Question

You are driving a 2420.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car's wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 26.8 m.
What is the coefficient of kinetic friction between your tires and the wet road?

Answer 1

Answer:

$U_k$ = 0.3731

Explanation:

First we will identify the important data of the question.

M = 2420 kg

V = 14 m/s

d = 26.8 m

$U_k =$ ?

So, we will use the law of the conservation of energy, it says that:

$E_i - E_f = W_f$

therefore:

$E_i = \frac{1}{2}MV^2\\E_f = 0\\W_f = F_kd$

where $F_k$ is the friction force

Replacing on the first equation, we get:

$\frac{1}{2}MV^2 -0 = F_kd$

$F_k$ is also equal to $U_kN$

where N is the normal force and Uk is the coefficient of kinetic friction.

solving the equation:

$\frac{1}{2}(2420)(14)^2 = U_kN(26.8)$

Before solve for $U_k$ we need to know the value of N, so we use the law of newton as:

∑$F_y$ = N - (2420)(9.8m/s) = 0

N = 23716

Finally, just solve for $U_k$ as:

$U_ k = \frac{\frac{1}{2}(2420)(14)^2 }{(26.8)(23716)}$

$U_k$ = 0.3731