The problem asks for the radius of a circular path that an ice skater travels on, given their speed and the coefficient of static friction. Here's how to solve it:

Given:

• Speed of the skater, $v = 4.25 \, \text{m/s}$
• Coefficient of static friction, $\mu_s = 0.05$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

Formula:

The centripetal force that keeps the skater in a circular path is provided by the static friction. The maximum static friction force is given by: $f_s = \mu_s \cdot N$ Where $N$ is the normal force, and on a flat surface, $N = mg$, so: $f_s = \mu_s \cdot m \cdot g$

This frictional force provides the centripetal force necessary to keep the skater in a circular motion: $f_s = \frac{m v^2}{r}$ Where $r$ is the radius of the path, and $m$ is the mass of the skater (which cancels out). Equating the friction force to the centripetal force: $\mu_s \cdot m \cdot g = \frac{m v^2}{r}$ Simplifying: $\mu_s \cdot g = \frac{v^2}{r}$

Now, solve for $r$: $r = \frac{v^2}{\mu_s \cdot g}$

Substituting the values:

$r = \frac{(4.25)^2}{0.05 \cdot 9.8}$ $r = \frac{18.0625}{0.49} \approx 36.86 \, \text{m}$

Thus, the radius of the circular path is approximately 36.86 meters.

Would you like more details or further explanations?

Related Questions:

1. How does the mass of the skater affect the calculation of the radius?
2. What would happen to the radius if the coefficient of friction increased?
3. How would the radius change if the skater increased their speed to 5 m/s?
4. What is the relationship between static friction and the ability to make tight turns?
5. Can the skater maintain a circular path if the ice had no friction at all?

Tip:

When solving problems involving friction and circular motion, always check if the forces balance correctly, particularly when friction is the only source of centripetal force.