10.2.4 Motion with Gravity and Friction

What if there’s friction? You learned in 《 SECTION 9.5 that friction increases the thermal energy of the system—defined to include both objects—by $Δ E_{th} = f_{k} Δ s$ . For an isolated system with both gravitational potential energy and friction, the energy principle becomes

Δ K + Δ U_{G} + Δ E_{th} = 0

(10.13)

or, equivalently, the energy conservation statement is

K_{i} + U_{Gi} = K_{f} + U_{Gf} + Δ E_{th}

(10.14)

Mechanical energy $K + U_{G}$ is not conserved if there is friction. Because $Δ E_{th} > 0$ (friction always makes surfaces hotter, never cooler), the final mechanical energy is less than the initial mechanical energy. That is, some fraction of the initial kinetic and potential energy is transformed into thermal energy during the motion. Friction causes objects to slow down, and motion ceases when all the mechanical energy has been transformed into thermal energy. Mechanical energy is conserved only when there are no dissipative forces and thus $Δ E_{th} = 0$ .

NOTE We can write the energy principle in terms of initial and final values of the kinetic energy and the potential energy, but not the thermal energy. Objects always have thermal energy—the atoms are constantly in motion—but we have no way to know how much. All we can calculate is the change in thermal energy.

Although mechanical energy is not conserved, the system’s energy is. Equation 10.13 tells us that the sum of kinetic, potential, and thermal energy—the energy of the system—does not change as the object moves on a surface with friction. The initial mechanical energy does not disappear; it’s merely transformed into an equal amount of thermal energy.

EXAMPLE 10.4 Skateboarding up a ramp

During the skateboard finals, Isabella encounters a 6.0-m-long, $15 °$ upward ramp. Isabella’s mass, including the skateboard, is 55 kg, and the coefficient of rolling friction between her wheels and the ramp is 0.025. With what speed must she start up the ramp to reach the top at 2.5 m/s? What percentage of her mechanical energy is “lost” to friction?

MODEL Let the system consist of the earth (including the ramp) and Isabella on the skateboard.

VISUALIZE FIGURE 10.10 shows a before-and-after pictorial representation. We’ve used the ramp’s dimensions and trigonometry to determine Isabella’s final position $y_{1}$ .

The figure illustrates the pictorial representation of the initial and final positions of Isabella skating up a ramp inclined at 15 degrees.

FIGURE 10.10 Full Alternative Text

SOLVE Isabella’s kinetic energy is transformed into potential energy as she gains height, but some of her kinetic energy is also transformed into increased thermal energy of her wheels and the ramp because of rolling friction. The energy principle including friction is

\begin{array}{l} K_{i} + U_{Gi} & = \frac{1}{2} m v_{0}^{2} + 0 \\ = K_{f} + U_{Gf} + Δ E_{th} = \frac{1}{2} m v_{1}^{2} + m g y_{1} + f_{r} Δ s \end{array}

where we’ve used rolling friction $f_{r}$ rather than the kinetic friction of sliding. Rolling friction is $f_{r} = μ_{r} n$ , and recall—from Chapter 6—that the normal force of an object on a slope is $n = m g cos θ$ . (Draw a free-body diagram if you’re not sure.) Thus

\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v_{1}^{2} + m g y_{1} + μ_{r} m g Δ s cos θ

The mass cancels. Solving for Isabella’s speed at the bottom of the ramp, we find

v_{0} = \sqrt{v_{1}^{2} + 2 g y_{1} + 2 μ_{r} g Δ s cos θ} = 6.3 m / s

Isabella’s initial mechanical energy is entirely kinetic energy: $K_{0} = \frac{1}{2} m v_{0}^{2} = 1090 J$ . The thermal energy of the ramp and her wheels increases by $Δ E_{th} = μ_{r} m g Δ s cos θ = 78 J$ . Thus the percentage of mechanical energy transformed into thermal energy as Isabella ascends the ramp is

\frac{78 J}{1090 J} \times 100 = 7.2 %

This energy is not truly lost—it’s still in the system—but it’s no longer available for motion.

REVIEW The ramp is $1.55 m \approx 5 ft$ high. Starting up the ramp at $6.3 m / s \approx 12 mph$ in order to reach the top at $2.5 m / s \approx 5 mph$ seems reasonable.