For the three cases shown below, assuming the displacement is 4.0 m to the right. Match the answers with questions. 10 N 8N 6N v the work done by the 10-N force A. 24 J v the work done by the 8-N force B. OJ the work done by the 6-N force C. 40 J D. 32 J E. 16 J
For the three cases shown below, assuming the displacement is 4.0 m to the right. Match the answers with questions. 10 N 8N 6N v the work done by the 10-N force A. 24 J v the work done by the 8-N force B. OJ the work done by the 6-N force C. 40 J D. 32 J E. 16 J
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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For the three cases shown below, assuming the displacement is 4.0 m to the right. Match the answers with questions.

Transcribed Image Text:**For the three cases shown below, assuming the displacement is 4.0 m to the right. Match the answers with questions.**
The diagram consists of three cases, each involving a force acting on an object with a displacement of 4.0 meters to the right, represented by a horizontal blue arrow labeled \(\vec{d}\).
1. **Case 1**: A vertical red arrow labeled 10 N, acting perpendicular to the displacement \(\vec{d}\).
2. **Case 2**: A red arrow labeled 8 N, at an angle of 60° to the displacement \(\vec{d}\).
3. **Case 3**: A red horizontal arrow labeled 6 N, acting in the same direction as the displacement \(\vec{d}\).
**Questions:**
- The work done by the 10-N force
- The work done by the 8-N force
- The work done by the 6-N force
**Possible Answers:**
A. 24 J
B. 0 J
C. 40 J
D. 32 J
E. 16 J
**Explanation of Diagram:**
- In Case 1, the 10 N force is perpendicular to the displacement, hence no work is done (0 J).
- In Case 2, work done is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \). Here, \( F = 8 \, \text{N} \), \( d = 4 \, \text{m} \), and \( \theta = 60^\circ \).
- In Case 3, the force is in the direction of displacement, so work done is \( W = F \cdot d = 6 \, \text{N} \times 4 \, \text{m} = 24 \, \text{J} \).
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