Spring Problem Involving Variables and Constants Only

[Argonaut]

Homework Statement

An experimental apparatus with mass $m$ is placed on a vertical spring of negligible mass and pushed down until the spring is compressed a distance $x$. The apparatus is then released and reaches its maximum height at a distance $h$ above the point where it is released. The apparatus is not attached to the spring, and at its maximum height it is no longer in contact with the spring. The maximum magnitude of acceleration the apparatus can have without being damaged is $a$, where $a > g$. (a) What should the force constant of the spring be? (b) What distance $x$ must the spring be compressed initially?

Relevant Equations

$$F=ma$$
$$U_{\text{grav}}=mgh$$
$$U_{\text{el}}=\frac{1}{2}kx^2$$

Here is my attempt at the solution:

a) The apparatus may only experience acceleration $a > g$ while in contact with the spring. Since the spring exerts the greatest force when it is the most compressed, the apparatus will undergo the greatest acceleration at that point. So Newton's second law gives
$$\sum F = ma$$
$$kx-mg = ma$$
Therefore, the force constant of the spring should be $$k = \frac{m(a+g)}{x}$$.

b) There are only conservative forces in the system, so energy is conserved. Let point 1 (with $y=0$) be the point where the apparatus is released and let point 2 be the point where it reaches height $h$. Then
$$U_1=U_2$$
$$\frac{1}{2}kx^2 = mgh$$
Expressing $x$
$$x=\sqrt{\frac{2mgh}{k}}$$

However, the official solution at the back of the book is
a)
$$k = \frac{m(g+a)^2}{2gh} $$
b)
$$x = \frac{2gh}{g+a} $$

I could 'reverse-engineer' both solutions. However, I don't understand how I should have known to express $k$ in terms of $m$, $a$, $g$ and $h$, and not $x$. Is it because of part b? Because essentially, both $k$ and $x$ are target variables and only the rest are known?

 

[TSny]

$$k = \frac{m(a+g)}{x}$$ $$x=\sqrt{\frac{2mgh}{k}}$$

These look good. Can you combine them so that $k$ is expressed in terms of $m,g, a$ and $h$ instead of $m, g, a$ and $x$?

 

[haruspex]

how I should have known to express k in terms of m, a, g and h, and not x.

The question ought to have stated, in part a, that the answer should be in terms of m, g, a and h.
I suppose you might have noticed that your answers expressed x in terms of k, then k in terms of x, in such a way that each could be expressed without the other; and since x usually refers to an unknown to be found, and you know k is to be found…

 

[Argonaut]

These look good. Can you combine them so that $k$ is expressed in terms of $m,g, a$ and $h$ instead of $m, g, a$ and $x$?

Yes and they give the book solution.

The question ought to have stated, in part a, that the answer should be in terms of m, g, a and h.
I suppose you might have noticed that your answers expressed x in terms of k, then k in terms of x, in such a way that each could be expressed without the other; and since x usually refers to an unknown to be found, and you know k is to be found…

Got it.

Thanks, both. It makes more sense now that I typed it up and pondered some more.