Dynamics Example 5 © 2000 D.P. Maloney

A skier who is gliding along at 6 m/s starts down an incline. She descends through a vertical distance of 60 m. If the skier has a mass of 55 kg, the angle of the incline is 40°, and her final speed is 30 m/s, what is the coefficient of friction between the skis and the snow?

Physical Principles and/or Ideas: Conservation of energy, work done against frictional

force, definition of kinetic energy and gravitational potential energy and definition of frictional force, i.e. relationship between the frictional force and the gravitational force.

Solution: We are going to solve this problem using the principle of conservation of energy taking account of the fact that work is a form of energy. If the situation were frictionless, then the total energy of the skier at the top would equal the skier’s total energy after descending the 60 m. But there is some friction between the skis and the snow, so some of the skier’s initial total energy will go into doing work against friction. So, we need to determine how much energy went into work. That is where we apply the conservation of energy principle. The difference between the skier’s total energy at the top and her total energy after descending the 60 m is the work done against friction. Notice carefully that it is the total energy that is conserved. At the top, the skier’s total energy is part kinetic and part potential. To make our job simpler, we will choose the lower level as the reference level for the potential energy. That means the skier will have zero potential energy at that point, so her total energy at the lower level is all kinetic.

Calculating these, we get

for the total energy at the top, and

for the total energy at the lower level. The difference between these two is the work done against friction.

Now the work done against friction is the product of the frictional force and the distance through which it acted. The frictional force is the product of the coefficient of friction and the normal force. The normal force exerted by the incline surface is equal in magnitude (which is all we really need here) to the perpendicular component of the gravitational force, i.e. mgCOS 40°. (How do we know this? Hint: how does the skier move in the direction perpendicular to the surface?) So the frictional force is

Next, we have to find the distance the skier moves in the direction of the frictional force, i.e. along the incline. This is not the 60 m, because that is a vertical distance. The diagram below shows the situation.

Using some trig, we know that . Applying this to this situation, we get:

Finally, substituting all of this back into our equation, we have

where the mass cancels out since it appears in all of the terms. Since we know everything in this equation except the coefficient of friction, we can solve for it. The value we get is: