You are driving 2500 Kg car at a constant speed of 14 m/s along an icy, but stright and lvel road. While approaching a traffic light, it turns red. You slam on the brakes. You wheels lock, the tires begin skidding and the car slides to a halt in a distance of 25 m. What is the coefficient of sliding friction between the tires and the icy road?
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Calculations:
m = 2500 Kg
Vi = 14 m/s
Vf = 0 m/s
deltaD = 25 m
a = (vf^2 - vi^2)/(2 x deltaD)
= -14 m/s / 50
= 2.8 m/s^2
W = (2500) (9.81)
= 24525 N
Fnet = (2500)(-0.28)
= -700 N
I`m not sure what to do after this.
dont forget its 14^2/50 - -3.92m/s^2
(friction force) x (distance of skid) = Kinetic energy loss
Use that equation to get the friction force, f.
The coefficient of kinetic friction, Uk, can be obtained from
Uk = f/(M*g)
oh ok thanks :)
Well, it seems like you're stuck in an icy situation! Let me help you slide towards a solution with a little humor.
To find the coefficient of sliding friction between the tires and the icy road, we need to use the formula:
Fnet = m * a
Where Fnet is the net force acting on the car, m is the mass of the car, and a is the acceleration. Now, since the car is stopping and the tires are locked, the frictional force is the only force acting on the car in the opposite direction. So we can write:
Fnet = -μ * m * g
Where μ is the coefficient of sliding friction and g is the acceleration due to gravity. Now let's plug in the values we have:
-700 N = -μ * 2500 kg * 9.81 m/s^2
Now it's time for some math fun! Let's solve for μ:
μ = -700 N / (2500 kg * 9.81 m/s^2)
μ = -0.028
Negative mass and negative friction coefficient? That's just my way of clowning around! So, when we remove the negatives, we get a coefficient of sliding friction between the tires and the icy road of approximately 0.028.
Remember, this answer is just an approximation based on the given information. So, stay safe on slippery roads and don't forget to bring your clown nose!
To find the coefficient of sliding friction between the tires and the icy road, you need to use the equation of motion for constant acceleration.
First, calculate the net force acting on the car using Newton's second law: Fnet = ma, where Fnet is the net force, m is the mass of the car, and a is the acceleration. In this case, the net force is the force due to friction, which opposes the motion of the car.
Given that the net force is -700 N (negative because it opposes motion), and the mass of the car is 2500 kg, you can rearrange the equation to solve for the acceleration:
a = Fnet / m = (-700 N) / (2500 kg) = -0.28 m/s^2
Now, use this acceleration value to calculate the coefficient of sliding friction (μ) between the tires and the icy road. The coefficient of friction is given by the equation μ = a / g, where g is the acceleration due to gravity (approximately 9.81 m/s^2).
μ = a / g = (-0.28 m/s^2) / (9.81 m/s^2) ≈ -0.0284
However, since the coefficient of friction cannot be negative, you should take the absolute value of μ.
μ ≈ 0.0284
Therefore, the coefficient of sliding friction between the tires and the icy road is approximately 0.0284.