Mг©canique 1re Annг©e Mpsi-pcsi-ptsi Apr 2026

Em=12k(x0−l0)2cap E sub m equals one-half k open paren x sub 0 minus l sub 0 close paren squared Cours : Physics 1: Elements of Mechanics

x(t)=l0+(x0−l0)cos(kmt)x open paren t close paren equals l sub 0 plus open paren x sub 0 minus l sub 0 close paren cosine open paren the square root of k over m end-fraction end-root t close paren 3. Determine the Mechanical Energy The total mechanical energy Emcap E sub m is the sum of kinetic and potential energy: MГ©canique 1re annГ©e MPSI-PCSI-PTSI

Em=Ec+Ep=12mẋ2+12k(x−l0)2cap E sub m equals cap E sub c plus cap E sub p equals one-half m x dot squared plus one-half k open paren x minus l sub 0 close paren squared Em=12k(x0−l0)2cap E sub m equals one-half k open

: Kinetic energy theorem, potential energy, and the conservation of mechanical energy. and the conservation of mechanical energy.