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!

 

[member 553083]

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.