Question

A 0.328-kg mass is attached to a spring with a force constant of 52.7 N/m. Part A If the mass is displaced 0.270 m from equilibrium and released, what is its speed when it is 0.134 m from equilibrium

Answer 1

Given:- А. B U20 m m = 0.320kg K= 52.7 N/m x=0.270 m v=?, at 0.134m from equilibriune. solution:- K= 52.7 N/m There are 3 positions. mm "Å → Equilibrium 'B' → mass displaced point H0.27-* C where speed will be zero (u =o) 'C' Position 0.134m from equilibrium. where speed is v = ? G 6. 136m 0.134

force Requivecſ to displaced the mass AB fe-kx 52.7 x 0.27 F=-14.229 N When the mass released from 'B', initial speed (4=0), distance of B toc is S = 0-136m, and acceleration will be F=ma a = 14.229 - 43.38 m/s2 0-328 uzo, a=43:38 m/s², S=0.136 m from motion of equ- v²=u²+ zas U = 0 + 2x 43.30x0.136 V=3.435 m/s ANS V = 11.799 At point c'

Negative force means it will tends in opposite to the displacement. There are 3 positions A, B and C. At C where the speed is required to find, is 0.134m from A or 0.136m from B.