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December 26th, 2018, 07:54 AM   #1
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Question How can I find the spring's maximum compression in this problem?

I'm still baffled by this problem which seems relatively easy and although by checking in the back the answers check I'm still unsure if what I did was conceptually or theoretically right or not. Therefore I hope somebody could give me the necessary clarification over this problem:

The problem is as follows:

A spring with $k=25.0 \frac{N}{m}$ is oriented vertically with one end fixed to the ground. A 0.100-kg mass on top of the spring compresses it. Find the spring s maximum compression in each of these cases: (a) You hold the mass while you gently compress the spring,and when you release the mass it sits at rest atop the spring. (b) You place the mass on the uncompressed spring and release it. (c) You drop the mass from $\textrm{10.0 cm}$ above the spring.

What I tried to do was this:

(a) Actually I am not sure how to interpret or understand in an equation what was meant by holding it, compressing and releasing it so it rest atop the spring. My guess was that holding the object would null the gravitational potential energy and by leaving it rest atop the spring would only make effect the gravitation force to act therefore it will be just a weight acting over the spring and since, $F=kx$ (again this part is where I'm unsure, is it a right assumption?)

Therefore the weight would be $mg$;

$mg=kx$

$x=\frac{mg}{k}= \frac{0.1\,kg\times9.8\,\frac{m}{s^2}}{25\,N}=0.03 92\,m$

to which results in $0.0392\,m$

This part seems to check, but again I don't know if this was the "right" interpretation as it would seem contradictory with the following question in part b.

(b) This is where I'm also stuck at as, just placing it over the spring wouldn't it had the same effect as the previous case?.

But then I thought, what if meant that the energy in this case is related to Initially there is gravitational potential energy (being released) and finally there is the elastic potential energy due the spring. (This part I'm unsure)

Anyways, this would be translated into:

$E_{initial}=E_{final}$ as conservation of energy

$mgh=\frac{1}{2}kx^2$

This part I understood the height which the mass travels during fall is the same which would be compressed therefore (in other words $h=x$);

$mgx=\frac{1}{2}kx^2$

$mg=\frac{1}{2}kx$

$x=\frac{2mg}{k}$

$x=\frac{2\times 0.1\,kg \times 9.8\, \frac{m}{s^2}}{25\,N}$

$x= 0.0784\,m$

To where I obtained $0.0784\,m$

This part checks with the answer in the back of my book

(c) Finally for this part I assumed that dropping from a certain height would meant to add up for the height the ball is falling plus the height that will be compressed by the spring;

So again referring to the conservation of energy:

$E_{initial}=E_{final}$ as conservation of energy

$mgh=\frac{1}{2}kx^2$

$mg\left(10\times 10^{-2}m+x\right)=\frac{1}{2}kx^2$

By rearranging a bit (because this would result in a second degree equation):

$\frac{1}{2}kx^2-mgx-0.1mg=0$

by mutiplying by 2 to both sides:

$kx^2-2mgx-0.2mg=0$

Then by applying the solution method for these equations:

$$x_{1,2}=\frac{+2mg\pm\sqrt{\left(2mg\right)^{2} + 4 \times 0.2kmg}}{2\times k}$$

Then inserting the values given:

$x_{1,2}=\frac{+2\times 0.1\,kg \times 9.8\,\frac{m}{s^2} \pm\sqrt{\left(2\times 0.1\,kg\,\times 9.8\,\frac{m}{s^2}\right)^2+4\times 0.2 \times 25 \, N \times 0.1\,kg\times 9.8\,\frac{m}{s^2}}}{2\times 25\,N}$

$x_{1}= 0.136\,m$
$x_{2}=-0.05763\, m$

Therefore I discarded the negative result. So that the maximum compressed reached by the spring would be $0.136\,m$

Which checks with the answer in the back. However the problem was below Work done by a variable force and to solve this problem I had to resort to conservation of energy. So is there another way to solve this problem without using conservation law?.

Last edited by Chemist116; December 26th, 2018 at 08:00 AM. Reason: separated answers for clearer displaying
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December 26th, 2018, 01:02 PM   #2
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part (a) is just finding the equilibrium position for the mass as it rests on the spring.
$\displaystyle \sum F = 0 \implies mg = kx$ is correct.


In part (b), the mass accelerates downward to the point of equilibrium where its velocity will be a maximum, and past equilibrium to a position below it where it comes to a stop. In this situation, the equation of motion for the mass is

$a = \dfrac{dv}{dt} = -kx-mg$

note that $x = 0$ is the spring's rest position ... the mass position will always be negative as it moves downward, hence $-kx \ge 0$

prior to reaching equilibrium, $|-mg| > |-kx| \implies a < 0$

at equilibrium, $|-mg|=|-kx| \implies a = 0$

and past the equilibrium position, $|-kx| > |-mg| \implies a > 0$

Conservation of energy is the simpler method to solve for the lowest position. It can also be done using calculus ...

$\dfrac{dv}{dt} \cdot v = -(kx+mg) \cdot \dfrac{dx}{dt}$

$v \, dv = -(kx+mg) \, dx$

$\dfrac{v^2}{2} = -\left(\dfrac{kx^2}{2} + mgx\right) + C$

since $v = 0$ when $x = 0$, then $C = 0$

$v = - \sqrt{-kx^2 - 2mgx}$

note that velocity is negative since the mass is moving downward ...

Velocity is zero when $-kx^2 - 2mgx = 0 \implies -x(kx + 2mg) = 0$

$x = 0$ is at the start ... the mass is initially released from rest

$x = -\dfrac{2mg}{k}$ ... the mass has reached its lowest position.


Part (c) can be done the same way by breaking the motion into two parts ... (1) free fall motion, and

(2) motion once it hits the spring. Note the initial velocity of the mass when it hits the spring will be a negative value instead of zero as in part (b).
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December 27th, 2018, 07:19 PM   #3
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Quote:
Originally Posted by skeeter
part (a) is just finding the equilibrium position for the mass as it rests on the spring ...
Interesting explanation, but would you mind to offer a "precalculus" alternative for this problem?. I do have some knowledge of what was meant in your steps but what I intended to ask was if there was an alternative without requiring "advanced" mathematics.

Anyways, I'd like to know how did you came to the conclusion to:

$\frac{dv}{dt}\cdot v = - \left(kx+mg\right)\cdot \frac{dx}{dt}$

But before that where does it comes from $a=\frac{dv}{dt}=-kx-mg$?

Will it be due a force analysis meaning that $ma=-kx-mg$ due the fact that is minus because the spring is being compressed and the force is pointing upwards while the weight is going downwards? But yet where the mass is going?

Part (c) looks straightforward when you say breaking up the motion in two parts as you mentioned, $(x+h)$ which seems to validate what I was doing, and going from that on would be pluggin these values in the equation you had derived. I'd be very glad if you could help with my doubts from above and if possible to offer a solution without calculus.
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December 28th, 2018, 06:42 AM   #4
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Quote:
... but would you mind to offer a "precalculus" alternative for this problem?. I do have some knowledge of what was meant in your steps but what I intended to ask was if there was an alternative without requiring "advanced" mathematics.
The only "precalculus method" I know to deal with this problem is energy conservation.

You asked for an alternative method ... I erred by assuming you were familiar with the calculus required to deal with a case of applying Newton's 2nd law of motion with non-uniform acceleration.
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