One more problem overcoming friction

[tenchick19]

I just need help with one more problem..

The most massive care ever built was the official car of the General Secretary of the Communist Party in the former Soviet Union. Suppose the car is moving down a ten degree slope when the driver suddenly applies the brakes. The net force acting on the car as it stops is -2.00 x 10^4N. If the coefficient of kinetic friction between the cars tires and the pavement is 0.797, what is the cars mass? What is the magnitude of the normal force that the pavement exerts on the car?

Thanks again!

 

[Nate810]

The car's mass can be calculated using the equation F = ma, where F is the net force of -2.00 x 10^4N, m is the unknown mass of the car, and a is the acceleration due to gravity of 9.8 m/s^2. Solving for m, we get m = -2.00 x 10^4 N / (9.8 m/s^2) = -2.04 x 10^3 kg. The magnitude of the normal force that the pavement exerts on the car can be calculated using the equation Ff = μN, where Ff is the net force of -2.00 x 10^4N, μ is the coefficient of kinetic friction of 0.797, and N is the normal force. Solving for N, we get N = -2.00 x 10^4 N / 0.797 = -2.52 x 10^4 N.

 

[quicknote]

I am glad to help you with this problem. It seems like you are facing a challenge in overcoming friction, which is a common issue in many mechanical systems. In this particular scenario, we are dealing with the friction between the tires of a car and the pavement.

To solve this problem, we need to use the formula for net force, which is Fnet = ma, where Fnet is the net force, m is the mass of the object, and a is the acceleration. In this case, the net force is given as -2.00 x 10^4N, and the acceleration is 0 m/s^2 since the car is coming to a stop. Therefore, we can rearrange the formula to solve for the mass, which gives us m = Fnet/a.

Plugging in the values, we get m = (-2.00 x 10^4N)/(0 m/s^2) = 0 kg. This result seems incorrect, as the car must have some mass to move down the slope. This is where the coefficient of kinetic friction comes into play.

The coefficient of kinetic friction is a measure of the amount of friction between two surfaces in motion. In this case, it tells us that the frictional force between the tires and the pavement is 0.797 times the normal force (Ff = μN). We can use this information to solve for the normal force, which is the force exerted by the pavement on the car.

We know that the normal force and the weight of the car (mg) are equal in magnitude, as the car is not accelerating vertically. Therefore, we can set the normal force equal to the weight of the car, giving us N = mg. Substituting this into the formula for friction, we get Ff = μmg.

Now, we can solve for the mass of the car by equating the frictional force to the net force, which gives us μmg = -2.00 x 10^4N. Rearranging the equation, we get m = (-2.00 x 10^4N)/(μg). Plugging in the values, we get m = (-2.00 x 10^4N)/(0.797 x 9.8 m/s^2) = 2587 kg.

Therefore, the mass of the car is 2587 kg, and the magnitude of the normal force exerted by