What is the minimum stopping distance for a car on a wet road?

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To determine the minimum stopping distance for a car on a wet road with a coefficient of static friction of 0.100 while traveling at 50.0 mi/h, the mass of the car is not required as it cancels out in the calculations. The stopping distance can be calculated using the formula that relates friction, acceleration, and velocity. For a dry surface with a coefficient of static friction of 0.600, the stopping distance will be significantly shorter due to higher friction. The discussion emphasizes the importance of understanding the relationship between friction and stopping distance without needing all variables explicitly stated. Proper application of physics principles allows for accurate calculations in both scenarios.
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A car is traveling at 50.0 mi/h on a horizontal highway. If the coefficient of static friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop? What is the stopping distance when the surface is dry and the coefficient of static friction is 0.600?

I'm uncertain how to solve this question because the only relationships between the coefficient of friction and acceleration that I'm aware of require knowing the normal force, and this question does not give the normal force or the mass of the object so the normal force can be calculated.

Any guidance would be appreciated.

Steve
 
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Don't give up so easily. Call the mass of the car "m" and see what happens. If information is not given, perhaps it's not needed. :wink:
 
Figured out "m" cancels out in the equation. Got the correct answer now. Thanks Doc!

Steve
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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