FOBoi1122
Dec7-03, 10:25 PM
A particle of mass m is attatched to 2 identical springs on a horizontal frictionless table. both springs have spring constant k and an unstretched length L the particle is pulled a distance x along a direction perpindicular to the initial configuration of the springs, show that the force exerted on the particle due to the springs is
F= -2kx(1-(L/(x^2+L^2)^.5))i
please help us, we've tried to use f=-kxcos(theta), where cos(theta) = x/(x^2+L^2) but we are unable to determine why x = (x^2+L^2)^.5 - L which is needed to solve the problem
F= -2kx(1-(L/(x^2+L^2)^.5))i
please help us, we've tried to use f=-kxcos(theta), where cos(theta) = x/(x^2+L^2) but we are unable to determine why x = (x^2+L^2)^.5 - L which is needed to solve the problem