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.
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to

where

is the charge density

is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
If the scientist repeats the experiment over and over and gets the same results. Also if the scientist peer reviews the experiment to make sure there is no bias in his or her results.
Potential energy at top:
PE = mgh
PE = 40 x 9.81 x 12
P.E = 4,708.8 J
Kinetic energy at bottom:
KE = 1/2 mv²
KE = 1/2 x 40 x 10²
K.E = 2,000 J
P.E = K.E + Frictional losses
Frictional losses = 4708 - 2000
Frictional losses = 2708 J
The answer is D.