Mechanics Question

**ly667** · April 15th 2012, 09:49 AM

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!

**biffboy** · April 15th 2012, 10:21 AM

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

**skeeter** · April 15th 2012, 10:42 AM

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

**biffboy** · April 15th 2012, 11:03 AM

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

**ly667** · April 15th 2012, 02:23 PM

No it is definitely just tan!

**skeeter** · April 15th 2012, 02:48 PM

Originally Posted by ly667

No it is definitely just tan!

huh? $\tan^{-1}(x) = \arctan(x)$ , the inverse tangent function.

**ly667** · April 15th 2012, 02:59 PM

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

**ly667** · April 15th 2012, 03:10 PM

$Mechanics Question-mechanics-question.png$

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

**skeeter** · April 15th 2012, 03:54 PM

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.

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