A traveler pulls on a suitcase strap at an angle of 36° above the horizontal. If 908 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap? 75 N 61 N 85 N 92 N
A traveler pulls on a suitcase strap at an angle of 36° above the horizontal. If 908 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap? 75 N 61 N 85 N 92 N
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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![### Physics Problem: Tension in a Suitcase Strap
A traveler pulls on a suitcase strap at an angle of 36° above the horizontal. If 908 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap?
#### Multiple-Choice Options:
- 75 N
- 61 N
- 85 N
- 92 N
To solve this problem, we need to understand the relationship between work, force, and distance in conjunction with the direction of the applied force.
The general formula for work is:
\[ W = F \cdot d \cdot \cos(\theta) \]
Where:
- \( W \) is the work done (908 J)
- \( F \) is the force (tension in the strap, which we need to find)
- \( d \) is the distance (15 m)
- \( \theta \) is the angle above the horizontal (36°)
Rearranging the formula to solve for \( F \):
\[ F = \frac{W}{d \cdot \cos(\theta)} \]
Substituting the known values:
\[ F = \frac{908 \, \text{J}}{15 \, \text{m} \cdot \cos(36^\circ)} \]
Calculating the cosine of 36 degrees:
\[ \cos(36^\circ) \approx 0.809 \]
Thus:
\[ F = \frac{908 \, \text{J}}{15 \, \text{m} \cdot 0.809} \]
\[ F \approx \frac{908}{12.135} \]
\[ F \approx 74.86 \, \text{N} \]
Approximating to the closest given option:
\[ F \approx 75 \, \text{N} \]
Therefore, the tension in the strap is:
\[ \boxed{75 \, \text{N}} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd99dc5ee-0347-4b7c-8511-9a91331fbdb4%2Fec7d70d7-76ee-4882-9313-48e6bfcbd865%2Fn0vda29_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Physics Problem: Tension in a Suitcase Strap
A traveler pulls on a suitcase strap at an angle of 36° above the horizontal. If 908 J of work are done by the strap while moving the suitcase a horizontal distance of 15 m, what is the tension in the strap?
#### Multiple-Choice Options:
- 75 N
- 61 N
- 85 N
- 92 N
To solve this problem, we need to understand the relationship between work, force, and distance in conjunction with the direction of the applied force.
The general formula for work is:
\[ W = F \cdot d \cdot \cos(\theta) \]
Where:
- \( W \) is the work done (908 J)
- \( F \) is the force (tension in the strap, which we need to find)
- \( d \) is the distance (15 m)
- \( \theta \) is the angle above the horizontal (36°)
Rearranging the formula to solve for \( F \):
\[ F = \frac{W}{d \cdot \cos(\theta)} \]
Substituting the known values:
\[ F = \frac{908 \, \text{J}}{15 \, \text{m} \cdot \cos(36^\circ)} \]
Calculating the cosine of 36 degrees:
\[ \cos(36^\circ) \approx 0.809 \]
Thus:
\[ F = \frac{908 \, \text{J}}{15 \, \text{m} \cdot 0.809} \]
\[ F \approx \frac{908}{12.135} \]
\[ F \approx 74.86 \, \text{N} \]
Approximating to the closest given option:
\[ F \approx 75 \, \text{N} \]
Therefore, the tension in the strap is:
\[ \boxed{75 \, \text{N}} \]
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