T2a1s.pdf - Physics 102{1 Assignment 1 solutions(Spring Term 13.7(a The spring constant of each band is k= Fs 15 N = = 1.5 \u00d7 103 N m x 1.0 \u00d7 10-2 m

T2a1s.pdf - Physics 102{1 Assignment 1 solutions(Spring...

This preview shows page 1 - 3 out of 3 pages.

Physics 102 Assignment 1 solutions (Spring Term){1} 13.7(a) The spring constant of each band is 3-215 N1.510 N m1.010msFkx===××Thus, when both bands are stretched 0.20 m, the total elastic potential energy is ()()223121.510Nm0.20 m2sPEkx==×=60 J(b) Conservation of mechanical energy gives ()()ssfiKEPEKEPE+=+, or 210060 J2mv+=+, so ()-3260 J5010kgv==×49 m s{2} 13.11At and conservation of energy gives , 0xAv==2102sEKEPEkA=+=+or 22EAk=(a) At 2xA=, the elastic potential energy is 22212288sEAkkPEkAk====4EFrom the energy conservation equation, the kinetic energy is then 4sEKEEPEE===34E(b) When sKEPE=, conservation of energy yields 2ssEKEPEPE=+=or 2sPEE=. Since we also have 22sPEkx=, this yields ()()22222skAEPEExkkkk=====2A
{3} 13.16(a) , so 0 at KExA==2102sEKEPEkA=+=+, or the total energy is ()()2211250 Nm0.035 m22EkA===0.15 J(b) The maximum speed occurs at the equilibrium position where .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture