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

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3 years ago

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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.

Physics

1 answer:

Kruka [31] · Answer 1 · 2022-02-21T01:27:26+03:00

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
<span>F = kx
270 N/m x 0.38 m = 102.6 N
</span>
b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.