Sled test on a salted slope: is the safety barrier adequate?
A loaded sled with a total mass of 40 kg (sled plus rider) is released from rest at the top of a straight snowy slope 12 m long, inclined at 25° to the horizontal. The top 8.0 m is hard-packed snow with a coefficient of kinetic friction of 0.02; the bottom 4.0 m has been salted and is wet, with a coefficient of 0.25. Both coefficients were measured by students beforehand and vary by about ±10% between runs. At the foot of the slope the ground becomes horizontal packed snow with a coefficient of friction of 0.10, and the sled slides 6.0 m across this run-out before striking a wooden barrier. The resort claims the barrier is safe because sleds arrive at less than 3.0 m/s. Use g = 9.8 m/s², sin 25° = 0.423 and cos 25° = 0.906. The class also notices that the sled's measured speed at the foot of the slope hardly changed when the salted section was made longer, which surprised them.
1. Calculate the sled's speed at the point where the smooth section ends and the salted section begins. State your assumptions and say whether the ±10% uncertainty in the friction coefficients matters.
Read the worked answer
Work with energy from the start to that point.
- Vertical drop: h = 8.0 × sin 25° = 8.0 × 0.423 = 3.38 m
- Potential energy lost: ΔPE = mgh = 40 × 9.8 × 3.38 = 1.33 × 10³ J
- Normal force: N = mg cos 25° = 40 × 9.8 × 0.906 = 355 N
- Friction force: f = μN = 0.02 × 355 = 7.1 N, so work removed = 7.1 × 8.0 = 57 J
- Kinetic energy: KE = 1330 − 57 = 1273 J
- v = √(2 × 1273 / 40) = √63.7 = 8.0 m/s
Assumptions: the sled behaves as a particle, there is no air resistance, the slope is straight and its surface is uniform, N = mg cosθ because there is no acceleration perpendicular to the slope, and no energy is stored in the snow.
Uncertainty: 10% of 57 J is 6 J, which is 0.5% of 1273 J. Since v ∝ √KE, the speed changes by about 0.25%, roughly 0.02 m/s. The answer is 8.0 m/s to two significant figures whatever happens, so the result is robust.
2. The class expected that lengthening the salted, rough section would slow the sled noticeably, but the measured speed at the foot hardly changed. Explain using an energy account, then calculate the speed at the foot for the 4.0 m rough section described.
Read the worked answer
From the 8.0 m point to the foot:
- Extra drop: h = 4.0 × 0.423 = 1.69 m, so ΔPE = 40 × 9.8 × 1.69 = 663 J is added
- Rough-section friction: f = 0.25 × 355 = 88.8 N, so work removed = 88.8 × 4.0 = 355 J
- KE at the foot = 1273 + 663 − 355 = 1581 J
- v = √(2 × 1581 / 40) = √79.1 = 8.9 m/s
Why lengthening the rough patch barely matters: the component of gravity down the slope is mg sin 25° = 40 × 9.8 × 0.423 = 166 N, which is larger than the rough-surface friction of 88.8 N. The sled therefore accelerates everywhere on the slope; there is no terminal speed. Moving one metre from the smooth section into the rough section swaps 7.1 J of friction loss for 88.8 J, a net extra loss of about 82 J, while every metre still supplies 166 J of potential energy. So the net gain stays positive at about 84 J per metre - lengthening the rough strip does slow the sled, but by only a few per cent of a 1600 J total. The sled would only stop accelerating if the rough coefficient exceeded tan 25° = 0.47, which it does not.
3. Using 8.9 m/s at the foot, test the resort's claim that sleds arrive at the barrier at less than 3.0 m/s. Calculate the run-out distance that would actually be needed and recommend a change.
Read the worked answer
On the horizontal run-out, N = mg, so friction f = 0.10 × 40 × 9.8 = 39.2 N. Over 6.0 m it removes 39.2 × 6.0 = 235 J.
KE at the barrier = 1581 − 235 = 1346 J, so v = √(2 × 1346 / 40) = √67.3 = 8.2 m/s.
The claim fails by nearly a factor of three - the barrier is struck at about 8 m/s, which is a serious injury risk.
Distance needed: at 3.0 m/s the sled would carry ½ × 40 × 3.0² = 180 J, so 1581 − 180 = 1401 J must be dissipated. d = 1401 / 39.2 = 36 m. If the run-out coefficient were also 10% wrong, the answer would be 33 m (μ = 0.11) or 40 m (μ = 0.09), so the conclusion is unchanged.
Recommendation: extend the run-out to at least 40 m, or make the run-out surface rougher, or fit a barrier that deforms and absorbs energy rather than a rigid wooden one. Note that the calculation treats sled and rider as one object; in a real impact the rider keeps moving after the sled stops, so rider restraint must be assessed separately.