Spring constant formlula for Force and Work Done

[danielsmith123123]

Homework Statement

A spring with a spring constant of 185 N/m hangs vertically from a surface, as shown in the image below. A mass of 200 g is then hung from the spring and the spring stretches to a new length as shown at right. How much (Δx) has this spring been stretched from its equilibrium length?

Relevant Equations

F_spring = -kx
WD_ spring = (1/2)kx^2

Finding x by force formula
- only force acting is gravity
ma/-k = x
(0.2)(-9.8)/185 = x
0.010594594 = x

Finding x by wd formula
WD_ spring = (1/2)kx^2
F x = (1/2)kx^2
2(mg)/k = x
[2(0.2)(-9.8)]/ 185 = x
0.021189189 = x

how come the work done and force formulas produce different values for x. I noticed without the 1/2 in the WD formula, I would get the same answer but isn't this the standard?

 

[kuruman]

At the equilibrium point, which is Δx below the relaxed end of the spring, the mass is at rest and the spring stretched by Δx. If you try to find Δx using conservation of energy, when the mass is Δx below the point of release it is still moving.

The standard here is to use a free body diagram and set the elastic force on the mass equal in magnitude and opposite in direction to the gravitational force exerted by the Earth.

Also, you don't need that many significant figures. They will not make your answer "more correct."

 

[danielsmith123123]

Thank you

 

[MatinSAR]

At the equilibrium point, which is Δx below the relaxed end of the spring, the mass is at rest and the spring stretched by Δx. If you try to find Δx using conservation of energy, when the mass is Δx below the point of release it is still moving.

The standard here is to use a free body diagram and set the elastic force on the mass equal in magnitude and opposite in direction to the gravitational force exerted by the Earth.

Also, you don't need that many significant figures. They will not make your answer "more correct."

Sorry, Can you please explain why the forces should have same magnitude and opposite direction? I've read somewhere that at equilibrium point(which is the lowest point) the net force that acts on the object is not zero.

However I dont see the related picture so maybe I misunderstood the problem.

 

[PhanthomJay]

I think the problem is not clearly worded, but maybe not since an image was not attached. There is a difference if the mass is hung from the spring and slowly lowered to its at rest equilibrium position, as opposed to being hung from the spring and quickly released. I believe the problem assumes the first case, where the net force on the mass is zero as kuruman has pointed out. In the second case, when the spring extends its full extent, its velocity is momentarily zero, but it is accelerating upward, so the net force is not zero in that case.

 

[kuruman]

I've read somewhere that at equilibrium point(which is the lowest point) the net force that acts on the object is not zero.

An object is at equilibrium when the sum of all the forces acting on it is zero. You probably confused the lowest point of the potential energy function (i.e. its minimum) with the lowest position of the mass. Study the figure below. It shows the potential energy function for the harmonic oscillator. The equilibrium point is at y = 0. To the left of it is the lowest position marked by the arrow.

 

[MatinSAR]

In the second case, when the spring extends its full extent, its velocity is momentarily zero, but it is accelerating upward, so the net force is not zero in that case.

An object is at equilibrium when the sum of all the forces acting on it is zero. You probably confused the lowest point of the potential energy function (i.e. its minimum) with the lowest position of the mass.

So I was thinking about 2nd case which is not happening in this problem. Thank you.

 

[Steve4Physics]

Finding x by wd formula
WD_ spring = (1/2)kx^2
F x = (1/2)kx^2
2(mg)/k = x
[2(0.2)(-9.8)]/ 185 = x
0.021189189 = x

In addition to the other replies can I add this...

The calculation shown above is incorrect. That's because the force exerted by the spring ($F$) while it is being stretched is not constant (not always equal to $mg$).

The force, best written as $F(x)$, is zero when the stretching begins and increases until it equals $mg$. Since $F$ is proportional to extension (Hooke's law), the average value of $F(x)$ during the extension is $\frac {0 + mg}2$. So the force used in the calculation should be $\frac {mg}2$ not $mg$.

It's worth noting that the work done by the spring is represented by the area under the $F(x)$ vs. $x$ graph. If you know a bit of calculus, this is $\int_L^{L+\Delta x} F(x)dx$ where $L$ is the unstretched length.

 

[nasu]

I think that by "F" in that equation he means the weight.

 

[Steve4Physics]

I think that by "F" in that equation he means the weight.

Assuming 'that equation' is
F x = (1/2)kx^2
then F is not the weight; it is half the weight - see Post #8.

 

[nasu]

He means to equate the work done by weight (mgx) with the change in elastic potential energy. I did not say that he did it right. Just that the error is not in assuming constant force for the work of elastic force.

 

[erobz]

The issue is for it to be in static equilibrium at deflection $x$, another unmodeled force (e.g. a force from a hand ) must have been aiding it on its journey ( or it oscillated under the action of non-conservative forces until its kinetic energy was lost as heat), if it hadn't the system would be oscillating around $x$. Correct?

 

[kuruman]

Here is how to do it using mechanical energy conservation. It takes longer than using a FBD but works fine.

Problem
A spring of constant $k$ hangs vertically. Mass $m$, held by a disembodied hand, is attached to the spring and released from rest. Find how much the spring stetches in terms of the given quantities and the acceleration of gravity.

Solution
We define the initial position of the mass as a triple point:
It is the origin of coordinates. The instantaneous position of the mass $y$ is referred to it. In this case $y\leq 0.$
It the reference point where the gravitational potential $U_g(y)$ is zero. In this case $U_g(y)=mgy$ and $U_g(y)\leq 0.$
It is the reference point where the elastic potential $U_{\text{el}}(y)$ is zero. In this case $U_{\text{el}}(y)=\frac{1}{2}ky^2$ and $U_{\text{el}}(y)\geq 0.$

We note that once released, the mass executes simple harmonic motion with amplitude $A$ and returns to the triple point after an integer number of oscillations. Thus, the maximum distance below the triple point is $2A.$ It follows that the equilibrium point is $A$ below the triple point.

The mechanical energy at the triple point $y=0$ is $$ME_1=K+U_g(0)+U_{\text{el}}(0)=0+0+0=0.$$ The mechanical energy at the lowest point $y=-2A$ is
$$ME_2=K+U_g(-2A)+U_{\text{el}}(-2A)=0+mg(-2A)+\frac{1}{2}(-2A)^2=-2mgA+2kA^2.$$Mechanical energy conservation requires that $$ME_1=ME_2\implies 0= -2mgA+2kA^2 \implies A=\frac{mg}{k}.$$ Answer
The equilibrium point is at $y=-\dfrac{mg}{k}.$