Energy problem studio

A block fired by a spring across a rough strip into a vertical loop is four states joined by three stages, and each stage moves energy between exactly two accounts. A to B, compressed spring to natural length: ½kx² becomes K, and the block leaves the spring at B because an unattached spring cannot pull. B to C, across the strip: K pays μₖmgd into thermal energy. C to D, bottom of the loop to its top: K pays mg(2R) into gravitational potential energy. With m = 0.25 kg, k = 400 N m⁻¹, x = 0.12 m, μₖ = 0.20, d = 0.50 m and R = 0.40 m the ledger reads K = 2.88 J at B, 2.88 − 0.245 = 2.635 J at C and 2.635 − 1.962 = 0.673 J at D. Draw the boundary before the first line: block, spring and Earth inside, so the spring and gravity appear only as U; the wall does no work because its contact point never moves, the smooth track does none because it pushes perpendicular to the motion, and friction appears only as ΔEₜₕ.

On a slope or a vertical launcher gravity keeps doing work while the spring compresses, so both change in the same line of the ledger. A 2.0 kg crate slides from rest 1.20 m down a smooth 30° incline onto a spring with k = 1500 N m⁻¹. Along the slope it travels a further x while compressing the spring, so mg sin 30° × (1.20 + x) = ½kx², which in SI units is 750x² − 9.81x − 11.772 = 0, giving x = 0.132 m. Drop the extra x and you get 0.125 m, short by exactly the energy gravity delivered during the compression. Nor is the crate fastest where it meets the spring. Its kinetic energy peaks where the total potential energy stops falling, dU/ds = 0, which is where the spring force balances the slope component of gravity: kx = mg sin 30°, so x = (9.81/1500) m = 6.5 mm into the compression. The force-from-potential relation locates the extremum; the stage boundary does not.

Kinetic friction removes μₖN for every metre slid, so its ledger entry is μₖN s, with s the path length rather than the displacement. N equals mg only on level ground. On a rough 25° ramp 0.50 m long, a 0.25 kg block with μₖ = 0.20 loses μₖmg cos 25° × 0.50 = 0.222 J, not the 0.245 J that μₖmgd would claim. In a rough dip N = mg cos φ + mv²/r depends on both the local slope angle φ and the speed, so the loss cannot be found from energy alone; it has to be given, measured, or integrated with forces. Friction also never pays back. A block that crosses a strip and returns pays twice, and one that shuttles between a spring and a smooth ramp keeps paying until all of its initial energy is thermal. That turns a many-bounce problem into one line: on level ground the total distance slid is E/(μₖmg), however many passes it takes.

A smooth track pushes perpendicular to the velocity at every instant, so its normal force does no work and the track's shape drops out of the speed. That same normal force is what bends the path, though, and only Newton's second law along the radius can find it. At the top of a loop of radius R, N and mg both point toward the centre: N + mg = mvₜₒₚ²/R. A track can push but not pull, so N ≥ 0 requires vₜₒₚ² ≥ gR, and the ledger from the bottom adds 4gR: v_b² ≥ 5gR. For R = 0.40 m that is v_b ≥ 4.43 m s⁻¹, or a smooth-track start 1.00 m up. At exactly that speed the bottom of the loop pushes with N_b = m(g + v_b²/R) = 6mg. Short of it, contact fails where N reaches zero on the upper half: with α measured up from the centre's level, mg sin α = mv²/R and v² = v_b² − 2gR(1 + sin α), so sin α = (v_b² − 2gR)/(3gR).

Energy compares states and is blind to the path between them, so read the question before choosing. A speed, height, compression or stopping distance between two known states: energy, in one line, whatever the track's shape. A force, an acceleration, a direction, or whether contact holds: Newton's second law, usually fed a v² that energy supplied. A time under a constant force: Newton plus constant-acceleration kinematics, the shortest route of all. A time under a varying force: energy gives v(x) and one integral finishes, t = ∫ dx/v. A 0.25 kg block pushed out by a 400 N m⁻¹ spring has v(x) = √(k/m) √(A² − x²) at compression x, and integrating from A down to 0 gives (π/2)√(m/k) = 0.0393 s for any starting compression A. Through a collision, or a block dropped onto a sticky surface, kinetic energy turns into deformation the ledger has no line for; those stages need momentum, which the next section builds.

Three checks catch most wrong answers. A negative kinetic energy means the state is never reached: send the 2.635 J block from the first section toward a smooth ramp 1.20 m high and the ledger claims K = 2.635 − 2.943 = −0.31 J at the top, which really says the block turns back where U = E, at 2.635 J ÷ mg = 1.07 m. A thermal term must come out positive, because friction only removes energy; a negative ΔEₜₕ means a sign or a pair of states has been swapped. And the limits must behave: set μₖ = 0 and the strip must vanish from the answer; make R small and the loop condition must become easy to meet. Read the potential-energy diagram the same way. Turning points sit where the horizontal line at E meets U(x), and with friction that line steps down by μₖmgd at every crossing of the strip, so the turning points creep inward pass by pass.