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

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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.
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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