Two Blocks on a Rough Incline
A 4.0 kg block rests on a 30° incline and is connected by a light, frictionless pulley at the top of the incline to a hanging 3.0 kg block. The coefficient of kinetic friction between the 4.0 kg block and the incline is 0.20, and the coefficient of static friction is 0.35. Take g = 10 m/s². A student releases the system from rest and observes that the 4.0 kg block moves up the incline. A second student claims the system should not move at all because the hanging mass is not large enough. The incline is 2.0 m long, and the hanging block descends toward the floor.
1. Determine whether the system actually moves when released, and if so, in which direction. Show the force comparison explicitly.
Read the worked answer
Compare the driving force with the maximum static friction before deciding.
Driving force along the incline from the hanging block: T = m₂g = 3.0 × 10 = 30 N. Component of the 4.0 kg block's weight down the incline: m₁g sin30° = 4.0 × 10 × 0.5 = 20 N.
If the 4.0 kg block tends to move up the incline, friction acts down the incline and the net driving force is 30 − 20 = 10 N. Maximum static friction: μs N = μs m₁g cos30° = 0.35 × 4.0 × 10 × 0.866 ≈ 12.1 N.
Since 10 N < 12.1 N, static friction can hold the system. The system does not move; the second student is correct, though the first student's observation of motion would require an external push or a lower coefficient.
Key judgement: the direction of impending motion fixes the direction of friction, and the comparison must use static friction, not kinetic friction.
2. Suppose the system is given a brief push so that the 4.0 kg block starts moving up the incline. Find the acceleration and the tension in the string while it slides.
Read the worked answer
Once sliding, kinetic friction acts down the incline (opposing upward motion).
For the 4.0 kg block (up the incline positive): T − m₁g sin30° − μk m₁g cos30° = m₁a T − 20 − 0.20 × 4.0 × 10 × 0.866 = 4.0a T − 20 − 6.93 = 4.0a → T − 26.93 = 4.0a
For the hanging 3.0 kg block (downward positive): m₂g − T = m₂a → 30 − T = 3.0a
Adding: 30 − 26.93 = 7.0a → a ≈ 0.44 m/s². T = 30 − 3.0 × 0.44 ≈ 28.7 N.
The acceleration is small and positive, so the block continues up the incline but slowly. Judgement point: kinetic friction replaces static friction once motion starts, and the tension is less than the hanging weight because the hanging block accelerates downward.
3. The 4.0 kg block travels the full 2.0 m of the incline. Using energy ideas, estimate its speed at the top and explain why the work done against friction matters.
Read the worked answer
Use work–energy for the whole system over the 2.0 m displacement.
Change in gravitational potential energy: the 4.0 kg block rises 2.0 sin30° = 1.0 m, gaining 4.0 × 10 × 1.0 = 40 J; the 3.0 kg block descends 2.0 m, losing 3.0 × 10 × 2.0 = 60 J. Net PE change = −20 J (system loses PE).
Work done against friction: f k × d = 6.93 × 2.0 ≈ 13.9 J, dissipated as heat.
Kinetic energy gained: 20 − 13.9 ≈ 6.1 J. Total mass 7.0 kg: ½ × 7.0 × v² = 6.1 → v² ≈ 1.74 → v ≈ 1.3 m/s.
Judgement: friction removes mechanical energy, so the speed is far below what a frictionless calculation (v ≈ 2.4 m/s) would give. The energy method avoids solving for tension and is a useful cross-check on the acceleration result.