A coefficient of kinetic friction, k , is needed to stop a car in a distance D on dry pavement with the wheels locked if the car starts from speed V_0.
What coefficient of kinetic friction (in terms of k ) is needed to stop this car in the same distance if it starts from an initial speed of 3V_0?
9K
Correct. It varies as V^2. The force must increase by 9 times to dissipate nine times as much energy as "work done against friction" in the same distance.
To solve this problem, we can start by understanding the relationship between the coefficient of kinetic friction (k), the distance (D), and the initial speed (V₀) required to stop a car on dry pavement.
We know that the force of kinetic friction (Fk) can be calculated using the equation:
Fk = k * N
where N is the normal force exerted by the car on the pavement. In this case, since the car's wheels are locked, the force of kinetic friction is responsible for bringing the car to a stop.
The normal force (N) can be calculated as the product of the car's mass (m) and the acceleration due to gravity (g):
N = m * g
Next, we need to determine the relationship between the force of kinetic friction and the car's stopping distance D.
The work done by the force of kinetic friction (Wk) is equal to the product of the force (Fk) and the displacement (D).
Wk = Fk * D
Also, work can be defined as the change in kinetic energy of the car:
Wk = (1/2) * m * (V₀² - 0)
Equating these two expressions for work, we have:
Fk * D = (1/2) * m * V₀²
Now we can substitute the expressions for Fk and N into the equation:
(k * N) * D = (1/2) * m * V₀²
Substituting N = m * g:
(k * m * g) * D = (1/2) * m * V₀²
Simplifying, we find:
k * g * D = (1/2) * V₀²
Finally, we can determine the relationship between the required coefficient of kinetic friction (k') when the initial speed is 3V₀ and the original coefficient of kinetic friction (k).
To find k', we need to substitute V₀ with 3V₀ in the equation:
k' * g * D = (1/2) * (3V₀)²
k' * g * D = (1/2) * 9V₀²
k' = (1/2) * 9
k' = 4.5
Therefore, the coefficient of kinetic friction (k') needed to stop the car in the same distance when it starts from an initial speed of 3V₀ is 4.5 times the original coefficient of kinetic friction (k).