Barron's 2nd Edition DT Q29

Barron's Book

29) When an object is traveling in a circular path, we call the force acting to the center, the centripetal force. This question says that there is friction which allows the car to travel in a circular path. Let's draw a visual of this.

Where f is frictional force

The diagram above shows that at any point on the path, the frictional force acts as the centripetal force. This means we can set up an equation equating the two.

Where F[c] is centripetal force, f is frictional force, m is mass, v is speed, r is radius, N is normal force, μ is coefficient of friction, and g(≈ 10 m/s^2) is the acceleration due to gravity

All we need to do left is plug in values and solve for speed(v).

Where r is radius, g is acceleration due to gravity, μ is the coefficient of friction, and v is speed.

As a note, we need to take g as positive because we were actually using it for measure normal force, which is in the opposite direction of gravity.

Answer: B