Question

questions 4 - 5: a car of mass m slides across a patch of ice at a speed v with its brakes locked. it hits dry pavement and skids to a stop in a distance d. the coefficient of kinetic friction between the tires and the dry road is μ. 4. if the car had a mass of 2m, it would have skidded a distance of (a) 0.5d (b) d (c) 1.41d (d) 2d

Explanation:

Step1: Calculate the frictional force

The frictional force \(F_f=\mu N\). Since \(N = mg\) (normal force equals weight for a horizontal surface), \(F_f=\mu mg\).

Step2: Use Newton's second law to find acceleration

By \(F = ma\), \(\mu mg=ma\), so \(a=\mu g\). The acceleration \(a\) is independent of mass \(m\) (because \(a = \mu g\), mass cancels out).

Step3: Use the kinematic equation \(v^{2}=v_{0}^{2}+2ad\)

The car stops, so \(v = 0\). Then \(0 = v_{0}^{2}-2ad\) (negative sign because it's decelerating), and \(d=\frac{v_{0}^{2}}{2a}\). Since \(a=\mu g\) and \(v_0\) (initial speed) is the same in both cases (mass - change situation), and \(a\) is independent of \(m\), the stopping - distance formula \(d=\frac{v_{0}^{2}}{2\mu g}\) shows that \(d\) does not depend on the mass of the car.

Answer:

B. \(d\)