NapkinCalc

Physics 1 — Motion & Energy

Energy conservation

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

Lift a mass and you bank m·g·h of potential energy; drop it and the bank converts to kinetic energy ½·m·v². No air resistance means nothing leaks — the check below proves the books balance.

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

Where next: insert the momentum & impulse or centripetal force formulas from the Σ library, or change m_box and notice v_bottom does not care — Galileo was right.