A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictionless tabletop. If the maximum speed the mass can have without breaking the string is 30 m/s. What is the maximum tension that the string can withstand without breaking? 450 N 400 N 350 N 300 N O O O O
A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictionless tabletop. If the maximum speed the mass can have without breaking the string is 30 m/s. What is the maximum tension that the string can withstand without breaking? 450 N 400 N 350 N 300 N O O O O
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: Circular Motion and Tension**
A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictionless tabletop. If the maximum speed the mass can have without breaking the string is 30 m/s, what is the maximum tension that the string can withstand without breaking?
- 450 N
- 400 N
- 350 N
- 300 N
**Explanation:**
To solve this problem, we need to calculate the maximum tension in the string, which acts as the centripetal force required to keep the toy moving in a circle. The formula for centripetal force is:
\[ F = \frac{mv^2}{r} \]
Where:
- \( F \) is the centripetal force (tension in this context),
- \( m \) is the mass of the toy (0.50 kg),
- \( v \) is the velocity (30 m/s),
- \( r \) is the radius of the circle (1.0 m).
Substitute the known values:
\[ F = \frac{0.50 \times (30)^2}{1.0} \]
Calculate:
\[ F = \frac{0.50 \times 900}{1.0} \]
\[ F = 450 \, \text{N} \]
Thus, the correct answer is **450 N**. The maximum tension that the string can withstand without breaking is 450 N.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4e5d9b3-811d-494f-a46a-3fa1730e8ad6%2Ffad6499f-52f8-476f-8640-0d4d58c12204%2Fy4mw5a_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem: Circular Motion and Tension**
A 0.50-kg toy is attached to the end of a 1.0-m very light string. The toy is whirled in a horizontal circular path on a frictionless tabletop. If the maximum speed the mass can have without breaking the string is 30 m/s, what is the maximum tension that the string can withstand without breaking?
- 450 N
- 400 N
- 350 N
- 300 N
**Explanation:**
To solve this problem, we need to calculate the maximum tension in the string, which acts as the centripetal force required to keep the toy moving in a circle. The formula for centripetal force is:
\[ F = \frac{mv^2}{r} \]
Where:
- \( F \) is the centripetal force (tension in this context),
- \( m \) is the mass of the toy (0.50 kg),
- \( v \) is the velocity (30 m/s),
- \( r \) is the radius of the circle (1.0 m).
Substitute the known values:
\[ F = \frac{0.50 \times (30)^2}{1.0} \]
Calculate:
\[ F = \frac{0.50 \times 900}{1.0} \]
\[ F = 450 \, \text{N} \]
Thus, the correct answer is **450 N**. The maximum tension that the string can withstand without breaking is 450 N.
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