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# College Physics

Science and Technology

## Nonconservative Forces

Tác giả: OpenStaxCollege

# Nonconservative Forces and Friction

Forces are either conservative or nonconservative. Conservative forces were discussed in Conservative Forces and Potential Energy. A nonconservative force is one for which work depends on the path taken. Friction is a good example of a nonconservative force. As illustrated in [link], work done against friction depends on the length of the path between the starting and ending points. Because of this dependence on path, there is no potential energy associated with nonconservative forces. An important characteristic is that the work done by a nonconservative force adds or removes mechanical energy from a system. Friction, for example, creates thermal energy that dissipates, removing energy from the system. Furthermore, even if the thermal energy is retained or captured, it cannot be fully converted back to work, so it is lost or not recoverable in that sense as well.

# How Nonconservative Forces Affect Mechanical Energy

Mechanical energy may not be conserved when nonconservative forces act. For example, when a car is brought to a stop by friction on level ground, it loses kinetic energy, which is dissipated as thermal energy, reducing its mechanical energy. [link] compares the effects of conservative and nonconservative forces. We often choose to understand simpler systems such as that described in [link](a) first before studying more complicated systems as in [link](b).

# How the Work-Energy Theorem Applies

Now let us consider what form the work-energy theorem takes when both conservative and nonconservative forces act. We will see that the work done by nonconservative forces equals the change in the mechanical energy of a system. As noted in Kinetic Energy and the Work-Energy Theorem, the work-energy theorem states that the net work on a system equals the change in its kinetic energy, or ${W}_{\text{net}}=\text{ΔKE}$. The net work is the sum of the work by nonconservative forces plus the work by conservative forces. That is,

${W}_{\text{net}}={W}_{\text{nc}}+{W}_{\text{c}},$

so that

${W}_{\text{nc}}+{W}_{c}=\text{Δ}\text{KE},$

where ${W}_{\text{nc}}$ is the total work done by all nonconservative forces and ${W}_{\text{c}}$ is the total work done by all conservative forces.

Consider [link], in which a person pushes a crate up a ramp and is opposed by friction. As in the previous section, we note that work done by a conservative force comes from a loss of gravitational potential energy, so that ${W}_{\text{c}}=-\text{Δ}\text{PE}$. Substituting this equation into the previous one and solving for ${W}_{\text{nc}}$ gives

${W}_{\text{nc}}=\text{Δ}\text{KE}+\text{Δ}\text{PE.}$

This equation means that the total mechanical energy $\left(\text{KE + PE}\right)$ changes by exactly the amount of work done by nonconservative forces. In [link], this is the work done by the person minus the work done by friction. So even if energy is not conserved for the system of interest (such as the crate), we know that an equal amount of work was done to cause the change in total mechanical energy.

We rearrange ${W}_{\text{nc}}=\text{Δ}\text{KE}+\text{Δ}\text{PE}$ to obtain

${\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}+{W}_{\text{nc}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}\text{.}$

This means that the amount of work done by nonconservative forces adds to the mechanical energy of a system. If ${W}_{\text{nc}}$ is positive, then mechanical energy is increased, such as when the person pushes the crate up the ramp in [link]. If ${W}_{\text{nc}}$ is negative, then mechanical energy is decreased, such as when the rock hits the ground in [link](b). If ${W}_{\text{nc}}$ is zero, then mechanical energy is conserved, and nonconservative forces are balanced. For example, when you push a lawn mower at constant speed on level ground, your work done is removed by the work of friction, and the mower has a constant energy.

# Applying Energy Conservation with Nonconservative Forces

When no change in potential energy occurs, applying ${\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}+{W}_{\text{nc}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}$ amounts to applying the work-energy theorem by setting the change in kinetic energy to be equal to the net work done on the system, which in the most general case includes both conservative and nonconservative forces. But when seeking instead to find a change in total mechanical energy in situations that involve changes in both potential and kinetic energy, the previous equation ${\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}+{W}_{\text{nc}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}$ says that you can start by finding the change in mechanical energy that would have resulted from just the conservative forces, including the potential energy changes, and add to it the work done, with the proper sign, by any nonconservative forces involved.

Calculating Distance Traveled: How Far a Baseball Player Slides

Consider the situation shown in [link], where a baseball player slides to a stop on level ground. Using energy considerations, calculate the distance the 65.0-kg baseball player slides, given that his initial speed is 6.00 m/s and the force of friction against him is a constant 450 N.

Strategy

Friction stops the player by converting his kinetic energy into other forms, including thermal energy. In terms of the work-energy theorem, the work done by friction, which is negative, is added to the initial kinetic energy to reduce it to zero. The work done by friction is negative, because $\mathbf{\text{f}}$ is in the opposite direction of the motion (that is, $\theta =\text{180º}$, and so $\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =-1$). Thus ${W}_{\text{nc}}=-\text{fd}$. The equation simplifies to

$\frac{1}{2}{{\text{mv}}_{i}}^{2}-\text{fd}=0$

or

$\text{fd}=\frac{1}{2}{{\text{mv}}_{i}}^{2}\text{.}$

This equation can now be solved for the distance $d$.

Solution

Solving the previous equation for $d$ and substituting known values yields

$\begin{array}{lll}d& =& \frac{{{\text{mv}}_{i}}^{2}}{2f}\\ & =& \frac{\left(\text{65.0 kg}\right)\left(6\text{.}\text{00 m/s}{\right)}^{2}}{\left(2\right)\left(\text{450 N}\right)}\\ & =& \text{2.60 m.}\end{array}$

Discussion

The most important point of this example is that the amount of nonconservative work equals the change in mechanical energy. For example, you must work harder to stop a truck, with its large mechanical energy, than to stop a mosquito.

Calculating Distance Traveled: Sliding Up an Incline

Suppose that the player from [link] is running up a hill having a $5\text{.}\text{00º}$ incline upward with a surface similar to that in the baseball stadium. The player slides with the same initial speed. Determine how far he slides.

Strategy

In this case, the work done by the nonconservative friction force on the player reduces the mechanical energy he has from his kinetic energy at zero height, to the final mechanical energy he has by moving through distance $d$ to reach height $h$ along the hill, with $h=d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}5.00º$. This is expressed by the equation

$\text{KE}{}_{\text{i}}\text{}+{\text{PE}}_{\text{i}}+{W}_{\text{nc}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}\text{.}$

Solution

The work done by friction is again ${W}_{\text{nc}}=-\text{fd}$; initially the potential energy is ${\text{PE}}_{i}=\text{mg}\cdot 0=0$ and the kinetic energy is ${\text{KE}}_{i}=\frac{1}{2}{{\text{mv}}_{i}}^{2}$; the final energy contributions are ${\text{KE}}_{f}=0$ for the kinetic energy and ${\text{PE}}_{f}=\text{mgh}=\text{mgd}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ for the potential energy.

Substituting these values gives

$\frac{1}{2}{{\text{mv}}_{i}}^{2}+0+\left(-\text{fd}\right)=0+\text{mgd}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\mathrm{\theta .}$

Solve this for $d$ to obtain

$\begin{array}{lll}d& =& \frac{\left(\frac{1}{2}\right){{\text{mv}}_{\text{i}}}^{2}}{f+\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta }\\ & =& \frac{\text{(0.5)}\left(\text{65.0 kg}\right)\left(\text{6.00 m/s}{\right)}^{2}}{\text{450 N}+\left(\text{65.0 kg}\right)\left({\text{9.80 m/s}}^{2}\right)\phantom{\rule{0.25em}{0ex}}{\text{sin (5.00º)}}^{}}\\ & =& \text{2.31 m.}\end{array}$

Discussion

As might have been expected, the player slides a shorter distance by sliding uphill. Note that the problem could also have been solved in terms of the forces directly and the work energy theorem, instead of using the potential energy. This method would have required combining the normal force and force of gravity vectors, which no longer cancel each other because they point in different directions, and friction, to find the net force. You could then use the net force and the net work to find the distance $d$ that reduces the kinetic energy to zero. By applying conservation of energy and using the potential energy instead, we need only consider the gravitational potential energy $\text{mgh}$, without combining and resolving force vectors. This simplifies the solution considerably.

# Section Summary

• A nonconservative force is one for which work depends on the path.
• Friction is an example of a nonconservative force that changes mechanical energy into thermal energy.
• Work ${W}_{\text{nc}}$ done by a nonconservative force changes the mechanical energy of a system. In equation form, ${W}_{\text{nc}}=\text{Δ}\text{KE}+\text{Δ}\text{PE}$ or, equivalently, ${\text{KE}}_{\text{i}}+{\text{PE}}_{\text{i}}+{W}_{\text{nc}}={\text{KE}}_{\text{f}}+{\text{PE}}_{\text{f}}$.
• When both conservative and nonconservative forces act, energy conservation can be applied and used to calculate motion in terms of the known potential energies of the conservative forces and the work done by nonconservative forces, instead of finding the net work from the net force, or having to directly apply Newton’s laws.

# Problems & Exercises

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-m-high rise as shown in [link]. Find her final speed at the top, given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)

9.46 m/s

(a) How high a hill can a car coast up (engine disengaged) if work done by friction is negligible and its initial speed is 110 km/h? (b) If, in actuality, a 750-kg car with an initial speed of 110 km/h is observed to coast up a hill to a height 22.0 m above its starting point, how much thermal energy was generated by friction? (c) What is the average force of friction if the hill has a slope $2\text{.}5º$ above the horizontal?