To get started find x when mass hangs freely We know in this position F=5g So solve 5g=8tan^-1(x/3)
A body of mass 5kg is attached to one end of a nonlinear spring which does not
obey Hooke’s law. The other end of the spring is ﬁxed, and no other forces act on the
particle, which moves in one dimension. The force exerted by the spring is given in
F = 8tan^-1(x/3)ex
where x is the extension of the spring beyond its natural length measured in metres,
and ex is a unit vector pointing in the direction of increasing x.
The spring is extended by a distance of 4 m; the body is held at rest for a moment
and then released. By considering the potential V (x) corresponding to the force F, or
otherwise, determine the speed at which the body is moving when it reaches the point
x = 0.
[Hint: integration by parts may be useful in order to determine V (x).]
ex is really e and a small x slightly below the e but i couldnt figure out how to insert here.
I havent a clue where to start nor do any of my classmates! Help!