Question

A motor cyclist along with the machine(vehicle) weighs 160 kg. While driving at 72Kmh-¹, he stops his machine(vehicle) over a distance of 8m. Find the retarding force of the brakes.

a) 4000dynes
b) 4000N
c) 3000N
d) 3000dynes

Answer 1

Given:-

• Mass of motor cycle ,m = 160kg

• Initial velocity ,u = 72×5/18 = 20m/s

• Final velocity ,v = 0m/s

• Distance,s = 8m

To Find:-

• Retardation Force ,F

Solution:-

Firstly we calculate the acceleration of the Motor bike.

By using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = 20² + 2×a×8

→ 0 = 400 + 16×a

→ -400 = 16×a

→ a = -400/16

→ a = -25m/s²

Here Negative sign show retardation

Now, Calculating the Force

• F = m×a

Substitute the value we get

→ F = 160 ×(-25)

→ F = -4000 N

Here negative sign show Force is acting in opposite directions.

Therefore, the Force acting if the brake is 4000 Newton.

Hence, Option (b) is correct.

Answer 2

ʳᵉᶠᵉʳ ᵗᵒ ᵗʰᵉ ᵃᵗᵗᵃᶜʰᵐᵉⁿᵗ !!

ʰᵒᵖᵉ ⁱᵗ ʰᵉˡᵖˢ :)

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