The car drives at a speed of 36 km / h on a smooth road. Having driven 150 m with the engine running

The car drives at a speed of 36 km / h on a smooth road. Having driven 150 m with the engine running, the car stopped. Determine the coefficient of friction and the braking time.

V0 = 36 km / h = 10 m / s.

V = 0 m / s.

g = 10 m / s2.

S = 150 m.

μ -?

With the engine off, the car moves horizontally under the influence of the friction force Ffr. 2 Newton’s law will have the form: m * a = Ftr.

The friction force Ffr is expressed by the formula: Ffr = μ * m * g, where μ is the coefficient of friction, m * g is the force of gravity.

The acceleration of the car a during braking is expressed by the formula: a = (V0 ^ 2 – V ^ 2) / 2 * S. Since the car stopped at the end of braking, its final speed V = 0 m / s, the formula will take the form: a = V0 ^ 2/2 * S.

m * V0 ^ 2/2 * S = μ * m * g.

The friction coefficient μ will be determined: μ = V0 ^ 2/2 * S * g.

μ = (10 m / s) ^ 2/2 * 150 m * 10 m / s2 = 0.03.

Answer: the coefficient of friction of the wheels on the road is μ = 0.03.