1. ## Mechanics Question

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
newtons by
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!

2. ## Re: Mechanics Question

To get started find x when mass hangs freely We know in this position F=5g So solve 5g=8tan^-1(x/3)

3. ## Re: Mechanics Question

Originally Posted by ly667
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
newtons by
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!
you sure the spring force equation is not $F = -8 \arctan\left(\frac{x}{3}\right)$ ? (force should be opposite to displacement)

$\frac{dV}{dx} = -F$

$V = 8 \int \arctan\left(\frac{x}{3}\right) \, dx$

integrating by parts ...

$u = \arctan\left(\frac{x}{3}\right) , dv = dx$

$du = \frac{3}{9+x^2} \, dx , v = x$

$8 \int \arctan\left(\frac{x}{3}\right) \, dx = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) - \int \frac{3x}{9+x^2} \, dx \right]$

$V = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) - \frac{3}{2} \log(9+x^2) + C \right]$

since $V(0) = 0$ ...

$V = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) - \frac{3}{2} \log(9+x^2) + \frac{3}{2} \log(9) \right]$

$V = 8\left[x \cdot \arctan\left(\frac{x}{3}\right) + \frac{3}{2} \log \left(\frac{9}{9+x^2} \right) \right]$

set $V(4) = \frac{1}{2} mv^2$ to find the speed of the mass as it passes equilibrium

4. ## Re: Mechanics Question

I was taking it to be moving vertically so the forces were F upwards and 5g downwards. On reading it again I realize it's moving along an x axis

5. ## Re: Mechanics Question

No it is definitely just tan!

6. ## Re: Mechanics Question

Originally Posted by ly667
No it is definitely just tan!
huh? $\tan^{-1}(x) = \arctan(x)$ , the inverse tangent function.

7. ## Re: Mechanics Question

Sorry I thought you meant in me typing out the question. What about the other value in that given equation, the ex?

8. ## Re: Mechanics Question

That is the entire question, my lamda value is 4.

9. ## Re: Mechanics Question

Originally Posted by ly667
Sorry I thought you meant in me typing out the question. What about the other value in that given equation, the ex?
$e_x$ is a unit vector in the direction of the spring's displacement. The negative sign in front of the force expression makes the force vector opposite in direction to the given unit vector.