Block on a Rough Incline: Two Stages, Two Accelerations
A block of mass 2 kg is projected up a rough plane inclined at 30° to the horizontal with an initial speed of 8 m/s. The coefficient of friction between the block and the plane is 0.25 and g = 9.8 m/s². A student models the motion in two stages but writes a single acceleration of 7.02 m/s² acting throughout, then concludes that the block returns to its starting point with exactly the speed it left with. The student's speed–time graph is a symmetric V-shape, and the time going up is set equal to the time coming down. You are asked to check the model before it is submitted. The examiner wants to see whether the block slides back at all, how the friction force behaves when the direction of motion reverses, and why the two journeys take different times even though the distance covered is the same in each. Treat the block as a particle and ignore air resistance.
1. Find the acceleration of the block while it moves up the plane and the distance it travels before momentarily stopping. Then decide, with justification, whether it slides back down.
Read the worked answer
Resolve along the plane. While the block moves up, friction acts down the plane, so
a_up = g sin30° + μ g cos30° = 9.8(0.5) + 0.25(9.8)(0.866) = 4.90 + 2.12 = 7.02 m/s²,
directed down the plane.
Stopping distance. Using v² = u² − 2as with v = 0:
s = 8²/(2 × 7.02) = 64/14.04 = 4.56 m.
Will it slide back? At the highest point the component of weight down the plane is mg sin30° = 2(9.8)(0.5) = 9.8 N. The maximum friction available is μmg cos30° = 0.25(2)(9.8)(0.866) = 4.24 N. Since 9.8 N > 4.24 N, the block does slide back. Equivalently, tan30° = 0.577 > μ = 0.25, which is the standard test for sliding on a rough incline. The student's single acceleration is therefore wrong: once the block stops, the friction reverses and the net force along the plane changes.
2. Find the speed with which the block returns to its starting point, and the time taken in each direction. Account for the difference in times.
Read the worked answer
Coming down. Friction now acts up the plane, opposing the motion, so
a_down = g sin30° − μ g cos30° = 4.90 − 2.12 = 2.78 m/s².
Starting from rest over the same 4.56 m: v² = u² + 2as gives v² = 2(2.78)(4.56) = 25.3, so v = 5.03 m/s. The block returns more slowly than it left because friction has done net negative work over the round trip - it opposes the motion in both directions.
Times. Up: t = (8 − 0)/7.02 = 1.14 s. Down: from s = ½at², t = √(2 × 4.56/2.78) = √3.28 = 1.81 s. Total journey 2.95 s.
Why they differ. The distance is the same, but the acceleration on the way down (2.78 m/s²) is much smaller than on the way up (7.02 m/s²), because friction and gravity oppose one another. For a fixed distance starting from rest, t = √(2s/a), so the smaller acceleration must take the longer time. The student's symmetry assumption is the error.
3. Describe the correct speed–time graph and use an energy check to confirm the return speed.
Read the worked answer
Graph. Two straight segments with different gradients: speed falls from 8 m/s to 0 in 1.14 s at −7.02 m/s², then rises from 0 to 5.03 m/s in 1.81 s at +2.78 m/s². The graph is not a symmetric V: the second limb is shallower and longer.
Errors in the student's graph. (i) A single acceleration ignores the reversal of friction when the block stops. (ii) Equal return speed would only hold on a smooth plane; friction removes mechanical energy. (iii) The equal-time assumption follows from the false symmetry.
Energy check. Loss of kinetic energy = ½(2)(8² − 5.03²) = 64 − 25.3 = 38.7 J. Work done against friction over the round trip = μmg cos30° × (2s) = 4.24 × 9.11 = 38.6 J. The two agree, so the computed return speed is consistent with the work–energy principle, and the discrepancy with the student's answer is entirely due to friction.