The Trolley on the Ramp: Choosing Between Kinematics and Energy
A 2.0 kg laboratory trolley is released from rest at the top of a 1.20 m long inclined ramp set at 15° to the horizontal. A student measures the time to reach the bottom three times and gets 1.62 s, 1.58 s and 1.60 s. The trolley's wheels have negligible friction, but a light string attached to the trolley passes over a pulley at the top of the ramp and drags a 0.30 kg mass along the ramp surface behind it. The string is taut throughout and the hanging 0.30 kg mass is not on the ramp-it hangs vertically and is pulled upward as the trolley descends. The student calculates acceleration using s = ½at² and gets 0.94 m/s², then claims this proves mechanical energy is not conserved.
1. Evaluate the student's reasoning. Is mechanical energy actually non-conserved here, or is the calculation flawed? Support your judgement with a calculation.
Read the worked answer
The student's conclusion does not follow. Using s = ½at² with s = 1.20 m and t = 1.60 s gives a = 2s/t² = 2(1.20)/(2.56) = 0.94 m/s², which agrees with the measurement. But this value is not evidence against energy conservation; it is simply the actual acceleration of the system.
Treat the trolley plus hanging mass as one system. The only external force doing work is gravity on the trolley along the ramp: F = mg sin15° = 2.0 × 9.8 × 0.259 = 5.08 N. The hanging mass adds inertia but contributes no net external work if we account for its weight correctly-actually the hanging mass weight (0.30 × 9.8 = 2.94 N) opposes motion, so net driving force = 5.08 − 2.94 = 2.14 N. Total mass = 2.30 kg, so a = 2.14/2.30 = 0.93 m/s². This matches the measured 0.94 m/s² to within experimental uncertainty.
Mechanical energy is conserved for the trolley–mass–Earth system; the student simply misidentified the acceleration as anomalous rather than checking it against Newton's second law. The judgement is that the data are consistent with conservation, and the claim is unsupported.
2. The student now argues: 'If energy were conserved, the trolley should accelerate at g sin15° = 2.5 m/s².' Explain why this reasoning is wrong and identify the conceptual error.
Read the worked answer
The student has assumed that the only mass in the system is the trolley, so that the net force mg sin15° acts on 2.0 kg alone. That would give a = g sin15° = 2.54 m/s². But the hanging 0.30 kg mass is coupled to the trolley by the taut string, so both accelerate together and the total inertia is 2.30 kg, not 2.0 kg.
More importantly, the hanging mass's weight (2.94 N) acts opposite to the direction of motion of the trolley, reducing the net driving force to 2.14 N. Dividing by the total mass gives 0.93 m/s².
The conceptual error is treating the trolley in isolation while ignoring the tension force from the string. In a connected system, tension is an internal force; the correct approach is either (a) analyse the whole system with external forces only, or (b) analyse each body separately with tension included and solve the simultaneous equations. Either method gives the same acceleration. The student's 2.5 m/s² would only apply to a trolley with no attached mass, and even then only if friction were truly negligible.