Account for energy when friction acts

Begin with the missing mechanical energy

A 2 kg block moves right at 4 m/s on a horizontal track. Its initial kinetic energy is ½ × 2 × 4² = 16 J. Suppose the track exerts a constant 4 N sliding resistance. Over the first 2 m, friction does −4 × 2 = −8 J of work on the block’s translational motion. The block therefore has 8 J of kinetic energy left.

The mechanical energy has decreased. That does not mean 8 J has ceased to exist. In the teaching model, contact transforms that energy into internal energy of the block and track together. The combined increase is 8 J. We assume negligible energy transfer to the surroundings and neglect sound and wear. Those assumptions must be reconsidered in a real investigation.

Before moving the controls, predict the energy allocation after 0, 2 and 4 m. Use resistance 4 N and save the three states. At those distances the kinetic energies are 16, 8 and 0 J; combined internal-energy increases are 0, 8 and 16 J. At 4 m the block stops. Waiting afterward does not increase sliding distance or produce further frictional dissipation in this model.

Choose the boundary before writing an equation

For the block’s translational motion, use Kfinal − Kinitial = Wfriction. K is kinetic energy in joules and Wfriction is the signed work associated with the friction force along the displacement. Here Wfriction = −fd, where f is the positive resistance magnitude in newtons and d is forward traveled distance in metres. The negative sign follows from force opposing displacement.

For an enlarged system containing the block and track, use Kfinal + ΔEinternal = Kinitial. ΔEinternal is the combined increase in their internal energy, not its absolute value. The track is fixed, so its bulk kinetic energy is neglected. No height change or spring compression occurs, so there is no changing gravitational or elastic store in this scenario.

These equations express the same modeled transfer from different viewpoints. Do not subtract 8 J of friction work and then subtract another 8 J of internal-energy increase from the same initial 16 J when calculating final K. That counts one transfer twice. State which quantities belong inside the boundary and which describe transfer across it.

The OpenStax energy treatment distinguishes mechanical conservation from cases involving nonconservative work. The numerical investigation here is an original synthetic scenario with an explicitly enlarged energy ledger.

Connect resistance to the contact model

The normal force is 20 N because the 2 kg block has no vertical acceleration, no extra vertical applied force, and the teaching gravitational acceleration is 10 m/s². The kinetic-friction coefficient is μ = f/N: positive resistance divided by normal-force magnitude. At f = 4 N, μ = 4/20 = 0.20. The coefficient has no units.

The controls choose resistance directly, and the ledger shows the implied coefficient. Changing resistance compares separate track conditions with the same initial block speed. It is not a measured material property or a simulation of microscopic contact. The OpenStax friction model describes the distinction between sliding and static contact and the approximate coefficient rule.

At target 2 m, change resistance to zero. K remains 16 J, speed remains 4 m/s and internal-energy increase is zero. This ideal comparison has no finite stopping point. It is not a claim that a real track has exactly zero resistance.

Read the graph as a journey

The solid line gives K along the actual forward journey. The dashed line gives ΔEinternal along that same journey. At every physical point their heights add to 16 J. The fixed energy scale allows fair comparisons when resistance changes. A circle and square identify the evaluated state, so the two quantities do not depend on color alone.

The dotted vertical line marks the requested target. If it lies beyond the stop, there is no physical journey to that line. The curves end at the stop instead of extending to negative kinetic energy. The ledger labels the actual stop separately and leaves target speed and target time unavailable. A zero-energy point at the stop is not evidence that a farther target was reached.

With resistance 6 N, the stop occurs at 16/6 ≈ 2.667 m. Request 3 m and inspect the warning. The formal expression 16 − 6 × 3 = −2 J rejects the target. The actual stopped state has K = 0 and ΔEinternal = 16 J, at 2.667 m. The next lesson develops this stopping prediction.

Energy increase is not a thermometer reading

The calculation gives an energy increase for the two contacting bodies together. It does not determine which body receives which fraction, the temperature distribution near contact, or either body’s temperature rise. Those questions require additional information such as heat capacity, participating mass, conduction, contact time and transfer to the environment.

Calling the entire 8 J increase ‘heat in the block’ would obscure both the system boundary and the mechanism. Heat is energy transferred because of a temperature difference; frictional conversion is not automatically such a transfer. The model records internal-energy increase caused by the loss of bulk motion and does not resolve later thermal transfers.

Transfer the accounting

A different block starts with 30 J of kinetic energy. It slides 3 m against a constant 5 N resistance, with the same closed block–track assumptions. Predict remaining kinetic energy and combined internal-energy increase. Then decide what changes if a later measurement shows that 4 J has left the combined system for the surroundings.

Check the dissipative-energy transfer

The slide reduces kinetic energy by 5 × 3 = 15 J, leaving K = 15 J and initially increasing combined internal energy by 15 J. If 4 J subsequently leaves without changing the bulk motion, the retained internal-energy increase is 11 J. The expanded balance is 15 J kinetic + 11 J retained internal increase + 4 J transferred out = 30 J. Neither balance specifies a temperature or a split between block and track.