Preview

Energy Skate Park: Basics

Q: A 60kg skater starts from 5m height. What is their speed at the bottom?

V = √(2gh) = √(2 × 9.8 × 5).

Q: With friction enabled, can the skater reach the same height on the other side?

Friction converts some mechanical energy to thermal energy.

Q: How does mass affect the maximum speed at the bottom of the ramp?

V = √(2gh) — mass cancels out!

Explore energy conservation on a skate ramp

A skater hangs motionless at the lip of a half-pipe, 4.0 m above the bottom, holding 2,352 joules of pure height. Let go and that number never changes — it only changes costume. Every joule of potential energy the skater sheds on the way down reappears as kinetic energy, so the speed at the bottom is decided before the ride begins: v = √(2gh), set by the vertical drop alone. The ramp's shape doesn't get a vote, and neither does the skater's mass. In this lab you ride a parabolic track while a live energy panel keeps the books in real time — blue kinetic, red potential, amber thermal, and a green total bar that refuses to move. Release from 9 m and the bottom speed climbs to 13.28 m/s; swap a 10 kg skater for a 100 kg one and the speed doesn't budge while the joules scale tenfold; dial in friction and watch mechanical energy leak into the amber bar while the green one still holds every joule. By the end, that flat green bar will read to you as what it is: a law of nature keeping accounts.

What you'll be able to do

Apply conservation of mechanical energy (KE + PE = const when only conservative forces act) to compute the bottom speed from release height, v = √(2gh), and verify it against the sim's live speed and energy readouts
Explain that the bottom speed depends only on the vertical drop — not on ramp steepness, path shape, or skater mass — because gravity is a conservative force whose work is path-independent, and mass cancels in ½mv² = mgh
Describe how friction converts mechanical energy into thermal energy so that KE + PE decreases while the grand total KE + PE + thermal remains constant, and read that bookkeeping off the sim's four energy bars

Formulas

E_{mech} = KE + PE = \frac{1}{2}mv^2 + mgh

Mechanical energy

E_{mech} = \text{const}

Conservation of energy (no friction)

v = \sqrt{2gh}

Speed at bottom from height h

Make a prediction

Default Frictionless setup: the 60 kg skater releases from rest at 4.0 m on the fixed parabolic half-pipe (g = 9.8 m/s²). Before pressing play — what will the Speed readout show as the skater crosses the bottom, and what decided that number?

No grading here — commit to a guess, then scroll down and test it yourself.

Loading the 3D lab…

Explore

Press play on the default Frictionless setup and keep your eyes on the energy panel through one full pass: the red PE bar drains from 2352.00 J into the blue KE bar and back while the green Total bar never leaves 2352.00 J. Catch the bottom crossing — disp-v reads ≈ 8.85 m/s exactly as KE peaks at ≈ 2352 J and PE hits ≈ 0 J.
Hit the High Drop preset (or drag Start Height to 9.0 m) and predict the bottom speed before the first crossing: v = √(2×9.8×9.0) ≈ 13.28 m/s. Verify it on disp-v, then try the extremes — 0.5 m gives ≈ 3.13 m/s and 10.0 m gives exactly 14.00 m/s. Height sets the speed; nothing else on screen does.
Back at 4.0 m, drag Mass from 10 kg all the way to 100 kg: the bottom speed stays ≈ 8.85 m/s every single time while the release Total climbs from 392.00 J to 3920.00 J. Ten times the skater, ten times the joules, identical speed — watch disp-m change and disp-v refuse to.
Hit the With Friction preset (μ = 0.15): the amber Thermal bar climbs on every pass, each swing peaks lower than the last, and the skater eventually parks at the bottom — yet the green Total bar still holds 2352.00 J, now split between PE and thermal. Mechanical energy left the ride; energy itself never did.

Scroll for the debrief ↓

Check

Did the lab agree with you?

Your prediction

You skipped the prediction — jump back to the preview and commit to a guess first; the comparison is the whole point.

Answer: ≈ 8.85 m/s — decided by the 4.0 m vertical drop alone: v = √(2gh) = √(2×9.8×4.0). A steeper or gentler ramp with the same drop would give exactly the same bottom speed.

The energy panel keeps the books in plain sight: at release E0 = mgh = 60 × 9.8 × 4.0 = 2352 J, all of it in the red PE bar, with the green Total bar pinned at 2352.00 J. The track is frictionless, so mechanical energy is conserved, and at the bottom (x = 0.00 m, h = 0) every joule has moved into KE: ½mv² = 2352 J gives v = √(2×2352/60) = √78.4 ≈ 8.85 m/s — exactly what the disp-v readout shows at the crossing. Mass cancels out of that algebra (½mv² = mgh ⇒ v = √(2gh)), which kills option C: drag Mass to 100 kg and the release energy climbs to 3920 J while the bottom speed stays ≈ 8.85 m/s, as the sim's own sidebar card and quiz Q3 insist. Option B is the targeted misconception — the steeper-is-faster instinct, documented as the chosen distractor in CU Boulder PER ramp diagnostics. Steepness does change the tangential acceleration along the way, but gravity is a conservative force: its work depends only on the endpoints, not the path (OpenStax College Physics 2e §7.4 — 'same final speed if it took the alternate path'), so the same 4.0 m of vertical drop buys the same √78.4 ≈ 8.85 m/s on any ramp shape — steep, gentle, or this fixed parabola y = 0.25x². Watch the green Total bar during the ride: it never flinches, and that stillness is the whole argument.

Quiz (0/3)

A 60kg skater starts from 5m height. What is their speed at the bottom?

With friction enabled, can the skater reach the same height on the other side?

How does mass affect the maximum speed at the bottom of the ramp?

You can now

Apply conservation of mechanical energy (KE + PE = const when only conservative forces act) to compute the bottom speed from release height, v = √(2gh), and verify it against the sim's live speed and energy readouts
Explain that the bottom speed depends only on the vertical drop — not on ramp steepness, path shape, or skater mass — because gravity is a conservative force whose work is path-independent, and mass cancels in ½mv² = mgh
Describe how friction converts mechanical energy into thermal energy so that KE + PE decreases while the grand total KE + PE + thermal remains constant, and read that bookkeeping off the sim's four energy bars