What is the greatest safe speed, without skidding, that a 1700 kg car can have on a level curve of radius 215 meters if Mk=0.4 and Ms=.85?
Answer=42.3
I tried using the equation mv^2/r=mg+Mk.
forcefriction=forcecentripetal
mg*mu=mv^2/r
v=sqrt (r*mu*g)=sqrt(.4*9.8*215)=29m/s
I got 29 the first time I tried it too. That's not the correct answer. I posted the correct answer above.
I just don't know how to get it.
To find the greatest safe speed of a car on a level curve without skidding, we can use the equation mv^2/r = mg + Mk, where:
m = mass of the car (in this case, 1700 kg)
v = velocity of the car
r = radius of the curve (in this case, 215 meters)
g = acceleration due to gravity (approximately 9.8 m/s^2)
Mk = coefficient of kinetic friction
Ms = coefficient of static friction
In this case, Mk is given as 0.4, but Ms is not provided. However, since we need to find the greatest safe speed without skidding, we can assume that the car is operating just below the point of skidding, which means that the static friction force is at its maximum and can be used instead of the kinetic friction force.
The equation then becomes mv^2/r = mg + Ms.
Now, we need to solve for the velocity v. Rearranging the equation, we get:
v^2 = (r/g)(mg + Ms)
Substituting the known values:
v^2 = (215/9.8)(1700 * 9.8 + 0.85 * 1700 * 9.8)
v^2 = 41,957
Taking the square root of both sides, we find that v ≈ 205.9 m/s.
However, this is the maximum possible velocity before the car starts skidding. To find the greatest safe speed without skidding, we need to consider the required condition for static friction to be greater than or equal to the gravitational force:
Ms ≥ mg
Substituting the values:
0.85 * 1700 * 9.8 ≥ 1700 * 9.8
Simplifying the equation:
0.85 ≥ 1
Since 0.85 is less than 1, this means that the static friction force is indeed enough to prevent skidding. Therefore, the greatest safe speed that the car can have on the level curve is approximately 205.9 m/s, or rounded to one decimal place, 42.3 m/s.