Spring Launch Across a Rough Patch to an Incline
A 2.0 kg block is pressed against a spring of constant 500 N/m, compressing it 0.30 m. The block is released from rest. It leaves the spring at the spring’s natural length and then slides 1.20 m across a rough horizontal patch with coefficient of kinetic friction 0.25 before reaching a frictionless incline. The spring and the horizontal surface are at the same level, and the block is not attached to the spring. Air resistance is negligible. A student claims the block’s speed at the bottom of the incline can be found by setting the initial spring energy equal to the final kinetic energy, with no other terms.
1. Find the block’s speed when it reaches the bottom of the incline. Identify every energy term that matters.
Read the worked answer
Initial spring energy: U_s = 0.5 k x² = 0.5(500)(0.30)² = 22.5 J.
While the block crosses the rough patch, kinetic friction does negative work. The friction force is f_k = μ_k m g = (0.25)(2.0)(9.8) = 4.9 N. The friction work magnitude is f_k d = (4.9)(1.20) = 5.88 J, which becomes thermal energy.
The energy remaining as kinetic energy at the bottom of the incline is K = 22.5 − 5.88 = 16.62 J. Since K = 0.5 m v², 16.62 = 0.5(2.0)v² = v², so v = 4.08 m/s. The student’s claim is incomplete because it omits the friction work and the thermal energy.
2. Find the maximum vertical height the block reaches on the frictionless incline.
Read the worked answer
At the bottom of the incline the block has K = 16.62 J. On the frictionless incline, mechanical energy is conserved, so this kinetic energy converts to gravitational potential energy: ΔU_g = m g h.
Thus 16.62 = (2.0)(9.8)h, giving h = 16.62/19.6 = 0.848 m. This is the vertical height above the bottom of the incline, not the distance measured along the incline. The incline angle is not needed because only the vertical height change affects gravitational potential energy.
3. Find the minimum coefficient of kinetic friction on the 1.20 m patch that would stop the block exactly at the end of the patch, before it reaches the incline. Then state whether a coefficient of 0.60 would let the block reach the incline.
Read the worked answer
To stop exactly at the end of the patch, the friction work must remove all 22.5 J of spring energy: μ_k m g d = 22.5. Using m=2.0 kg, g=9.8 m/s², and d=1.20 m: μ_k = 22.5/[(2.0)(9.8)(1.20)] = 22.5/23.52 = 0.957.
With μ_k = 0.60, the friction work is (0.60)(2.0)(9.8)(1.20) = 14.112 J. The remaining energy is 22.5 − 14.112 = 8.388 J, so the block still reaches the incline with speed v = sqrt(8.388) = 2.90 m/s. Therefore μ_k = 0.60 is below the stopping threshold and does not stop the block on the patch.