Forces such as friction and drag cause the macroscopic kinetic energy of a system to be dissipated as thermal energy. Hence these are called dissipative forces. FIGURE 9.23 shows how microscopic interactions are responsible for transforming macroscopic kinetic energy into thermal energy when two objects slide against each other. Because friction causes both objects to get warmer, with increased thermal energy, we must define the system to include both objects whose temperature changes—both the sofa and the floor.
For example, FIGURE 9.24 shows a box being pulled at constant speed across a horizontal surface with friction. As you can imagine, both the surface and the box are getting warmer—increasing thermal energy—but the kinetic energy is not changing. If we define the system to be box + surface, then the increasing thermal energy of the system is entirely due to the work being done on the system by tension in the rope: .
The work done by tension in pulling the box through distance is simply thus . Because the box is moving with constant velocity, and thus no net force, the tension force has to exactly balance the friction force: . Consequently, the increase in thermal energy due to the dissipative force of friction is
Notice that the increase in thermal energy is directly proportional to the total distance of sliding. Dissipative forces always increase the thermal energy; they never decrease it.
You might wonder why we didn’t simply calculate the work done by friction. The rather subtle reason is that work is defined only for forces acting on a particle. There is work being done on individual atoms at the boundary as they are pulled this way and that, but we would need a detailed knowledge of atomic-level friction forces to calculate this work. The friction force is an average force on the object as a whole; it is not a force on any particular particle, so we cannot use it to calculate work. Furthermore, increasing thermal energy is not an energy transfer—the definition of work—from the box to the surface or from the surface to the box; both the box and the surface are gaining thermal energy. The techniques used to calculate the work done on a particle cannot be used to calculate the work done by dissipative forces.
NOTE The considerations that led to Equation 9.27 allow us to calculate the total increase in thermal energy of the entire system, but we cannot determine what fraction of goes to the box and what fraction goes to the surface.
A rope with 30 N of tension pulls a 10 kg crate 3.0 m across a horizontal floor, starting from rest. The coefficient of kinetic friction between the crate and the floor is 0.20. What is the increase in thermal energy? What is the crate’s final speed?
MODEL Let the system consist of both the crate—modeled as an extended object—and the floor. The tension in the rope does work on the system, but the vertical normal force and gravitational force do not.
SOLVE The energy principle, Equation 9.26, is . The friction force on an object moving on a horizontal surface is , so the increase in thermal energy, given by Equation 9.27, is
The tension force does work . 59 J of this goes to increasing the thermal energy, so is the crate’s change of kinetic energy. Because . Using the definition of kinetic energy, we find that the crate’s final speed is
REVIEW The thermal energy of the crate and floor increases by 59 J. We cannot determine for the crate (or floor) alone.