physics i need help

Question

A 3,500kg car rests on rough horizontal ground. A rope is attached to the car and is pulled with a force of 12,000N to the left. As a result, the car accelerates at 2.9 m/s^2. The coefficient of sliding friction (μ) between the car and the ground is ___. 

Answer 1

There are 4 forces that act on the car

·       Weight mg – direction down

·       Normal reaction force N – direction up

·       External force F = 12 000 N – direction to the left

·       Frictional force f – direction to the right (opposite to the direction of motion)

Let’s select coordinate system with x-axis directed to the left and y-axis perpendicular to it up.

In scalar form Newton’s second law for

                                                                x:   F – f = ma                       (1)

                                                                 y:   N – mg = 0                     (2)

We know that friction f = ϻN, where ϻ is coefficient of kinetic (sliding) friction. Using this formula and equation (2) we can write:

                                                             f = ϻmg

Putting it in to equation (1) we get:

                                                               F - ϻmg = ma

Expressing coefficient of friction, we will obtain:

                                                                                    ϻ = (F – ma) / mg

Let’s plug in the numerical values of all variables, then

                                                          ϻ = ( 12 000 N – 3 500 kg x 2.9 m/s²) / 3 500 kg x10 m/s² = 0.05

                                                                                                Answer: 0.05