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

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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:
\[ \boxed{75 \, \text{N}} \]
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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