A ski patroller pulls a rescue toboggan across a flat snow surface by exerting a constant force of 50 pounds on a handle that makes a 22° angle with the horizontal (see figure). Determine the work done in pulling the toboggan 230 feet. (Round your answer to one decimal place.) ft-lb 22°
A ski patroller pulls a rescue toboggan across a flat snow surface by exerting a constant force of 50 pounds on a handle that makes a 22° angle with the horizontal (see figure). Determine the work done in pulling the toboggan 230 feet. (Round your answer to one decimal place.) ft-lb 22°
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Problem Statement:
**Topic:** Work and Energy in Physics
A ski patroller pulls a rescue toboggan across a flat snow surface by exerting a constant force of 50 pounds on a handle that makes a 22° angle with the horizontal (see figure). Determine the work done in pulling the toboggan 230 feet. (Round your answer to one decimal place.)
**Work done = __________ ft-lb**
### Explanation of the Diagram:
The figure shows a ski patroller pulling a rescue toboggan. The ski patroller is depicted on skis, pulling the toboggan using a handle. The handle forms an angle of 22° with the horizontal ground. Various trees, mountains, and clouds compose the background scenery, indicating a mountainous terrain covered with snow.
### Calculations:
To determine the work done, we use the formula:
\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]
where:
- **Force (F)** = 50 pounds
- **Distance (d)** = 230 feet
- **Angle (θ)** = 22°
First, convert the angle to radians if necessary. However, since typical calculators and computational tools can directly use degrees for cosine functions, we proceed with degrees:
\[ \theta = 22^\circ \]
Calculate the cosine of the angle:
\[ \cos(22^\circ) \]
Then, multiply the force, distance, and the cosine of the angle:
\[ \text{Work} = 50 \times 230 \times \cos(22^\circ) \]
Using a calculator to find \(\cos(22^\circ)\):
\[ \cos(22^\circ) \approx 0.927 \]
Substitute the values back in:
\[ \text{Work} = 50 \times 230 \times 0.927 \approx 10651.5 \, \text{ft-lb} \]
### Answer:
The work done in pulling the toboggan is approximately **10,651.5 ft-lb**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc23443b2-575f-41b5-b981-5d857bb6e2d8%2F6f5f4c08-7507-42fc-a2ac-152a04f22ff0%2Fgwzrlj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Topic:** Work and Energy in Physics
A ski patroller pulls a rescue toboggan across a flat snow surface by exerting a constant force of 50 pounds on a handle that makes a 22° angle with the horizontal (see figure). Determine the work done in pulling the toboggan 230 feet. (Round your answer to one decimal place.)
**Work done = __________ ft-lb**
### Explanation of the Diagram:
The figure shows a ski patroller pulling a rescue toboggan. The ski patroller is depicted on skis, pulling the toboggan using a handle. The handle forms an angle of 22° with the horizontal ground. Various trees, mountains, and clouds compose the background scenery, indicating a mountainous terrain covered with snow.
### Calculations:
To determine the work done, we use the formula:
\[ \text{Work} = \text{Force} \times \text{Distance} \times \cos(\theta) \]
where:
- **Force (F)** = 50 pounds
- **Distance (d)** = 230 feet
- **Angle (θ)** = 22°
First, convert the angle to radians if necessary. However, since typical calculators and computational tools can directly use degrees for cosine functions, we proceed with degrees:
\[ \theta = 22^\circ \]
Calculate the cosine of the angle:
\[ \cos(22^\circ) \]
Then, multiply the force, distance, and the cosine of the angle:
\[ \text{Work} = 50 \times 230 \times \cos(22^\circ) \]
Using a calculator to find \(\cos(22^\circ)\):
\[ \cos(22^\circ) \approx 0.927 \]
Substitute the values back in:
\[ \text{Work} = 50 \times 230 \times 0.927 \approx 10651.5 \, \text{ft-lb} \]
### Answer:
The work done in pulling the toboggan is approximately **10,651.5 ft-lb**.
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