meerany
A flat (unbanked) curve on a highway has a radius of 220.0
Ill. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum
coefficient of friction that will prevent sliding? (b) Suppose the highway is
icy and the coefficient of friction between the tires and pavement is only
one-third what you found in part (a). What should be the maximum speed of
the car so it can round the curve safely?
can some one help me in this question..
Centripetal force
=mv²/r
friction force
=μmg
Equate and solve for μ
μ=mv²/r / mg
=v²/gr
For the second part, reduce μ to one-third, and try the reverse calculation.
i just had to do this for my one class, so i’ll try to explain it.
using the equation, substitute the numbers from the equation in.
so you should have 25 / 220 * 9.8.
multiply 220 by 9.8 to get 2,156. then divide 25 by 2,156 to get 0.01159.
ANSWER: 0.011
Certainly! Let's break down each part of the question step by step.
(a) To find the minimum coefficient of friction that will prevent sliding, we can use the centripetal force equation:
F = (mv^2) / r
Where:
F = Centripetal force
m = Mass of the car
v = Velocity of the car
r = Radius of the curve
In this case, the centripetal force is provided by the friction force between the car's tires and the road. So, we can rewrite the equation as:
F_f = (mv^2) / r
Now, we need to solve for the coefficient of friction (μ) using the formula:
F_f = μN
Where:
F_f = Friction force
μ = Coefficient of friction
N = Normal force (equal to the weight of the car)
Since the curve is flat, the normal force is equal to the weight of the car (mg).
Combining the two equations, we have:
F_f = μmg
Setting the two expressions for F_f equal to each other:
(μmg) = (mv^2) / r
Now, we can cancel out the mass (m) from both sides of the equation:
μg = (v^2) / r
Finally, we can solve for the coefficient of friction (μ):
μ = (v^2) / (rg)
Plugging in the given values:
v = 25.0 m/s
r = 220.0 m
μ = (25.0^2) / (220.0 × g)
We can assume acceleration due to gravity (g) is approximately 9.8 m/s^2.
μ = (625.0) / (2156.0) ≈ 0.290
Therefore, the minimum coefficient of friction that will prevent sliding is approximately 0.290.
(b) Now, let's calculate the maximum speed of the car that can safely round the curve on icy pavement. We are given that the coefficient of friction on icy pavement is only one-third of what we found in part (a).
So, the maximum speed can be determined using the same equation as before:
μ = (v^2) / (rg)
Now, we substitute the new value for μ:
1/3 * 0.290 = (v^2) / (220.0 × g)
Simplifying this equation:
v^2 = (1/3 * 0.290 * 220.0 × g)
v^2 ≈ 18.84 × g
Taking the square root of both sides:
v ≈ √(18.84 × g)
Plugging in the value of g (approximately 9.8 m/s^2):
v ≈ √(18.84 × 9.8)
v ≈ √184.872
v ≈ 13.60 m/s
Therefore, the maximum speed at which the car can safely round the curve on icy pavement is approximately 13.60 m/s.