Question

show all you work, including units, on separate paper. follow the \problem solving method\
this final exam is cumulative and covers material from the entire course.
from the list of choices, select the one best answer.
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question 29 of 62
a stone with a mass of 0.45 kg is attached to one end of a string 0.90 m long. the string will break if its tension exceeds 52 n. the stone is spun around in a horizontal circle about its end. what is the maximum speed the stone can attain without breaking?
6.0 m/s
8.0 m/s
10 m/s
12 m/s

Explanation:

Step1: Identify the centripetal force formula

The tension in the string provides the centripetal force for the circular motion. The formula for centripetal force is \( F = \frac{mv^2}{r} \), where \( F \) is the centripetal force (tension here), \( m \) is the mass, \( v \) is the velocity, and \( r \) is the radius of the circle.

Step2: Rearrange the formula to solve for velocity

We need to solve for \( v \). Rearranging the formula \( F = \frac{mv^2}{r} \) gives \( v = \sqrt{\frac{Fr}{m}} \).

Step3: Substitute the given values

We know that \( F = 52 \, \text{N} \), \( m = 0.45 \, \text{kg} \), and \( r = 0.90 \, \text{m} \). Substituting these values into the formula:
\( v = \sqrt{\frac{52 \times 0.90}{0.45}} \)

Step4: Calculate the value inside the square root

First, calculate the numerator: \( 52 \times 0.90 = 46.8 \). Then divide by the mass: \( \frac{46.8}{0.45} = 104 \).

Step5: Take the square root

Now, take the square root of 104: \( v = \sqrt{104} \approx 10.2 \, \text{m/s} \), which is approximately 10 m/s.

Answer:

10 m/s (corresponding to the option "10 m/s")