Sled Run on a Nagano Slope: Where Did the Energy Go?
On a winter field trip in the mountains of Nagano, a class measures a straight sled run. The top of the run is 12.0 m above the bottom, and the slope is 60.0 m long. A student of mass 30.0 kg rides a 5.0 kg plastic sled. Three runs give speeds at the bottom of 9.0 m/s, 9.1 m/s and 9.0 m/s. Before the trip the class used g = 9.8 m/s² and predicted about 12.6 m/s, assuming an average friction force of 22 N between sled and snow. On the day, the upper half of the run is packed and fast, but the lower half has soft, deep snow in which the runners sink a few centimetres. One student argues that the prediction 'must have been calculated wrong' and wants to write in the report that energy was not conserved on the slope.
1. Show, using energy, that an average friction force of 22 N does give a speed of about 12.6 m/s, and explain the measured 9.0 m/s without claiming that energy was destroyed.
Read the worked answer
Total mass = 30.0 + 5.0 = 35 kg.
Energy in: mgh = 35 × 9.8 × 12.0 = 4116 J. Energy removed by friction: 22 N × 60.0 m = 1320 J. Kinetic energy at the bottom: 4116 − 1320 = 2796 J. v = √(2 × 2796 / 35) = √159.8 ≈ 12.6 m/s.
A force check gives the same answer: the component of gravity along the slope is mg sinθ = 35 × 9.8 × (12.0/60.0) = 68.6 N, so the net force is 68.6 − 22 = 46.6 N, a = 1.33 m/s², and v² = 2 × 1.33 × 60 ≈ 160.
So the arithmetic is right; the assumption is what fails. Energy is not destroyed: the 4116 J is still accounted for, but about 2700 J ends up as thermal energy in the snow and runners, plus a little sound and the kinetic energy of snow pushed aside. The correct statement is 'friction was larger than we assumed', not 'energy was not conserved'.
2. Estimate the average friction force that actually acted along the run, and judge whether soft snow can plausibly account for the difference.
Read the worked answer
Kinetic energy at 9.0 m/s: ½ × 35 × 9.0² = 1418 J.
Energy lost to friction = 4116 − 1418 = 2698 J. Average friction force = 2698 J ÷ 60.0 m ≈ 45 N.
So the real average is roughly double the assumed 22 N. This is plausible for two reasons:
- Sinking runners deform snow, and permanently deforming snow costs far more energy than sliding over a hard, packed surface.
- The force is not constant. On the packed upper half it may be near 20 N; on the soft lower half it must be well above 45 N to give that average. A single 'average' hides a large change along the run.
A judgement worth stating: the 9.0 m/s result is not evidence of a bad calculation, it is evidence that friction depends strongly on snow condition, which is why the same sled behaves differently on different days.
3. At the bottom the slope flattens out. Taking the measured 9.0 m/s and a coefficient of friction of 0.30 on the flat, how far does the sled travel before stopping, and what happens to its kinetic energy?
Read the worked answer
On the flat, the normal force equals the weight, so friction = μmg = 0.30 × 35 × 9.8 = 103 N.
Kinetic energy to be removed = 1418 J. Distance = E / F = 1418 ÷ 103 ≈ 13.8 m (about 14 m).
Equivalently, deceleration = μg = 0.30 × 9.8 = 2.94 m/s², and d = v²/(2a) = 81 ÷ 5.88 ≈ 13.8 m.
All of the kinetic energy becomes thermal energy in the sled runners and the snow surface, with a small amount as sound. The temperature rise is tiny because the energy is spread over tens of kilograms of material, which is why a sled does not feel hot.
Judgement point: the answer is sensitive to μ. If the flat is icier (μ = 0.15), the sled travels about 28 m - double. A single number quoted without stating the friction assumption is not a reliable prediction.