle between the force and the direction of movement. In this case, since the force exerted by the forklift is vertically upward and the distance moved is also vertically upward, \( \theta \) = 0 degrees. Therefore, \( \cos (0) \) = 1.
Thus, the work done:
\[ \text{Work} = (2.00 \times 10^3 \text{ N}) \times (5.00 \text{ m}) \times 1 = 10.00 \times 10^3 \text{ J} \]
\[ \text{Work} = 1.00 \times 10^4 \text{ J} \]
The amount of work done by the forklift is therefore \( 1.00 \times 10^4 \) J. Hence, the selected answer (0.00 J) is not correct.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa8dc949f-7f1d-4c1b-9c4e-dfcc47554f56%2Ffe68c690-eac8-482f-bac6-a91ffeebebd6%2F2g493o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Work Calculation Problem
**You may use a calculator.**
---
**Problem:**
A forklift exerts an upward force of \( 2.00 \times 10^3 \) N on a box as it moves the box 5.00 m higher. How much work does the forklift do?
---
**Answer Choices:**
- \( \circ \ 1.00 \times 10^4 \) J
- \( \circ \ 500 \) J
- \( \circ \ 0.00 \) J
- \( \circ \ 1.00 \times 10^3 \) J
**Selected Answer:**
- \( \bullet \ 0.00 \) J
---
**Explanation:**
In this problem, we are asked to calculate the work done by a forklift. Work can be calculated using the formula:
\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]
Where:
- Force (\(F\)) = \( 2.00 \times 10^3 \) N
- Distance (\(d\)) = 5.00 m
- \( \theta \) is the angle between the force and the direction of movement. In this case, since the force exerted by the forklift is vertically upward and the distance moved is also vertically upward, \( \theta \) = 0 degrees. Therefore, \( \cos (0) \) = 1.
Thus, the work done:
\[ \text{Work} = (2.00 \times 10^3 \text{ N}) \times (5.00 \text{ m}) \times 1 = 10.00 \times 10^3 \text{ J} \]
\[ \text{Work} = 1.00 \times 10^4 \text{ J} \]
The amount of work done by the forklift is therefore \( 1.00 \times 10^4 \) J. Hence, the selected answer (0.00 J) is not correct.
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