The coefficient of friction between the tires of Shirley's 1985 Ford Coupe and the dry pavement is 0.79. She doesn't drive the car often because it's becoming an antique, but she took it out for a gentle drive yesterday.
While driving, she has to suddenly and forcefully brake when the light turned red. She probably shouldn't have been texting and driving! Her car (mass = 1.37x103 kg) skids to a stop with an acceleration of _____ m/s2. Use the approximation g ≈ 10 m/s2.
10^3*
To find the acceleration of the car while skidding to a stop, we can use Newton's second law of motion, which states that the force exerted on an object is equal to its mass multiplied by its acceleration. In this case, the force is due to the friction between the tires and the pavement.
The frictional force can be calculated using the formula:
Frictional Force = coefficient of friction × normal force
The normal force is the force exerted by the surface on the object perpendicular to the surface. In this case, it is equal to the weight of the car, which can be calculated as:
Weight = mass × acceleration due to gravity
Plugging in the given values:
Weight = (1.37 × 10^3 kg) × (10 m/s^2)
Weight = 1.37 × 10^4 N
Now, we can calculate the frictional force:
Frictional Force = (0.79) × (1.37 × 10^4 N)
Frictional Force = 1.08 × 10^4 N
Since the car is skidding to a stop, the frictional force opposes the car's initial motion. Therefore, the net force acting on the car is equal to the frictional force. According to Newton's second law, this net force is equal to the mass of the car multiplied by its acceleration:
Frictional Force = mass × acceleration
1.08 × 10^4 N = (1.37 × 10^3 kg) × acceleration
Solving for acceleration, we get:
acceleration = (1.08 × 10^4 N) / (1.37 × 10^3 kg)
acceleration ≈ 7.88 m/s^2
Therefore, the car skids to a stop with an acceleration of approximately 7.88 m/s^2.