A car moving at a speed of 80 km / h, after turning off the engine, passes 40 m to a stop.
| June 20, 2021 | education
A car moving at a speed of 80 km / h, after turning off the engine, passes 40 m to a stop. Determine the coefficient of friction.
Data: Vн (speed with which the car was moving) = 80 km / h (22.22 m / s); S (way to stop) = 40 m.
Reference data: g (acceleration due to gravity) ≈ 10 m / s2.
Since the work of the friction force is equal to the change in the kinetic energy of the car, the coefficient of friction is expressed from the equality: ΔEk = Atr.
0.5 * m * (Vn ^ 2 – Vk ^ 2) = Ftr. * S.
0.5 * m * Vn ^ 2 – 0 = μ * N * S.
0.5 * m * Vn ^ 2 = μ * m * g * S.
μ = 0.5 * Vn ^ 2 / (g * S).
μ = 0.5 * 22.22 ^ 2 / (10 * 40) ≈ 0.62.
Answer: The coefficient of friction is 0.62.
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