Friction Calculation During Vehicle Braking

This solution determines the coefficient of kinetic friction ($\mu_k$) for a vehicle experiencing emergency braking.

Determine Vehicle Deceleration

Using the kinematic equation $v^2 = u^2 + 2as$ with initial velocity ($u = 12$ m/s), final velocity ($v = 0$ m/s), and distance ($s = 15$ m):

$ v^2 = u^2 + 2as $

$ 0^2 = (12 \text{ m/s})^2 + 2 \times a \times (15 \text{ m}) $

$ 0 = 144 \text{ m}^2/\text{s}^2 + 30a \text{ m} $

Solving for acceleration ($a$):

$ a = \frac{-144 \text{ m}^2/\text{s}^2}{30 \text{ m}} = -4.8 \text{ m/s}^2 $

The magnitude of the vehicle's deceleration is $4.8$ m/s$^2$. The negative sign indicates the acceleration opposes the velocity.

Calculate Coefficient of Kinetic Friction

The deceleration is caused by the kinetic friction force ($F_f$). The force is related to the coefficient of kinetic friction ($\mu_k$) and the normal force ($N$) by $F_f = \mu_k N$. On a level road, $N = mg$, where $m$ is the mass and $g$ is the acceleration due to gravity.

According to Newton's second law, the net force equals mass times acceleration ($F_{net} = ma$). Here, the friction force provides the net force, acting opposite to motion:

$ ma = -F_f $

$ ma = -\mu_k mg $

Cancelling mass ($m$) from both sides, we get the relationship between deceleration magnitude and friction:

$ |a| = \mu_k g $

Now, we calculate $\mu_k$ using the determined deceleration ($|a| = 4.8$ m/s$^2$) and the given value of $g = 10$ m/s$^2$:

$ \mu_k = \frac{|a|}{g} $

$ \mu_k = \frac{4.8 \text{ m/s}^2}{10 \text{ m/s}^2} $

$ \mu_k = 0.48 $

The coefficient of kinetic friction is $0.48$.