I am back on intro physics forum Question on springs

[flyingpig]

Homework Statement

It feels good to be back after spending a month in the calculus and beyond section.

Here is my question

[PLAIN]http://img97.imageshack.us/img97/3883/unleduh.jpg

Imagine a spring was pushed back by some distance $$x_0$$ and then it ends up at some distance 5m up the ramp where the dashed line, the equilibrium point is at at the top of the ramp.

Now my question is, what is x? The final distance from the equilibrium point to 5m?

My concern is that is the x in F = -kx the distance or the $$5cos\phi$$ distance because of $$\vec{F} \cdot \vec{dx}$$?

What if I want to find work done by spring? Let's say $$x_0 = -4m$$ from the equilibrium point.

Is it

$$\int_{x_0 = -4}^{x = 5m} \vec{F_s} \cdot \vec{dx}$$ or
$$\int_{x_0 = -4}^{x = 5cos\phi\;m} \vec{F_s} \cdot \vec{dx}$$

 

[wbandersonjr]

I am confused by the question and drawing. It appears that some object is 5m up the ramp and the spring is on the horizontal surface with its unstretched length at the dotted line, the bottom of the ramp. Where is x$_{o}$ on your drawing? Is the spring compressed and released to send the object 5m up the ramp?

 

[flyingpig]

X₀ isn't on the picture, it doesn't matter where x₀ is in the piccture. It was compressed and released and the object ended up 5m up the ramp

 

[wbandersonjr]

and then it ends up at some distance 5m up the ramp where the dashed line, the equilibrium point is at at the top of the ramp.

Now my question is, what is x? The final distance from the equilibrium point to 5m?

Ok, sounds good. Now, you need to find the distance, x, from the 5m point to where? the dotted line, the top of the ramp, the bottom of the ramp?

 

[flyingpig]

5m up the ramp, the equilibrium point is just before you hit the ramp

 

[wbandersonjr]

If you want to find the work the spring does on the object, I would look at work and energy. Think about how work relates to energy, specifically elastic potential, kinetic, and gravitational potential energies.

What if I want to find work done by spring? Let's say x0=−4m from the equilibrium point.

Is it

∫x=5mx0=−4Fs⃗⋅dx⃗ or
∫x=5cosϕmx0=−4Fs⃗⋅dx⃗

If you want to find the work done by the spring, you set up the right integral, but with the wrong limits. If the spring is compressed to -4m, with 0 being the unstretched position of the spring, then the integral should have limits from -4 to 0.

 

[flyingpig]

No I am talking about the 5m. The block ended up 5m up the ramp. So my x_final going to be just x_final = 5 or x_final = 5cos(theta)

 

[wbandersonjr]

In the integral for the definition of work the limits are the initial and final position of the object through which the force in question is doing work. So where does the spring begin doing work and where does it stop doing work? Hint: the spring stops "pushing" the object at a certain point and the object "coasts" from then on.

 

[wbandersonjr]

Maybe i am totally misunderstanding what you are asking

My concern is that is the x in F = -kx the distance or the 5cosϕ distance because of F⃗⋅dx⃗?

=The x in F=-kx is actually a $\Delta$x, the correct equation is F=-k$\Delta$x, which is the difference between the current length of the spring and the equilibrium length.

What if I want to find work done by spring? Let's say x0=−4m from the equilibrium point.

Is it

∫x=5mx0=−4Fs⃗⋅dx⃗ or
∫x=5cosϕmx0=−4Fs⃗⋅dx⃗

Then, if you wanted to set up the integral, yes, it would be $\int$F$_{s}$$\cdot$dx. But, the limits are incorrect in both cases because the spring does not continue exerting a force on the object while it moves up the ramp.

 

[flyingpig]

What is the correct answer then?

 

[fluidistic]

What is the correct answer then?

$\Delta x = 4m$.
For the integral limits, see post #8. I think it's correct.

P.S.: we have almost the same number of posts :D

 

[flyingpig]

$\Delta x = 4m$.
For the integral limits, see post #8. I think it's correct.

P.S.: we have almost the same number of posts :D

But post #8 doesn't have the integral...

Oh please I will beat your post count soon

 

[fluidistic]

The limits are -4 (lower) and 0 (upper). Because as soon as the mass reaches x=0 it detaches from the spring and the mass goes over the ramp without being attached to the spring. So the spring exert a force on the mass from $x_0=-4m$ up to $x=0m$.

If you want you can complicate a bit the problem and assume the mass keeps attached to the spring and it reaches an height h above the ground. You could task yourself to calculate the spring constant...
Well let's keep simple, unless you have finished all your homework and still want more .

 

[flyingpig]

The limits are -4 (lower) and 0 (upper). Because as soon as the mass reaches x=0 it detaches from the spring and the mass goes over the ramp without being attached to the spring. So the spring exert a force on the mass from $x_0=-4m$ up to $x=0m$.

If you want you can complicate a bit the problem and assume the mass keeps attached to the spring and it reaches an height h above the ground. You could task yourself to calculate the spring constant...
Well let's keep simple, unless you have finished all your homework and still want more .

OKay okay okay okay, what if the spring goes all the way up to the ramp? How do we now the spring releases it at x = 0? What happened to inertia?

 

[fluidistic]

OKay okay okay okay, what if the spring goes all the way up to the ramp? How do we now the spring releases it at x = 0? What happened to inertia?

The spring will have inertia if you consider it has a mass. In this problem it is assumed the spring is mass-less.
I do not "know" whether the spring goes all the way up to the ramp. I assumed the action of the spring here was just to push the mass, without being attached to it. Notice that it cannot push forward on the mass even without the ramp once the mass passes the equilibrium position of the spring. I think my assumption seems reasonable considering the difficulty of the problem.
If you consider the mass that's being attached to the spring then the problem is more complicated. Your "x" which should be in fact "$\Delta x$" would be the total distance the mass "slides without friction" (sorry I don't know the name in English) over the ground. But in that case $\vec F \cdot d\vec x$ isn't worth simply Fdx. The angle between $\vec F$ and $d\vec x$ would be a function of time and this is where the difficulty lies to me. I personally find this problem interesting and if I have time I'd love to try to solve it (maybe using the Lagrangian) but I don't think it's that easy. If you're interested, ask help for it... I'm personally busy with another course for now.
By the way I'm glad you posted here to understand more this problem. If I haven't been clear, ask for further help, I'll let others answer better than me. I hope I didn't mess up anywhere.