Physics, 18.01.2021 14:00 puppycity78

Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6. a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?

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Answer from: michaelmontoya64

a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.

b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.

Explanation:

a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:

$\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}$ (1)

$\Sigma F_{y} = N -m\cdot g = 0$ (2)

Where:

$\mu_{s}$ - Static coefficient of friction, dimensionless.

- Normal force from ground to the car, measured in newtons.

- Maximum speed of the car, measured in meters per second.

- Radius of the circular track, measured in meters.

- Mass, measured in kilograms.

- Gravitational acceleration, measured in meters per square second.

By applying (2) in (1):

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$ (3)

The force of static friction the road can apply in the car (), measured in newtons, is: ( $\mu_{s} = 0.6$ , $m = 1500\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ )

$f = \mu_{s}\cdot m \cdot g$

$f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$f = 8826.3\,N$

The force of static friction the road can apply before the car starts to move is 8826.3 newtons.

b) Then, we calculate the maximum speed of the car by (3):

$\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}$

$\mu_{s}\cdot g = \frac{v^{2}}{R}$

$v = \sqrt{\mu_{s}\cdot g\cdot R}$

If we know that $\mu_{s} = 0.6$ , $g = 9.807\,\frac{m}{s^{2}}$ and $R = 200\,m$ , then the maximum speed of the car can travel before it would start to slide is:

$v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}$

$v \approx 34.305\,\frac{m}{s}$

The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.

Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The S

Answer from: Quest

the answer is d. 4

hope this

also the more points you give the faster your questions get answered

Answer from: Quest

hello, i am papaguy.

i have found an answer to your question but use it as information.

information: because weak acids/bases have low concentration of hydrogen ions, the addition of water has a large impact on the ph. 2. a difference in 1 unit is a 10x difference in concentration for example; a liquid with ph of 3 is 10x more acidic than a liquid with a ph of 4. therefore, a liquid with a ph of 3 is stronger.

best of luck!

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