Lessons · energy

Conservation of energy: why the speed depends on the drop, not the path

Regime 2 — calculus does what algebra cannot. The acceleration isn't constant, so the algebra formulas don't apply; calculus is the only road in, and SymPy proves the closed form solves the equation of motion.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A $m = 2$ kg block is released from rest at the top of a frictionless ramp, $H = 20$ m above the bottom. As it slides down, potential energy turns into kinetic energy — but the total never changes. How fast is it going at the bottom, and would a steeper ramp change that? Drag the height: watch the KE and PE bars trade while the Total bar stays flat. The speed at any height depends only on how far it has dropped, not on the shape of the ramp.

Total energy

E = mgH

(conserved)

400 J

Speed at the bottom

v = \sqrt{2gH}

20 m/s

KE at half height

= \tfrac12 mgH

200 J

PE at the top

= mgH

(all potential)

400 J

Energy conservation is the first integral of

F = ma

m\,\frac{dv}{dt} = F \ \Longrightarrow\ m\,v\,dv = F\,dx \ \Longrightarrow\ \tfrac12 mv^2 = \int F\,dx

Start from Newton's second law and multiply by $dx$ : since $dv/dt\cdot dx = v\,dv$ , the equation becomes $mv\,dv = F\,dx$ . Integrating once gives $\tfrac12 mv^2 = \int F\,dx$ — the kinetic energy is the work done by the force. Energy is not a separate principle; it is the equation of motion, integrated.

check $\tfrac{d}{dh}(\text{KE}+\text{PE})=0$ ; check $\tfrac12 mv^2+mgh=mgH$ ; show $v=\sqrt{2g(H-h)}$ falls out; check $\text{KE}=\int_h^H mg\,dh'$ (path-independent)

✓ The total energy does not change with height: $\tfrac{d}{dh}(\text{KE}+\text{PE}) = 0$ . [structural]
✓ The memorized law holds at every height: $\tfrac12 mv^2 + mgh = mgH$ (the energy is conserved). [structural]
✓ The speed falls out of the kinetic energy: $\tfrac12 mv^2 = \text{KE}$ , so $v = \sqrt{2g(H-h)}$ . [structural]
✓ The kinetic energy is exactly the work gravity does over the descent: $\text{KE} = \int_h^H mg\,dh'$ — the first integral of $F=ma$ , independent of the path. [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “A steeper ramp gets the block to the bottom moving faster than a gentle ramp from the same height.”

It does not. By conservation of energy the speed at the bottom is $v = \sqrt{2gH}$ , set entirely by the height dropped — a steep ramp and a gentle ramp to the same depth give exactly the same bottom speed. The steeper ramp gets there in less time (and with a larger acceleration along the slope), but the final speed is identical, because gravity is conservative: the work it does, $mg(H-h)$ , depends only on the change in height, not the route. Drag the cursor and read the speed off the kinetic-energy bar — it tracks the height, nothing else.

Modeling assumptions — author-asserted, disclosed not discharged

Frictionless track, so no mechanical energy is lost to heat; the normal force does no work (it is perpendicular to the motion).
$g = 10$ m/s² as a magnitude, height $h$ measured upward from the bottom; the block starts from rest at $h = H$ .

The stacked graph, fully annotated

A static rendering (Matplotlib) at the default parameters — the interactive version is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):