A 3.00 kg object rests on a 25.0° frictionless incline and is connected by a light, inextensible string over a pulley to a 2.00 kg hanging mass. What is the acceleration of the system? ==Specific Expectation: B3.5==

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We start by identifying the forces and choosing a positive direction for the acceleration along the string. The 2.00 kg mass hanging vertically tends to move downward, pulling the 3.00 kg mass up the incline. For the 3.00 kg mass on the incline, the component of gravity along the incline is m1 g sin(25°). The string is light and inextensible, so both masses share the same magnitude of acceleration a.

Option a: 2.40 m/s². This value would imply a larger net driving force or a smaller total mass in the system. If you plug in a = 2.40, you would require a net force of (m1 + m2) a = 5 × 2.40 = 12 N. However, the actual net force along the system is m2 g − m1 g sin(25°) ≈ 19.6 − (12.41) ≈ 7.19 N, which corresponds to a ≈ 7.19/5 ≈ 1.44 m/s². Thus 2.40 m/s² is too large.

Option b: 1.87 m/s². This is closer but still too large. It would require a net force around 9.35 N, which contradicts the calculated net force of about 7.19 N. The discrepancy suggests an overestimation of the incline’s resistance (the sin component) or an underestimation of the system’s inertia.

Option c: 1.44 m/s². This matches the correct calculation. Compute sin(25°) ≈ 0.4226, so m1 g sinθ ≈ 3 × 9.8 × 0.4226 ≈ 12.41 N. The hanging mass provides m2 g ≈ 2 × 9.8 = 19.6 N. Net driving force ≈ 19.6 − 12.41 ≈ 7.19 N. Total mass = 3 + 2 = 5 kg, so a ≈ 7.19 / 5 ≈ 1.44 m/s², which is exactly option c.

Option d: 1.63 m/s². This value would require a net force around 8.15 N (since 5 kg × 1.63 m/s² ≈ 8.15 N), which is larger than the actual net force calculated from m2 g and m1 g sinθ. The mismatch arises from misplacing the incline component or miscomputing the weight difference; the incline’s parallel component reduces the driving force, making 1.63 m/s² too high.

In summary, the correct reasoning yields a = (m2 g − m1 g sin(25°)) / (m1 + m2) ≈ (19.6 − 12.41) / 5 ≈ 1.44 m/s², confirming option c. The other choices either overestimate the net driving force or misaccount for the incline’s resistive component.

SPH4U - Physics 12 (2025-26) - A

A 3.00 kg object rests on a 25.0° frictionless incline and is connected by a light, inextensible string over a pulley to a 2.00 kg hanging mass. What is the acceleration of the system? ==Specific Expectation: B3.5==

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