A 1000 kg car is going 8.90 m/s has a minimum braking distance of 6.0 meters. Find the average braking force exerted on the car during the stopping process.
a=v²/2s
F=ma
24
To find the average braking force exerted on the car during the stopping process, we can use Newton's second law of motion, which states that force is equal to the mass of an object multiplied by its acceleration. In this case, the acceleration is given by the change in velocity divided by the time taken to stop.
First, we need to find the deceleration of the car. We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (8.90 m/s), a is the acceleration (deceleration in this case), and s is the distance traveled (6.0 meters).
Rearranging the equation, we get a = (v^2 - u^2) / (2s). Plugging in the values, we have: a = (0^2 - 8.90^2) / (2 * 6.0) = -105.755 m/s^2. Since the car is decelerating, the acceleration is negative.
Now, we can calculate the average braking force using Newton's second law: F = m * a, where F is the force, m is the mass of the car (1000 kg), and a is the acceleration (-105.755 m/s^2).
Plugging in the values, we have: F = 1000 kg * (-105.755 m/s^2) = -105755 N. The negative sign indicates that the force is in the opposite direction to the motion of the car, as it is a braking force.
Therefore, the average braking force exerted on the car during the stopping process is 105,755 N.