Block on a Rough Incline with a Hanging Mass
A 2.0 kg block is placed on a rough plane inclined at 30.0° to the horizontal. The coefficients of friction are μs = 0.50 and μk = 0.35. The block is released from rest and a motion sensor records an acceleration of 1.2 m/s² down the incline. In a second trial, the block is connected by a light inextensible string over a frictionless pulley at the top of the incline to a hanging mass of 0.80 kg; the block is again released from rest. A student predicts that the block will accelerate up the incline because the hanging mass “pulls harder”. The laboratory has g = 9.8 m/s². The student is asked to reconcile the measured acceleration with the quoted coefficients, and then to decide whether the 0.80 kg hanging mass is enough to move the block at all.
1. Calculate the acceleration predicted using μk = 0.35 and compare it with the measured 1.2 m/s². What kinetic coefficient is actually implied?
Read the worked answer
With μk = 0.35: a = g(sin 30° − μk cos 30°) = 9.8(0.500 − 0.35 × 0.8660) = 9.8(0.500 − 0.3031) = 9.8 × 0.1969 = 1.93 m/s².
The measured 1.2 m/s² is lower, so the actual kinetic friction is larger. Solve: 1.2 = 9.8(0.500 − μk × 0.8660) 0.12245 = 0.500 − 0.8660μk μk = (0.500 − 0.12245)/0.8660 = 0.436.
The quoted 0.35 is not consistent with the motion sensor; a plausible cause is a rougher patch or a different block surface from the one used for the quoted coefficient.
2. With the 0.80 kg hanging mass, decide whether the block moves. Show the static-friction argument.
Read the worked answer
Tension from the hanging mass is T = m₂g = 0.80 × 9.8 = 7.84 N.
Without friction, the block’s weight component down the incline is m₁g sin 30° = 2.0 × 9.8 × 0.5 = 9.80 N. The net tendency is 9.80 − 7.84 = 1.96 N down the incline.
Maximum static friction is μs m₁g cos 30° = 0.50 × 2.0 × 9.8 × 0.866 = 8.49 N.
Since 8.49 N > 1.96 N, static friction can adjust to 1.96 N up the incline and hold the block at rest. The student’s prediction is wrong; the block does not move, so kinetic friction is not relevant.
3. Find the minimum hanging mass needed to start the block moving up the incline. Also find the minimum mass that would just prevent it from sliding down.
Read the worked answer
To start motion up the incline, the tension must overcome both the down-slope weight component and maximum static friction: m₂g > m₁g sin 30° + μs m₁g cos 30° = 9.80 + 8.49 = 18.29 N. m₂ > 18.29/9.8 = 1.87 kg.
With 0.80 kg the block stays at rest.
To just prevent sliding down, the hanging mass must supply enough tension that the required static friction does not exceed its maximum: m₂g ≥ m₁g sin 30° − μs m₁g cos 30° = 9.80 − 8.49 = 1.31 N. m₂ ≥ 1.31/9.8 = 0.134 kg.
Below about 0.13 kg the block would slide down the incline.