Two springs, with force constants k1=175N/m and k2=270N/m, are connected in series

Question

When a mass m=0.50kg is attached to the springs, what is the amount of stretch, x?

Answer 1

F = M*g = 0.50 * 9.8 = 4.9 N.

X = (1m/175N)*4.9N. + (1m/270N)*4.9N = 0.028 + 0.0181 = 0.0461 m.