Question

when brakes are applied to a car moving with a velocity of 54 km^-1 it comes to rest within 2 metre. calculate the frictional resistance if mass of the car is 150 kg​

Answer 1

Answer :-

Frictional resistance is 8437.5 Newtons .

Explanation :-

We have :-

→ Initial velocity (u) = 54 km/h

→ Distance covered (s) = 2 m

→ Mass of the car (m) = 150 kg

→ Final velocity (v) = 0 m/s

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Firstly, let's convert the initial velocity of the car from 'km/h' to 'm/s' .

⇒ 1 km/h = 5/18 m/s

⇒ 54 km/h = 54(5/18) m/s

⇒ (3 × 5) m/s

⇒ 15 m/s

Let's calculate the acceleration of the car (a) by using the 3rd equation of motion .

v² - u² = 2as

⇒ 0 - (15)² = 2a(2)

⇒ -225 = 4a

⇒ a = -225/4

⇒ a = -56.25 m/s²

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Now, we can calculate the frictional resistance (force) by using Newton's 2nd law of motion .

F = ma

⇒ F = 150(-56.25)

⇒ F = -8437.5 N

[Here, -ve sign represents resistive force ] .

Answer 2

Answer:

The frictional resistance if mass of the car is 150 kg​ = -8437.5N

Explanation:

We have :

→ Initial velocity (u) = 54 km/h

→ Distance covered (s) = 2 m

→ Mass of the car (m) = 150 kg

→ Final velocity (v) = 0 m/s

________________________________

Firstly, let's convert the initial velocity of the car from 'km/h' to 'm/s' .

We know that,

$\Rightarrow \boxed{\bold{ 1kmph}= \frac{5}{18}\:\bold{ m/s}}$

$\Rightarrow 54kmph = 54[\frac{5}{18}]\: m/s$

⇒ (3 × 5) m/s

⇒ 15 m/s

Let's calculate the acceleration of the car (a) by using the 3rd equation of motion .

$\bold{\underline{v^2-u^2 = 2as}}$

⇒ 0 - (15)² = 2a(2)

⇒ -225 = 4a

⇒ a = -225/4

⇒ a = -56.25 m/s²

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Now, we can calculate the frictional resistance (force) by using Newton's 2nd law of force

$\boxed{\bold{F=ma}}$

⇒ F = 150(-56.25)

⇒ F = -8437.5 N