Apply conservation of energy to systems with friction, multiple energy forms and measured efficiency

Learning goal: State the principle of conservation of energy for a defined system and time interval, translate between verbal descriptions, energy diagrams and equations, use free body diagrams and Newton's laws to find the friction force that removes energy, solve for final speeds and heights when several energy forms change at once, and calculate the efficiency of a real energy-transforming system.

The principle, stated carefully. For any defined system and any chosen time interval, the total initial energy of the system plus the energy that enters equals the total final energy plus the energy that leaves: E_initial + E_in = E_final + E_out. Energy comes in forms that look very different (kinetic, gravitational potential, elastic potential, thermal, and others), but every one of them is measured in joules, so the amount that disappears from one form can be matched against the amount that appears in another. The principle holds for any system and interval you care to define. The accounts in this lesson cover events with no nuclear changes, so they leave out one term, the energy tied up in mass itself; that term changes measurably only in nuclear processes, and the nuclear energy lesson adds it (energy is conserved there too). Choosing the system is a decision, not a fact of nature: you may define the system as the block plus ramp, and then the energy dissipated by friction stays inside as thermal energy of both surfaces; or you may track only the block's mechanical energy, and then the negative work of friction shows how much mechanical energy the block lost, part of which warms the block itself and part the ramp. Different choices give different bookkeeping and the same physics.

Three representations of one event. A situation can be described in words ("the cart rolls down, speeds up, and coasts to a stop on the carpet"), as an energy diagram (bar charts showing each form at chosen moments, whose total is constant unless energy crosses the system boundary), or as an equation (½mv₁² + mgh₁ + ½kx₁² + W_ext = ½mv₂² + mgh₂ + ½kx₂²), where W_ext is the work done on the system by external forces such as friction (negative when it removes energy). Being able to move between the three is the skill: a bar chart that does not add up signals that a form of energy or a transfer has been left out.

External forces and friction: find them with a free body diagram. Friction is the usual route by which mechanical energy leaves a system. To include it you need its size, and that comes from a free body diagram and Newton's laws, not from the energy equation. On a level floor the block does not accelerate vertically, so the normal force equals the weight, N = mg, and kinetic friction is f = μ_k N. On an incline this course does not calculate friction from a coefficient: the friction force is either given or measured directly on the ramp (pull the block up the ramp at constant speed with a spring scale held parallel to the surface; the free body diagram along the ramp gives pull = mg sin θ + f, so f = pull − mg sin θ). The work friction does over a distance d along the surface is −f d. Worked model: predicting where a block stops. A 0.40 kg block is released from rest at the top of a 1.0 m long ramp inclined at 30° (height 0.50 m) and slides onto a level floor. The friction force on the ramp has been measured as 0.68 N, and μ_k = 0.20 on the floor. Define the system as the block plus Earth, so friction is external. Energy leaving the field on the way down: mgh = (0.40)(9.8)(0.50) = 1.96 J. Work by friction on the ramp: −(0.68)(1.0) = −0.68 J. Kinetic energy at the bottom: 1.96 − 0.68 = 1.28 J, so v = √(2 × 1.28/0.40) = 2.5 m/s. On the floor: N = 3.92 N, f = 0.784 N, and the block stops when friction has removed the remaining 1.28 J: d = 1.28/0.784 = 1.6 m from the foot of the ramp. The prediction can be tested directly; discrepancies come from assumptions worth listing: the friction force may not be constant along the ramp or the floor, the transition at the foot of the ramp costs some energy, the block may tip or bounce there, and the measured lengths, friction force and coefficient carry uncertainty. How close counts as agreement depends on those uncertainties; as a suggested classroom guide (not a published standard), a prediction within about 10 % of the measured stopping point, judged against the spread of your own trials, supports the claim that the model captures the main transfers.

Efficiency. Real systems deliver only part of the energy put in as the useful form. Efficiency = useful energy out ÷ energy in, usually as a percentage. For the ramp as a device for turning field energy into kinetic energy at the bottom, efficiency = 1.28/1.96 = 65 %. Proposed refinements (a lubricant, a different surface material) may reduce the friction force; each is a prediction to test, and the efficiency before and after can be compared with the same calculation. (Reducing the contact area does not reduce friction in the simple model of friction.) The energy that is "lost" has not vanished: it is thermal energy in the block and the surfaces, which the mechanical bookkeeping treats as leaving the system.

Several forms at once. Problems may combine elastic, gravitational and kinetic energy. A spring with k = 200 N/m compressed 0.10 m launches a 0.050 kg ball straight up. Initial elastic energy: ½(200)(0.10)² = 1.0 J. Ignoring air resistance, at the top all of it is in the ball-Earth field: mgh = 1.0 J, so h = 1.0/((0.050)(9.8)) = 2.0 m above the ball's position at full compression. (Measure h from there, not from the spring's relaxed length: the ball has already risen 0.10 m while the spring was still pushing it, and that rise is part of mgh.) With air resistance the ball rises less, and the shortfall in mgh measures the energy the air carried away. Energy methods also let you find an unknown quantity: if a launcher spring whose force vs. compression graph is a straight line is compressed 0.080 m and launches a 0.020 kg ball straight up to a highest point 1.2 m above its position at full compression, then ½ k (0.080)² = (0.020)(9.8)(1.2) = 0.235 J gives k = 73.5 N/m, which you can compare with a direct measurement of k from the slope of that graph. (A rubber band is a poor choice for this: its graph curves and it gives back less energy than was put in, so it has no single k, only an effective stiffness for one particular stretch.) Stating the uncertainty in each measurement (a few millimeters in stretch is a several-percent change in k, because k depends on x²) is part of the answer.

Misconception: friction destroys energy. Friction converts ordered kinetic energy into disordered thermal energy of the two surfaces; the total is unchanged. The block stops, but the floor and block are slightly warmer. Conservation of energy is not violated; it is what lets you calculate where the block stops.

Investigation: predict the stopping point. Materials: two wooden blocks of different mass (or one block with a second face of a different material), a ramp with an adjustable angle (a board on books), a meter stick, a spring scale, a protractor. First measure μ_k on the floor: drag the block at constant speed with the spring scale held horizontal; f equals the scale reading, N equals the block's weight, μ_k = f/N; repeat three times and average. Measure the ramp angle and length, then measure the friction force on the ramp directly: pull the block up the ramp at constant speed with the scale parallel to the surface and subtract mg sin θ from the reading; repeat three times and average. Predict the stopping distance using the energy account above, then release the block from rest and measure where it stops; repeat five times and average. Repeat everything for the second sliding object. Record the prediction, the measurements, the percent difference and a list of assumptions that could explain the discrepancy. What the model cannot show: how the thermal energy is shared between block and floor, or what happens to an object that rolls. A can rolling without slipping is not slowed by sliding friction at all (it loses energy to a much smaller rolling resistance and also carries rotational kinetic energy), so this sliding model does not apply to it. Extension, finding an unknown: launch a small ball straight up with a spring launcher whose spring gives a straight force vs. compression graph, measure the compression, the ball's mass and the height reached above the ball's position at full compression, calculate k from ½ k x² = mgh, and compare it with k from the slope of the graph, stating the uncertainty of each. Safety: keep fingers away from the foot of the ramp; do not use a ramp higher than waist level; for the extension wear safety glasses and never aim the launcher toward anyone.

Worked example 1 — block down a ramp onto a floor. Given above: 0.40 kg, ramp 1.0 m at 30°, measured ramp friction 0.68 N, μ_k = 0.20 on the floor. Field energy released 1.96 J; ramp friction removes 0.68 J; kinetic energy at the bottom 1.28 J; speed 2.5 m/s; floor friction 0.784 N; stopping distance 1.6 m. Energy diagram: at the top one bar, 1.96 J gravitational; at the bottom, 1.28 J kinetic plus 0.68 J transferred out as thermal; at rest, 1.96 J thermal. Each column totals 1.96 J.

Worked example 2 — a roller coaster with friction. A 500 kg car passes the top of a 20 m hill at 2.0 m/s. Between the top and a point 5.0 m above the ground, friction does −8000 J of work on it. Energy at the top: ½(500)(2.0)² + (500)(9.8)(20) = 1000 + 98 000 = 99 000 J. At 5.0 m: gravitational term (500)(9.8)(5.0) = 24 500 J, so kinetic energy = 99 000 − 24 500 − 8000 = 66 500 J and v = √(2 × 66 500/500) = √266 = 16.3 m/s. Without friction it would be √(2 × 74 500/500) = 17.3 m/s; friction cost about 1 m/s.

Worked example 3 — spring, then height. k = 200 N/m compressed 0.10 m launches a 0.050 kg ball upward. Elastic energy 1.0 J becomes field energy mgh at the top: h = 1.0/(0.050 × 9.8) = 2.04 m ≈ 2.0 m, measured from the ball's position at full compression. If the ball is observed to reach only 1.8 m, then 1.0 − (0.050)(9.8)(1.8) = 0.12 J left the system as work against air resistance during the rise.

Worked example 4 — finding an unknown from an energy account. A launcher spring with a straight-line force vs. compression graph is compressed 0.080 m and launches a 0.020 kg ball vertically; its highest point is 1.2 m above where it sat at full compression. ½ k (0.080)² = (0.020)(9.8)(1.2) → k = 0.2352/0.0032 = 73.5 N/m. Suppose (an example value, not a real data set) that the slope of the force vs. compression graph gave 78 N/m: that is a 6 % difference, consistent with air resistance and friction in the launcher making the launch less efficient than the ideal model assumes.

Representation task: bar charts and an equation. For the roller-coaster example, draw energy bar charts at the top of the hill, at the 5.0 m point and at the bottom of the track (assume a further −4000 J of friction work between the 5.0 m point and the ground). Label kinetic, gravitational and "transferred out as thermal" bars in joules, check that every chart totals 99 000 J, and write the single equation that links the top to the bottom. Then describe the same event in three sentences of plain words with no numbers.

Independent performance: a stopping-point prediction report. Carry out the ramp investigation for two different sliding objects and write a report. Success criteria: a free body diagram of the block on the ramp showing the weight, the normal force and the friction force, with the friction force on the ramp measured directly (constant-speed pull up the ramp minus mg sin θ); μ_k on the level floor measured with the constant-speed method, three trials averaged, with the spring-scale reading uncertainty stated; the full energy equation from release to rest with every term in joules; predicted and measured stopping distances for each object with percent differences; an efficiency (kinetic energy at the foot of the ramp ÷ gravitational energy released) before and after one refinement you made to reduce friction; and a paragraph naming at least three assumptions that could account for the discrepancy, saying which one you think matters most for the second object and why.