Block on a Rough Incline, a Rough Patch, and a Spring
A 2.0 kg block is released from rest at the top of a rough incline that makes 30° with the horizontal and is 3.0 m long along the slope. The coefficient of kinetic friction between the block and the incline is 0.25. At the bottom, the block slides onto a horizontal surface. The first 1.5 m of this horizontal surface is rough with coefficient of kinetic friction 0.40; beyond that the surface is frictionless and ends in an unstretched spring of spring constant 800 N/m mounted on a wall. The block is observed to come momentarily to rest after compressing the spring. Take g = 9.8 m/s². A student claims that because friction acts on the incline, the speed at the bottom can be found by subtracting the friction force from the component of weight down the slope and then using v² = 2as with the net force. Another student says energy accounting is safer because the normal force changes on the horizontal part. You are asked to adjudicate and compute the motion.
1. Calculate the speed of the block at the bottom of the incline.
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Take the incline as the path. The height dropped is h = 3.0 sin30° = 1.5 m, so gravity does Wg = mgh = 2.0×9.8×1.5 = 29.4 J. The normal force on the incline is N = mg cos30° = 2.0×9.8×0.866 = 17.0 N, so kinetic friction is f = 0.25×17.0 = 4.24 N and its work is Wf = −4.24×3.0 = −12.7 J. The net work is 29.4 − 12.7 = 16.7 J. Since the block starts from rest, ½mv² = 16.7 J, giving v² = 16.7 (because m = 2.0 kg) and v = 4.08 m/s. The force-along-slope check gives mg sin30° − f = 9.8 − 4.24 = 5.56 N, a = 2.78 m/s², and v² = 2×2.78×3.0 = 16.7, so both accounts agree. The friction force is not subtracted from weight as vectors; only components along the slope enter.
2. Determine the maximum compression of the spring.
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The kinetic energy at the foot is 16.7 J. On the 1.5 m rough patch the friction force is μmg = 0.40×2.0×9.8 = 7.84 N, so the work removed is 7.84×1.5 = 11.76 J. The energy entering the spring section is 16.7 − 11.76 = 4.94 J. The spring section is stated frictionless, so all of this becomes elastic potential energy: ½kx² = 4.94 J, with k = 800 N/m. Thus x² = 2×4.94/800 = 0.01235 m² and x = 0.111 m. The block is momentarily at rest at maximum compression; it does not remain there because the spring then pushes it back. Using v² = 2as alone would not account for the spring's varying force, so energy is the appropriate method.
3. A student predicts that releasing the block from the midpoint of the incline will produce half the spring compression. Evaluate this prediction.
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At the midpoint the block travels 1.5 m down the incline. The height dropped is 0.75 m, so gravity supplies 2.0×9.8×0.75 = 14.7 J. Friction on the shorter incline removes 4.24×1.5 = 6.36 J, leaving 8.34 J of kinetic energy at the bottom. The horizontal rough patch requires 7.84×1.5 = 11.76 J to cross completely, which is more than 8.34 J. The block therefore stops after d = 8.34/7.84 = 1.06 m on the rough patch, before reaching the spring. The predicted half-compression is wrong: the actual compression is zero. The example shows that a linear scaling of release height does not produce a linear scaling of spring compression when fixed friction losses are present.