# Must one include friction and gravitational force in the work term?

Example problem:

"A sled has an initial velocity up a ramp of 5 m/s, the ramp has an angle of 20 degrees to the horizontal, and the coefficient of dynamic friction between the sled and the plane is 0.5. Find the maximum height the sled will travel to."

So this is a problem I gave my students, as an energy problem (yes you can do it with dynamics, but I'm focused on energy here) and I tell them that they need to include the friction in their conservation formula as work. Now, what I'm not 100% on is do you take the net work force on the sled or not? That is to say, in the work term, do you take build into it both the force down the ramp by the friction AND the force down the ramp by the earth, or do you just use the frictional force? My reasoning is, if you're including a potential energy term in the conservation of energy formula, then don't include it in the work term because you'll be counting it twice. Is this correct? My model for this problem assumes that the friction is a non-conservative energy, and the gravitational potential energy is both conservative and within the system of the sled.

Any help is greatly appreciated!

Yes and no, but I think in your context the answer is yes: only account for friction work.

This is a good place to consider that energy accounting depends on the choice of system. In introductory courses, potential energy is usually introduced as gravitational potential energy, $mgh$. IMO, this is dangerous because it can lead to confusion, as we see in your case. Kinetic energy is the energy associated with the motion of one object, while potential energy is the energy associated with the configuration or position of a pair of objects. The usual $E=1/2mv^2 + mgh$ implies that your system is the sled and the earth. It also implies that we are modeling our objects as point particles with no internal structure, and no internal energy. Friction causes a transfer of energy out of the system. (You might argue that the thermal energy generated is still in the system, but recall that our system is comprised of two point particles that have no means to carry internal energy.) The change in energy as the sled moves down the ramp is a transfer of energy from one form inside the system to another form inside the system; the work done by gravity is internal work. Transfer of energy into or out of the system is accounted for by heat or external work.

But you can work the problem another way. You can take the system to be the sled alone. This is one point particle. It can have kinetic energy but there is no notion of potential energy for a single particle. Now the work done by gravity is external work, as is the work done by friction. Working the problem this way gives you exactly the same answer, of course; nature doesn't care where we draw our imaginary boundary line defining what we mean by "the system".

Carefully defining the system, the energy of the system (in general: kinetic, potential, thermal, chemical, nuclear, etc, etc), internal work and external work provides a framework for setting up the problem and keeping thoughts organized. Also it's extensible. As you add more forms of energy, and as the concept of heat is introduced, you simply plug these new ideas into the framework.

• This is perfect, the process of defining what is in and out of the system is something that I neglected and is as you say crucial to defining what is going on. Thank you! – vannucci Nov 15 '15 at 14:59

When you have a non conserving force in the system(such as friction) an energy only approach only gives you an estimate of the results. It's not a particularly good way to look at the situation. The initial energy is going to be the kinetic energy.

The resulting energy will be the gravitational potential less the friction losses. The frictional losses will depend on the speed and time traveled, which will be the force applied (kinetic friction) times (instantaneous velocity). This is an an actual work.

To answer your question more directly, it depends on the individual what they define as work. I would reference the gravitational energy as a loss of GPE, but it's all energy.

As the total work on the system equals the change in kinetic energy: $$W_{Friction}+W_{Potential} = \Delta K$$

Taking: $$v_{i}=v_{f}=0$$ we can write

$$W_{Friction}+W_{Potential} = 0$$ $$W_{Friction}+\Delta V = 0$$ $$W_{Friction}-V_{i} + V_{f} = 0$$ $$W_{Friction} + V_{f}= V_{i}$$

## obs:

Work by definition is $$\int \vec{F}.d\vec{s}$$ from this equation one can derive my statement about the work of gravitational force is equal to the potential energy variation and the total work system equals the change in kinetic energy

I did not quite understand your problem in solving a simple question. But i think i get your doubt. The friction when ever taken in the work energy theorem is always taken as negative and as the work is done against the direction of the gravity the work is negative. here do not worry about the potential energy just calculate the work done

when you equate both the side you will get the answer