An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 n/m. two small charged objects are attached to this spring, one at each end. the charges on the objects have equal magnitudes. because of these charges, the spring stretches by 0.022 m relative to its unstrained length. determine (
a. the possible algebraic signs and (
b. the magnitude of the charges.

The solution you should use is Hooke's law: F=-kx

It should have the same signs because they repel due to the stretch of the spring. 

a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be 
F = kx 
270 N/m x 0.38 m = 102.6 N 

b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force. 

Sorry can’t u just tryna get points

An atom that has fewer electrons that protons is called a positive ion.