Non-Conservation of Energy

Section 8.8 Non-Conservation of Energy

We found above that if only conservative forces act on a system, then, energy is conserved during its motion. A conservative force was found to have work that has no path-dependence. Work by a non-conservative force, such as drag or friction, does depend on path. To handle non-conservative forces in an energy consideration, we separate net work in the work-energy theorem into contributions of conservative and non-conservative forces.

\begin{equation*} W_{if}^{\text{cons}} + W_{if}^{\text{non-cons}} = K_f - K_i, \end{equation*}

where the superscript “cons” refers to the conservative part and “non-cons” to the non-conservative part. We know that the work by conservative forces will be equal to the difference of initial and final potential energies.

\begin{equation*} \sum U_i - \sum U_f + W_{if}^{\text{non-cons}} = K_f - K_i, \end{equation*}

where \(\sum U \) stands for the sum of potential energies from all the conservative forces involved, not just one. Now, we can send the \(U\)’s to the right side, where we bundle them with the corresponding kinetic energy term.

\begin{equation*} W_{if}^{\text{non-cons}} = \left(K_f + \sum U_f\right) - \left(K_i+\sum U_i\right). \end{equation*}

The \(\left( K + \sum U \right)\) is energy, so we replace it by \(E \text{,}\) the symbol for energy, and rewrite this equation as

\begin{equation} W_{if}^{\text{non-cons}} = E_f - E_i.\tag{8.43} \end{equation}

That is, any change in energy, either increase or decrease, comes from the work by the nonconservative forces. Sometimes, we call this the general principle of conservation of energy.

Example 8.56. The Impact of Drag Force on the Energy of a Skier Going Down a Slope.

A \(70\text{-kg}\) skier starts from rest on top of a \(40\text{ m}\) high hill. When he reaches the bottom of the hill he is going at \(20\text{ m/s}\text{.}\) How much energy has he lost to air and friction?

Answer.

\(13,500\text{ J} \text{.}\)

Solution.

We look at the difference in energy.

\begin{align*} \Delta E \amp = (K_f + 0) - (0 +U_i) =\dfrac{1}{2}m\,v_f^2 - m\,g\,h_i \\ \amp = \dfrac{1}{2}\times 70\times 20^2 - 70\times 9.81\times 40 = -13,500\text{ J}, \end{align*}

where I have rounded off the answer to 3 digits. The negative value of \(\Delta E\) means a loss in energy. This is the work done by the nonconservative forces, which are air drag and friction.

Exercises Exercises

1. Determining Conservation of Energy in the Motion of a Skydiver.

A sky diver exits an air plane at almost zero speed. He falls a distance of \(50\text{ m}\text{.}\) His speed at that point is \(10\text{ m/s}\text{.}\) Show calculations to decide if his energy is conserved or not.

Hint.

Find the sign of \(\Delta E\text{.}\)

Answer.

Not conserved

Solution.

We compare energy at the later instant with the energy when the skydiver fell from the plane.

\begin{align*} \Delta E \amp = E_f - E_i = (K_f + 0) - (0 +U_i) \\ \amp = m\left( \dfrac{1}{2}\times 10^2 - 9.81\times 50 \right) = -440.5\, m\lt 0. \end{align*}

This says that the skydiver lost energy in his motion. That is, the energy of skydiver is not conserved.

2. Power Needed to Overcome Friction.

A crate is being pushed across a rough floor surface. If no force is applied on the crate, the crate will slow down and come to a stop. The friction is such that if the crate of mass \(50 \text{ kg}\) moving at speed \(8 \text{ m/s}\) comes to rest in \(10 \text{ sec}\) at constant deceleration. You need to continue to push on the crate if you want the motion to continue at constant speed \(8\text{ m/s}\text{.}\) What would be the work done by your force in one second, also called power of the force?

Hint.

In one second, the displacement \(d = v\text{.}\)

Answer.

\(320 \text{ J/s}\text{.}\)

Solution.

From the data on deceleration, we find that the friction force is

\begin{equation*} F_k = 50 \text{ kg}\times \dfrac{ 8 \text{ m/s} - 0 }{10 \text{sec}} = 40\text{ N}. \end{equation*}

We need to supply the energy at the rate equal to the rate at which energy is removed by this friction force. The distance traveled in \(1\text{ sec}\) will be equal to \(v\text{.}\) Therefore, work per second, i.e., power \(P\text{,}\) will be

\begin{equation*} P = F_k\times v = 40\text{ N} \times 8\text{ m/s} = 320\text{ J/s}. \end{equation*}

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