Question

The 2.5-Mg van is traveling with a speed of 100 km>h when the brakes are applied and all four wheels lock. If the speed decreases to 40 km>h in 5 s, determine the coefficient of kinetic friction between the tires and the road.

Answer 1

Answer:

0.34

Explanation:

2.5 Mg = 2500 kg

The change in speed from 100 km/h to 40 km/h is

$\Delta v = 100 - 40 = 60 km/h = 60 * 1000 / (60 * 60) = 16.67 m/s$

The deceleration caused by friction force is the change in speed per unit of time

$a = \Delta v / \Delta t = 16.67 / 5 = 3.33 m/s^2$

Using Newton 2nd law we can calculate the friction force that caused this deceleration:

F = ma = 2500 * 3.33 = 8333 N

Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient

$F = mg\mu$

$8333 = 2500*9.8\mu$

$\mu = \frac{8333}{2500 * 9.8} = 0.34$