NapkinCalc

Physics 1 — Motion & Energy

Energy conservation

continues from lesson 4 — values defined earlier in the course stay live here

ELI5: lift a mass and you bank energy as potential, m·g·h; let it fall and the bank pays out as kinetic energy, ½·m·v². With no air resistance nothing leaks — so the energy at the top exactly equals the energy at the bottom. That single idea skips all the timing details.

hdrop:=12mh_{drop} := 12 m = 12 m
Ep:=mbox∗g∗hdropE_{p} := m_{box} * g * h_{drop} = 1177.2 J potential energy at the top
vbottom:=sqrt(2∗g∗hdrop)v_{bottom} := sqrt(2 * g * h_{drop}) = 15.344 m / s speed at the bottom
Ek=12⋅mbox⋅vbottom2E_{k} = \frac{1}{2} \cdot m_{box} \cdot v_{bottom}^{2} = 1177.2 J kinetic energy at the bottom
✓ pass abs(Ek−Ep)<0.001Jabs(E_{k} - E_{p}) < 0.001 J energy in = energy out

Real-world hook: energy conservation runs roller coasters (the first hill must be the tallest), hydroelectric dams, and pole vaulting (running speed → height).

Try it yourself: drop the 10 kg box from 5 m. How fast is it going at the bottom? (Set ½·m·v² = m·g·h.)

vyou:=sqrt(2∗g∗5m)v_{you} := sqrt(2 * g * 5 m) = 9.9045 m / s ✏️ Your turn: find the speed at the bottom after a 5 m drop. The check verifies kinetic energy = potential energy (½mv² = mgh) — so it never states the speed.
✓ pass abs((1/2)∗mbox∗vyou2−mbox∗g∗5m)<0.1Jabs((1/2) * m_{box} * v_{you}^2 - m_{box} * g * 5 m) < 0.1 J green when kinetic energy equals the potential energy

Where next: Physics 2 lets objects interact — collisions, circular motion, and waves. (Notice m_box never entered v_bottom — Galileo was right: everything falls the same.)