A car travels around a circular turn of radius 50 meters while maintaining a constant speed of 15 m/s.

Find the minimum value for the coefficient of friction necessary to keep the car on the road.

I got .40 for the minimum value of friction.

In icy conditions, the coefficient of friction drops to 0.15. Now find the maximum safe speed for making the turn.

I got 8 m/s

Are these right?

centripetal force=mv^2/r

friction force=centripetal force
mg*mu= mv^2/r

mu=v^2/(rg)

b. again, friction force=centripetal force
mg*mu=mv^2/r

v^2=rg*mu
solve for v

To find the minimum value for the coefficient of friction necessary to keep the car on the road, we can start by using the concept of centripetal force. In this case, the centripetal force is provided by the friction force between the tires and the road.

Centripetal force (F_c) can be calculated using the equation:

F_c = (m * v^2) / r

where m is the mass of the car, v is the velocity, and r is the radius of the circular turn.

To find the minimum coefficient of friction (μ), we need to compare the maximum friction force (F_f) to the normal force (F_n). The normal force is equal to the weight of the car (m * g), where g is the acceleration due to gravity.

The maximum friction force can be calculated using:

F_f = μ * F_n

where μ is the coefficient of friction.

Now let's calculate the minimum value for the coefficient of friction:

F_c = F_f

(m * v^2) / r = μ * (m * g)

Simplifying the equation:

v^2 = μ * (r * g)

μ = v^2 / (r * g)

Substituting the given values:

μ = (15 m/s)^2 / (50 m * 9.8 m/s^2)

μ ≈ 0.47

Therefore, the minimum value for the coefficient of friction necessary to keep the car on the road is approximately 0.47. Since you obtained 0.40, your answer is incorrect.

Now let's move on to finding the maximum safe speed for making the turn in icy conditions, where the coefficient of friction drops to 0.15.

Using the same equation for centripetal force:

F_c = (m * v^2) / r

we can substitute the new value of the coefficient of friction (μ = 0.15) and solve for the maximum safe speed (v):

(μ * m * g) = (m * v^2) / r

0.15 * m * 9.8 m/s^2 = (m * v^2) / 50 m

Simplifying the equation:

0.15 * 9.8 m/s^2 = v^2 / 50

v^2 = (0.15 * 9.8 m/s^2) * 50

v^2 = 73.5 m^2/s^2

v ≈ 8.57 m/s

Therefore, the maximum safe speed for making the turn in icy conditions is approximately 8.57 m/s. Since you obtained 8 m/s, your answer is incorrect.