Question

scenario: consider a car of mass m moving with initial speed v0 on a straight, flat road. at time t = 0, the driver fully applies the brakes to avoid colliding with debris in the road in front of the car. the cars wheels lock, causing the car to slide on the roadway until the car stops, before running over the debris. the distance that the car slides is d. the coefficient of kinetic friction between the cars tires and the roadway is a constant value μk. using representations part a: the dot at right represents the car. draw a free - body diagram showing and labeling the forces (not components) exerted on the car, while the car slides to a stop. draw the relative lengths of all vectors to reflect the relative magnitudes of all the forces. each force must be represented by a distinct arrow starting on and pointing away from the dot. note that the car is moving to the right. argumentation part b: the stopping distance d depends on the value of v0 and μk. i. does the value of d increase or decrease with increasing initial speed v0? give a physical explanation why this is the relationship. increase decrease remains the same the value of d increases with ii. does the value of d increase or decrease with increasing coefficient of friction μk? give a physical explanation why this is the case. increase decrease remains the same the value of d decreases by

Explanation:

Step1: Analyze forces for free - body diagram

The car has weight \(F_g = mg\) acting downwards, normal force \(F_N\) acting upwards, and kinetic - friction force \(F_f=\mu_kF_N\) acting opposite to the motion (to the left as the car moves right). Since the car is on a flat road, \(F_N = F_g=mg\) and \(F_f=\mu_kmg\).

Step2: Analyze effect of initial speed on \(D\)

Using the kinematic equation \(v^2 = v_0^2+2aD\), where \(v = 0\) (car stops), and from Newton's second - law \(F = ma\), with \(F=-F_f=-\mu_kmg\), so \(a =-\mu_kg\). Substituting into the kinematic equation gives \(0 = v_0^2-2\mu_kgD\), or \(D=\frac{v_0^2}{2\mu_kg}\). As \(v_0\) increases, \(D\) increases because a higher initial speed means more kinetic energy \(\frac{1}{2}mv_0^2\) that needs to be dissipated by the friction force.

Step3: Analyze effect of \(\mu_k\) on \(D\)

From \(D=\frac{v_0^2}{2\mu_kg}\), as \(\mu_k\) increases, the denominator of the fraction increases. So, \(D\) decreases. A larger coefficient of kinetic friction means a stronger friction force, which dissipates the car's kinetic energy more quickly, resulting in a shorter stopping distance.

Answer:

PART A:

• Draw a dot to represent the car.
• Draw a vector \(F_g\) (weight) pointing downwards.
• Draw a vector \(F_N\) (normal force) pointing upwards of equal length to \(F_g\).
• Draw a vector \(F_f\) (kinetic - friction force) pointing to the left.

PART B:
I. Increase. Explanation: Higher initial speed means more kinetic energy to be dissipated by friction, so stopping distance increases.
II. Decrease. Explanation: Larger \(\mu_k\) gives stronger friction force, dissipating kinetic energy faster and reducing stopping distance.