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

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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.

**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 \( W = F \cdot d = 6 \, \text{N} \times 4 \, \text{m} = 24 \, \text{J} \).
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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