A car is moving with a velocity of 16 m/s when the brakes are applied and the wheels lock (stop spinning). The car then slides to a stop in 37 m. Find the coefficient of kinetic friction between the tires and the road.

Initial kinetic energy equals the work done against friction.

(1/2)MVo^2 = M*g*muk*X

muk = Vo^2/(2*g*X) is the kinetic friction coefficient you are looking for. You do the numbers.
X = 37 m
Vo = 16 m/s

To find the coefficient of kinetic friction between the tires and the road, we can use the following formula:

\(a = \frac{{v_f - v_i}}{{t}}\)

where:
\(a\) = acceleration
\(v_f\) = final velocity (0 m/s, since the car came to a stop)
\(v_i\) = initial velocity (16 m/s)
\(t\) = time interval (unknown)

Rearranging the equation to solve for time (\(t\)), we get:

\(t = \frac{{v_f - v_i}}{{a}}\)

Since the car is moving with a constant acceleration while sliding, the acceleration can be given by:

\(a = \frac{{\mu \cdot g}}{2}\)

where:
\(\mu\) = coefficient of kinetic friction
\(g\) = acceleration due to gravity (9.8 m/s\(^2\))

Plugging in the given values, we have:

\(0 = 16 - \frac{{\mu \cdot g}}{2} \cdot t\)

We are given that the car slides to a stop in 37 m, which gives us:

\(t = \frac{{37}}{{v_i}}\)

Now, we can substitute this value of \(t\) into the equation:

\(0 = 16 - \frac{{\mu \cdot g}}{2} \cdot \frac{{37}}{{v_i}}\)

Simplifying further:

\(\mu = \frac{{32 \cdot v_i}}{{37 \cdot g}}\)

Plugging in the values:

\(\mu = \frac{{32 \cdot 16}}{{37 \cdot 9.8}}\)

Solving this equation will give us the coefficient of kinetic friction between the tires and the road.