The coefficient of sliding friction between rubber and asphalt is about \(0.6\). What minimum distance, in meters, can an automobile slide on a horizontal surface if it is traveling at \(25 \mathrm{~m} / \mathrm{s}\) ? a) 38 c) 48 b) 43 d) 53 Thermodynamics

We are given the coefficient of sliding friction between rubber and asphalt (\(\mu = 0.6\)) and the initial speed of the automobile (\(v_0=25 \mathrm{~m}/\mathrm{s}\)). Our goal is to find the minimum distance the car can slide before stopping on a horizontal surface.

Since the car is sliding on a horizontal surface under the effect of friction, the force acting on the car is frictional force. To calculate the acceleration due to friction, first we need to calculate the frictional force acting on the automobile. The frictional force is given by the equation: \(F_f = \mu N\) Since the car is on a horizontal surface, the normal force (N) is equal to the weight of the car (mg). Therefore, the frictional force can be written as: \(F_f = \mu mg\) Now, we can calculate the acceleration due to friction using Newton's second law: \(F = ma\) So, \(F_f = ma\) Substituting the frictional force, \( \mu mg = ma\) Since \(m\) is present on both sides of the equation, we can divide both sides by \(m\) to cancel it out: \( \mu g = a\) Now substitute the given value of \(\mu\) and the gravitational acceleration (\(g=9.8~\mathrm{m/s^2}\)) to find the acceleration due to friction: \(a = 0.6 \times 9.8 = 5.88~\mathrm{m/s^2}\)

To find the stopping distance, we can use the following kinematic equation: \(v^2 =v_0^2 + 2a(x - x_0)\) Since the automobile comes to a stop (v = 0), the equation becomes: \(0 = v_0^2 + 2a(x - x_0)\) Rearrange the equation to solve for \(x - x_0\): \((x - x_0) = -\frac{v_0^2}{2a}\) Now plug in the initial velocity (\(v_0 = 25~\mathrm{m/s}\)) and the acceleration due to friction (\(a = 5.88~\mathrm{m/s^2}\)): \((x - x_0) = -\frac{(25~\mathrm{m/s})^2}{2(5.88~\mathrm{m/s^2})} = 53.06~\mathrm{m}\) This result gives the minimum distance the automobile can slide before coming to a stop: Minimum stopping distance = 53.06 meters (approximately) Comparing this result to our options, we find that it is closest to option d) 53. So, the correct answer is d) 53.