Final answer:
The coefficient of friction acting on the brakes of a 750kg car that stops from a speed of 25.0 m/s in 4.8 seconds is 0.53.
Step-by-step explanation:
To determine the coefficient of friction acting on the brakes of a 750kg car that comes to a stop from a speed of 25.0 m/s in 4.8 seconds, we can use the formula that relates force, mass, acceleration, and friction. First, we'll calculate the deceleration (negative acceleration) of the car using the change in velocity over time, which is (0 - 25.0 m/s) / 4.8 s = -5.21 m/s2. The force required to stop the car is then given by F = ma, so F = 750kg * -5.21 m/s2 = -3907.5 N.
The normal force (N) acting on the car is equal to the weight of the car, which is mass times gravity (N = mg), N = 750kg * 9.8m/s2 = 7350 N. The coefficient of friction (μ) is the ratio of the force of friction to the normal force, μ = Ff / N. Since the force required to stop the car is the force of friction, we can solve for μ: μ = -3907.5 N / 7350 N = -0.53. However, since the coefficient of friction cannot be negative, we consider the absolute value, thus μ = 0.53.