Air Resistance for Sliding Object

Any help would be appreciated. Thanks.

The force due to gravity is the mass of the object times the gravitational acceleration, 9.8 meters per second per second at the Earth's surface, times the sine of the plane's angle of incline.

The force due to sliding friction is the mass of the object times the gravitational acceleration times the cosine of the plane's angle of incline times the coefficient of kinetic friction.

The coefficient of kinetic friction depends on the nature of the two surfaces sliding over each other and must be determined experimentally or approximated by some reasonable guess. In the real world this coefficient is so sensitive to conditions that you may not get the same value on Thursday that you got on Sunday if you try to measure it.

The force due to air resistance may be approximated by a formula called Stokes law which says that air resistance force is proportional to density of the air times the cross sectional area of the object times the square of the velocity of the object.

The force of gravity aiding the motion is precise and constant. The force of sliding friction may be imprecise but at least it is constant. The force of air resistance is the source of the mathematical difficulties in the formula. It lacks precision but worse then that it depends on velocity which changes constantly throughout the experiment.

To calculate the distance the object covers before it comes to rest requires that we break the path up into small enough slices that the change in velocity over that distance may be neglected. Calculate the net force on the object using the velocity applicable at that instant and find the energy lost during that little slice of the path by multiplying the net force times the length of that path segment. When the sum of all such energy losses equals the energy available, the object will stop.

Due to the position dependence of the object's velocity, the formula you are looking for becomes a non-linear first order differential equation. The solution to such equations requires the application of lots of calculus and more than a little luck. If what you want is the answer, I recommend you use our DynaLab program . It will produce a graph of the motion for you. If what you want is to understand how to solve the problem. I have explained the physics for you. The math is outside the scope of a simple email. For that you will need to find a good differential equations text and work through it.

Please let me know how this explanation suits you. If I have missed the mark I will try again.

JDJones

M. Casco Associates