A Block on a Ramp and the Missing Joules
A 2.0 kg aluminium block is released from rest at the top of a 25° incline and slides 2.5 m along the surface to the bottom, where a light gate records its speed as 4.0 m/s. A student predicts the speed at the bottom using v² = 2gL·sinθ, obtains about 4.6 m/s, and claims the light gate must be miscalibrated. The teacher insists the discrepancy is real and asks where the missing energy went. Thermometers taped to the block and to the ramp both read the same before and after the run, to the nearest 0.1 °C.
Additional information: g = 9.8 m/s²; specific heat capacity of aluminium = 900 J kg⁻¹ K⁻¹; the block is in contact with the ramp for about 0.9 s.
1. Calculate the coefficient of kinetic friction and show where every joule of gravitational potential energy ended up.
Read the worked answer
Potential energy lost: the vertical drop is h = 2.5 × sin 25° = 2.5 × 0.4226 = 1.057 m, so ΔPE = mgh = 2.0 × 9.8 × 1.057 = 20.7 J.
Kinetic energy at the bottom: KE = ½mv² = 0.5 × 2.0 × 4.0² = 16.0 J.
Missing energy: 20.7 − 16.0 = 4.7 J, which is work done against friction.
Friction force: W = f·L, so f = 4.7/2.5 = 1.88 N.
Normal force: N = mg cos 25° = 2.0 × 9.8 × 0.9063 = 17.8 N.
Coefficient: μk = f/N = 1.88/17.8 ≈ 0.11.
Cross-check via acceleration: a = v²/2L = 16/(2 × 2.5) = 3.2 m/s², versus g sin 25° = 4.14 m/s² without friction. Then a = g(sin θ − μ cos θ) gives 3.2 = 9.8(0.4226 − 0.9063μ), so μ ≈ 0.11 - consistent.
Judgement: a coefficient of 0.11 is entirely ordinary for metal on a dry surface, so the value is credible and the light gate is fine. The assumption of constant friction is reasonable only because μk is velocity-independent for dry sliding; if the surface were lubricated or the speed high enough to heat the contact, that assumption would need testing.
2. The thermometers showed no change. Reconcile that with the energy accounting, and quantify what temperature rise you would actually expect if 60% of the friction work entered the block.
Read the worked answer
Distribution of the 4.7 J: work done by friction is not destroyed. It is degraded into internal energy, shared between the block, the ramp and a little airborne sound. How it is partitioned depends on the surfaces; taking the stated 60% for the block:
Q_block = 0.60 × 4.7 = 2.8 J.
Temperature rise: ΔT = Q/(mc) = 2.8/(2.0 × 900) = 2.8/1800 = 0.0016 K, i.e. about 0.002 °C.
Why the thermometers see nothing: the instrument resolves 0.1 °C, roughly 60 times larger than the predicted change. Aluminium's high specific heat and the block's 2 kg mass both work against a measurable rise. A null thermometer reading is therefore fully consistent with the energy balance - it is a statement about instrument resolution, not about energy conservation.
Second-law point: the reverse never happens. Sixteen joules of directed kinetic energy plus 4.7 J of thermal energy will not spontaneously regroup into 20.7 J of gravitational potential energy, because that would require a decrease in total entropy. The irreversibility is the physical content of the word 'friction' here, and it is why the process has a defined direction in time.
Check: repeating the run 300 times in the same insulation would give a rise of about 0.5 °C - detectable, and a good laboratory demonstration.
3. If the ramp were raised to 40° while keeping μk and the sliding distance the same, would the *fraction* of potential energy lost to friction rise or fall? Justify without relying on numerical intuition.
Read the worked answer
Set up the ratio. For a slide of length L on angle θ, the energy lost to friction is f·L = μmg cos θ · L, and the potential energy released is mgL sin θ. The fraction is therefore
friction/PE = (μmgL cos θ)/(mgL sin θ) = μ cot θ.
The mass and the length L cancel, so the answer depends only on μ and the angle.
Conclusion: cot θ decreases as θ increases, so steepening the ramp from 25° to 40° reduces the fraction of energy lost. Numerically, cot 25° = 2.14 and cot 40° = 1.19, so with μ ≈ 0.11 the loss fraction falls from about 23% to about 13%.
Physical reading, and the trap: a steeper ramp increases the normal force? No - the opposite. N = mg cos θ falls as θ rises, so friction is weaker along the steeper surface. Students often reason that steeper means faster and therefore more friction; friction here does not depend on speed, and the sliding distance is held fixed in this question. If instead the height were held fixed, the block on the steeper ramp would travel a shorter distance but with the same friction force, so the absolute loss would shrink even further. Either way, steeper means less fractional loss.