Question

On a snowy day, when the coefficient of friction μs between a car’s tires and the road is 0.50, the maximum speed that the car can go around a curve is 20 mph. What is the maximum speed at which the car can take the same curve on a sunny day when μs=1.0?

Answer 1

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

$F=\mu mg = m \frac{v^2}{r}$ (1)

where

$\mu$ is the coefficient of friction

m is the mass of the car

g = 9.8 m/s^2 is the acceleration due to gravity

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

$\mu=0.50\\v = 20 mph = 8.9 m/s$

So the radius of the curve is

$r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m$

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when $\mu=1.0$. We find:

$v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph$

Answer 2

The maximum speed at which the car can take the same curve on a sunny day is about 28 mph

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Further explanation

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

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Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

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Given:

coefficient of friction on a snowy day = μs₁ = 0.50

maximum speed of the car on a snowy day = v₁ = 20 mph

coefficient of friction on a sunny day = μs₂ = 1.0

Asked:

maximum speed of the car on a snowy day = v₂ = ?

Solution:

Firstly , we will derive the formula to calculate the maximum speed of the car:

$\Sigma F = ma$

$f = m \frac{v^2}{R}$

$\mu N = m \frac{v^2}{R}$

$\mu m g = m \frac{v^2}{R}$

$\mu g = \frac{v^2}{R}$

$v^2 = \mu g R$

$\boxed {v = \sqrt { \mu g R } }$

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Next , we will compare the maximum speed of the car on a snowy day and on the sunny day:

$v_1 : v_2 = \sqrt { \mu_1 g R } : \sqrt { \mu_2 g R }$

$v_1 : v_2 = \sqrt { \mu_1 } : \sqrt { \mu_2 }$

$20 : v_2 = \sqrt { 0.50 } : \sqrt { 1.0 }$

$20 : v_2 = \frac{1}{2} \sqrt{2}$

$v_2 = 20 \div \frac{1}{2} \sqrt{2}$

$v_2 = 20 \sqrt{2} \texttt{ mph}$

$\boxed{v_2 \approx 28 \texttt{ mph}}$

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Learn more

• Impacts of Gravity : https://brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
• The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Circular Motion