Block on a Rough Surface Attached to a Spring
A 0.50 kg block lies on a rough horizontal surface and is attached to a horizontal spring of stiffness 200 N/m. The coefficients of friction are μs = 0.60 and μk = 0.40. The spring is compressed by 0.150 m from its natural length and the block is released from rest. Take g = 10 m/s². The block moves along a straight line; the spring remains within its elastic limit and does not buckle. The surface is uniform, so the friction force has constant magnitude μk mg while the block is moving and can take any value up to μs mg when the block is instantaneously at rest. The block is never lifted off the surface. The block is small enough to be treated as a particle, and the spring mass is negligible.
1. Find the position of the block at its first turning point on the other side of the natural length, and decide whether the block remains at rest there.
Read the worked answer
Take x as displacement from natural length, positive to the right; the initial position is x = +0.150 m, where the spring is compressed and pushes left. While moving left, kinetic friction acts right with magnitude fk = μk mg = 0.40×0.50×10 = 2.0 N.
Energy from release to the first turning point x1: ½kx0² − fk(x0 − x1) = ½kx1² 2.25 − 2.0(0.150 − x1) = 100x1²
This gives 100x1² − 2x1 − 1.95 = 0, whose roots are x1 = 0.150 (the start) and x1 = −0.130 m. So the first turning point is 0.130 m on the other side of natural length.
At x1 = −0.130 m the spring is stretched and pulls right with force k|x1| = 200×0.130 = 26 N. Maximum static friction is μs mg = 0.60×5 = 3.0 N. Since 26 N > 3.0 N, the block does not stay at rest; it starts moving right.
2. Determine the final resting position of the block and the total distance it travels before stopping permanently.
Read the worked answer
Each half-cycle the amplitude decreases by 2fk/k = 2(2.0)/200 = 0.020 m. The turning-point magnitudes are therefore 0.150, 0.130, 0.110, 0.090, 0.070, 0.050, 0.030, 0.010 m.
Signs alternate: +0.150 → −0.130 → +0.110 → −0.090 → +0.070 → −0.050 → +0.030 → −0.010 m.
At the turning point of magnitude 0.010 m, the spring force is k(0.010) = 2.0 N, which is less than the maximum static friction 3.0 N. Hence the block stops there permanently, at x = −0.010 m.
Total distance: 0.280 + 0.240 + 0.200 + 0.160 + 0.120 + 0.080 + 0.040 = 1.12 m.
Energy check: initial spring energy 2.25 J; friction work 2.0×1.12 = 2.24 J; final spring energy 100(0.010)² = 0.010 J; sum 2.25 J.