A spring exerts a 17 N force after being stretched by 3.0 cm from its equilibrium length. By how much will the spring force increase if the spring is stretched from 5.0 cm away from equilibrium to 6.0 rm cm from equilibrium?
I don't know how to approach this at all or how to solve it.
physics - drwls, Thursday, July 28, 2011 at 3:48am
The spring constant k is 17/3 = 5.67 N/cm
The force increase going from x = 5 to x = 6 cm deflection is k*(6-5) = 5.67 N