4. During a road test, a driver brakes a 1420 kg car moving at 64.8 km/h [W]. The car slows down and comes to a stop after moving 729 m [W].
(a) Calculate the net force acting on the car. (ans: 316 N [E]
(b) What is the force of friction acting on the car while it is slowing down? Explain
your reasoning. [ans: 316 N [E]]

a. -312.4 N or 312.4 N [E] b. -312.4 N or 312.4 N [E]

Explanation:

a. Using v² = u² + 2as where u = initial velocity of car = 64.8 km/h = 64.8 × 1000 m/3600 s = 18 m/s, v = final velocity of car = 0 m/s since the car stops, a = deceleration of car and s = distance car moved = 729 m.

So, making the acceleration subject of the formula, we have

a = (v² - u²)/2s

substituting the values of the variables, we have

a = ((0 m/s)² - (18 m/s)²)/(2 × 729 m)

= - (18 m²/s²)/(2 × 729 m)

= -324 m²/s² ÷ 1458 m

= -0.22 m/s²

So, the net force , F acting on the car is F = ma where m = mass of car = 1420 kg and a = deceleration = -0.22 m/s².

So, F = ma = 1420 kg × -0.22 m/s² = 312.4 N or 312.4 N [E]

b. Since no forward force acts on the car while it is slowing down, the frictional force equals the force acting to stop the car, f = ma which is equal to the net force where m = mass of car = 1420 kg and a = deceleration = -0.22 m/s²

So, f = ma = 1420 kg × -0.22 m/s² = -312.4 N or 312.4 N [E]