Answer
given,
k = 250 N/m
q = 900 N/m³
(FSp)s=−kΔs−q(Δs)^3
work done = Force x displacement

limits are x = 0 to x = 0.15 m
work done

![W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15](https://tex.z-dn.net/?f=W%20%3D%20%5B%5Cdfrac%7Bkx%5E2%7D%7B2%7D%2B%5Cdfrac%7Bqx%5E4%7D%7B4%7D%2B%20C%5D_0%5E0.15)

W = 3.375 + 0.1139
W = 3.3488 J
b) % cubic term =
% cubic term =
It’s 1 that would be the lowest point potential energy
I think that would be the moon
"too small to clear objects that are in its orbital path" , which means that it probably not a planet.
hope this helps
In this case, you need the formula below where:
F = force
k = coulombs constant 8.99 x10^{9} N.m^{2} . C^{-2}
q1 = electric charge 1
q2 = electric charge 2
r = the distance between the charges

pls note: make sure your units are correct (in meters etc, not fm (<em>femto-meters</em>)).
Curiously, this question doesn't tell you what atom you are next to the nucleus of. Different numbers of protons in the nucleus of the atom will make for vastly different forces in your answer...
<span>b. weakens as 1/d, where d is the distance between objects.</span>