Question

a car moving at 40 km/h is to be stopped by applying brakes in the next 4m. if the car weighs 2000kg what average force must be applied on it?​

Answer 1

Answer:

-30,864 N.

Explanation:

1 km = 1000 m, 1 h = 3600 s, 1 km/h = 1000/3600 = 5/18 m/s

v = 40 * 5/18 = 100/9

This is approximately 11.1 m/s.

vf^2 = vi^2 + 2 * a * d, vf = 0, d = 4

0 = (100/9)^2 + 2 * a * 4

a = -(100/9)^2 ÷ 8

This is approximately -15.432 m/s^2

Average force = mass * acceleration = 2000 * -(100/9)^2 ÷ 8

This is approximately -30,864 N.

Answer 2

Answer:

Explanation:

Given :-

Initial Speed of car , u = 40 km/h = 40 × 5/18 = 11.11 m/s

Final Speed of car , v = 0 (Because breaks applied)

Distance traveled after breaks applied, S = 4 meter

Mass of the car, m = 2000 kg

To Find :-

Average force

Formula to be used :-

Motion 3rd equation i.e  v² - u² = 2as

Force Formula i.e Force = Mass × Acceleration

Solution :-

Putting all the values, we get

⇒ v² - u² = 2as

⇒ 0 - (200/18)² =-2a × 4

⇒ a = 15.42 m/s²

Putting Acceleration value, we get

Force = Mass × Acceleration

Force = 2000 × 15.42

Force = 30840 Newton

Force = 3.08 × 10⁴ Newton.

Hence, the average force must be applied is 3.08 × 10⁴ Newton.