Question

A 950 kg truck accelerates from rest to 15 m/s in 2.3 seconds as it drives down a rough road. If the forward force of the engine is 7000 newtons, determine the force of friction between the tires and the road. 804 N 6,196 N 7,250 N 14,250 N

Answer 1

Answer:

804 N

Explanation:

We are given;

Mass; m = 950 kg

Initial velocity; u = 0 m/s

Final velocity; v = 15 m/s

Time; t = 2.3 s

Engine force; F_e = 7000 N

From Newton's first equation of motion;

v = u + at

15 = 0 + 2.3a

a = 15/2.3

a = 6.5217 m/s²

From Newton's second law of motion and since friction always opposes motion, we have;

F_e - F_f = ma

Where F_f is frictional force.

Thus;

F_f = F_e - ma

F_f = 7000 - (950 × 6.5217)

F_f ≈ 804 N