Question

a car running with velocity 72 kilometres per hour on a level road is stopped after travelling a distance of 30 metres after disengaging its engine the coefficient of friction between the road on the types are​

Answer 1

Hey Dear,

◆ Answer -

μ = 0.67

● Explaination -

# Given -

v = 72 km/h = 20 m/s

s = 30 m

g = 10 m/s

μ = ?

# Solution -

Effective deceleration in stopping the car will be

a = -μg

a = -10μ

According to Newton's 3rd kinematic equation -

v^2 = u^2 + 2as

0^2 = 20^2 + 2(-10μ)×30

μ = 400 / 600

μ = 0.67

Therefore, coefficient of static friction between road and tyres of car is 0.67 .

Best luck...

Answer 2

Explanation:

Hope you understand by this