# Mechanics Question

• Apr 15th 2012, 09:49 AM
ly667
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!
• Apr 15th 2012, 10:21 AM
biffboy
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)
• Apr 15th 2012, 10:42 AM
skeeter
Re: Mechanics Question
Quote:

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 $\displaystyle F = -8 \arctan\left(\frac{x}{3}\right)$ ? (force should be opposite to displacement)

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

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

integrating by parts ...

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

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

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

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

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

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

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

set $\displaystyle V(4) = \frac{1}{2} mv^2$ to find the speed of the mass as it passes equilibrium
• Apr 15th 2012, 11:03 AM
biffboy
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
• Apr 15th 2012, 02:23 PM
ly667
Re: Mechanics Question
No it is definitely just tan!
• Apr 15th 2012, 02:48 PM
skeeter
Re: Mechanics Question
Quote:

Originally Posted by ly667
No it is definitely just tan!

huh? $\displaystyle \tan^{-1}(x) = \arctan(x)$ , the inverse tangent function.
• Apr 15th 2012, 02:59 PM
ly667
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?
• Apr 15th 2012, 03:10 PM
ly667
Re: Mechanics Question
Attachment 23610

That is the entire question, my lamda value is 4.
• Apr 15th 2012, 03:54 PM
skeeter
Re: Mechanics Question
Quote:

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?

$\displaystyle e_x$ 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.