Doing work against gravity (vertically) and against friction (horizontally) If I am raising an object vertically with a constant force which equals $mg$, the object rises with a constant velocity as gravity and my force offset. In this case, the mechanical energy (gravitational potential energy) increases. However, if I am pushing an object horizontally against friction (the friction equals my pushing force), the object also moves with constant velocity, but without an increase in mechanical energy. While it seems that in both cases I apply a constant force against another constant force, only in different directions, why is there an energy increase in the first case but not the second? Where does the work done by my pushing force go, since the kinetic energy of the object is not increased by it?
 A: This is all a matter of how you choose your system.  In addition, it illustrates some of the conceptual difficulties that arise when dealing with the work done by a frictional force.
First, let's re-analyze the first case, from two different perspectives.

*

*System = object being lifted.  Then, there are two external forces acting: the gravitational force exerted by the Earth, and the normal force exerted by the hand.  These two forces are equal and opposite in this case (since the object is being lifted at constant velocity), and so the works done are equal and opposite, so the net work done on the system is zero.  Therefore, there is no net change in energy of the system, which we already knew because its kinetic energy is constant. Mathematically,
$$
\Delta E_{\textrm{system}} = W_{\textrm{net external}}
= W_{G}+W_n = 0\,.
$$


*System = object being lifted + Earth.  In this case, the only external force is exerted by the hand.  This work is positive, so energy is being added to the system, and it is being added in the form of gravitational potential energy. Mathematically,
$$
\Delta E_{\textrm{system}} = W_{\textrm{net external}}
= W_n = F_n\Delta y\,,
$$
and therefore, since $\Delta K = 0$,
$$
\Delta U_{\textrm{grav}} = \Delta E_{\textrm{system}} = W_n>0\,.
$$
The point is, you need to be careful about how you choose your system.  The correct analogy between the lifted object and the dragged object requires you to choose the system as the object being lifted, because in that case, all of the forces are external, and so the net work done ends up being zero, and no energy is being added to the system.  If you change your perspective and include the Earth in the system, then you compute the change in potential energy instead of the work done by gravity, but you can't do this in the friction case, because friction is not a conservative force, and hence there's no associated potential energy.

Now, what about the case of friction?  In order to do this right, we need to include both the object, and the surface it's sliding against in the system. Then, the frictional force is an internal force, and the math goes like this:


*System = object + surface.  There is one external force, which is the force exerted by the hand.  Thus, there is a net work done on the system, equal to
$$
W_{\textrm{net, external}} = W_{\textrm{by hand}} = F\Delta x\,,
$$
where $\Delta x$ is the magnitude of the displacement of the object.  Thus, the energy increases according to
$$
\Delta E_{\textrm{system}} = W_{\textrm{net, external}} = F\Delta x >0\,.
$$
Since the object is being pulled at a constant velocity, $\Delta K = 0$, and so where is the energy going?  It's not potential energy, because none of these forces are conservative forces.

The answer is that the extra energy is added in the form of thermal energy: the object's (and surface's) temperatures increase as the molecules making up the object and the surface get rattled around as they're dragged along each other.  Microscopically, surface imperfections are "catching" on each other and "releasing", and therefore vibrating, which creates sound waves in the materials that then dissipate into thermal energy.


*If you try to take System = object, you will run into difficulties when thinking about the work done by friction, exactly because the increase in thermal energy vanishes. That is, if you naively perform the calculation
$$
W_{\textrm{net external}} = W_{\textrm{pull}} + W_{\textrm{friction}}\,,
$$
you will get zero, since the forces are equal and opposite, and the displacements are the same.  But this suggests that the net change in internal energy is zero!  The problem is that you have to be careful with the "type" of work you are computing. See the paper here.


To conclude, the (thermodynamically) correct statement of the principle of conservation of energy takes the form
$$
\Delta K + \Delta U + \Delta E_{\textrm{thermal}} = \Delta E_{\textrm{system}} = W_{\textrm{net, external}}
$$
in the absence of heating.
