Respuesta :
Answer:
50%
Explanation:
For the particular case we assume that the centripetal force is equal to the frictional force. Starting from this, we realize that while the car takes a curve, which allows it to not leave the track, it is the friction force that adheres it to the floor, in this way,
[tex]F_c = F_f[/tex]
The values of these equations are given by,
[tex]F_c = \frac{mv^2}{R}[/tex]
[tex]F_f = \mu N = \mu mg[/tex]
Equating the terms
[tex]\frac{mv^2}{R} = \mu mg[/tex]
[tex]v = \sqrt{\mu Rg}[/tex]
Both gravity and radius are values that during the trajectory with constants, so
[tex]v \approx \sqrt{\mu}[/tex]
The difference comes in winter and dry, so,
\mu_{dry} = \mu
as in winter it is a quarter, you have to,
\mu_{winter} = \frac{1}{4} \mu_{dry} = \frac{1}{4} \mu
Performing the proportion we have to
[tex]\frac{\mu_{winter}}{\mu_{dry}} = \frac{\sqrt{\mu/4}}{\mu}[/tex]
[tex]\frac{\mu_winter}{\mu_{dry}} = \frac{1}{2} = 0.5[/tex]
We can conclude that the new value for this speed is 50% of its value on a dry day.
The new value for the velocity in winter is 70% of its value on a dry day.
What is centripetal force?
Centripetal force is the force applied to the object due to its circular motion.
We know that during the Curve the friction force should be equal to the centripetal force of the car in order to keep the car on track. So,
[tex]F_C = F_f[/tex]
We know the values of friction force and the centripetal force,
[tex]\dfrac{mV^2}{R}= \mu N\\\\\dfrac{mV^2}{R}= \mu mg\\\\\dfrac{V^2}{R}= \mu g\\\\V^2 = \mu g R\\\\V = \sqrt{\mu g R}[/tex]
Since the value of the acceleration due to gravity is g and the radius of the curve will be constant, therefore,
[tex]V =\sqrt{\mu}[/tex]
The coefficient of friction between the road the tire is μ, therefore, the coefficient of friction in the winter is,
[tex]\mu_{dry} = \mu\\\\\mu_{winter} = \frac{1}{4} \mu_{dry} \\\\\mu_{winter}= \frac{\mu_{dry} }{4}\\[/tex]
Now calculate the ratio of the two velocities we get.
[tex]\dfrac{V_{dry}}{V_{winter}} = \sqrt{\dfrac{\mu_{dry}}{\mu_{winter}}}\\\\\\\dfrac{V_{dry}}{V_{winter}} = \sqrt{\dfrac{\mu_{dry}}{\frac{\mu_{dry}}{2}}}\\\\\\\dfrac{V_{dry}}{V_{winter}} = \sqrt{2}\\\\\dfrac{V_{dry}}{V_{winter}} = 1.4142\\\\V_{dry}0.7 = V_{winter}[/tex]
Hence, the new value for the velocity in winter is 70% of its value on a dry day.
Learn more about Centripetal Force:
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