In my physics class, the current chapter discusses kinetic energy. This is my very first physics course, and unfortunately, remote instruction has made it a little more difficult to understand certain concepts. I am struggling with things related to the spring constant ($k$). I am aware that the spring force is equal to $-kx$ with Hooke's law and I am growing to be more and more familiar with the integrals that explain how to find the work related to the spring force. However, in my attempts to practice, I've been confronted by a problem that I can't really figure out how to dissect.

The problem goes as such:

In the figure, we must apply a force of magnitude 84.0 N to hold the block stationary at $x = -3.00$ cm. From that position, we then slowly move the block so that our force does $+5.00$ J of work on the spring–block system; the block is then again stationary. What are the block's positions? ((a) positive and (b) negative)

It is accompanied by this image, which explains the motion of the block attached to the spring:


So far, I've tried using the integral from bounds $-3$ to $0$ of $-kx \text{ }dx$. This left me with $kx^2/d$. I attempted to plug in some numbers into what I thought was a general solution, using $(84/-3)/2*3^2$. However, I believe my struggles are with finding out what the spring constant will be in this situation. I understand the spring constant is equal to force over position, but I can't find a way to manage to piece that together correctly, having tried $84/-3$. I want to know how to be able to figure out the spring constant when it isn't given, as this is my first time finding a problem where it hasn't been a part of the given information.

Thank you very much in advance for your help! I really appreciate it. I hope you're all staying safe during these stressful and confusing times.

Best regards, Denise

  • $\begingroup$ You need to show us the figure. It's not clear if $x$ is simply the position of the block when the 84 N force is applied, or if its the change in length due to the force. In the equation $F=-kx$, $x$ is the change in length as a result of the applied force $F$. $\endgroup$
    – Bob D
    Commented Mar 31, 2020 at 1:04
  • $\begingroup$ I think I have fixed the problem with the figure. Can you see it now? Apologies for the technical difficulty. $\endgroup$ Commented Mar 31, 2020 at 1:41
  • $\begingroup$ OK now I think it's clear. Just to confirm my understanding, x = 3 cm is the change in length, and $x=d=3cm$, when $F$=-84 N$, correct? $\endgroup$
    – Bob D
    Commented Mar 31, 2020 at 1:57
  • $\begingroup$ the spring constant is given in newtons per cm, your integral from -3 to 0 is wrong, and the upper bounds of integration are the solution, so you can't be plugging them in. $\endgroup$
    – JEB
    Commented Mar 31, 2020 at 1:59
  • $\begingroup$ @BobD As far as I am able to interpret it, that is correct. Again, while I am struggling with trying to find the spring constant here, I think that I am also having some issues deciphering what exactly is given to us in this question. Thanks for reaching out so far! $\endgroup$ Commented Mar 31, 2020 at 2:02

1 Answer 1


So this problem gives us $k$ in a sneaky fashion, by describing a particular force at a particular distance:

$$ F(x=d) = F_0$$

with stated values of $d$ and $F_0$. It appears the problem tries to confuse us with a positive displacement drawing and a negative displacement value...but seriously, I am not going to worry about that. Springs are restorative, so force always points to equilibrium which is traditionally $x=0$, and if it is not: protest.

Fortunately, Hooke's Law is linear:

$$ F(x) = -kx $$

so that solving for $k$ is straightforward.

From there, the problem states that the spring is moved slowly. The slowly part means we can ignore kinetic energy, this problem is strictly about potential energy.

What is given is a total amount of done, $W_0$, and we know the definition of work going from $a$ to $b$ is

$$ W = \int_a^b F(x)dx $$

so plug in what we know:

$$ W_0 = \int_{a=-3\,\rm cm}^{b=?}{{-kxdx}}= \frac k 2[b^2-a^2] $$

which should have 2 real solutions, with $b$ on the left and right of $a$.

  • $\begingroup$ This has helped so much! Thank you very much for your help! I feel like I have a better understanding of the concept overall! $\endgroup$ Commented Mar 31, 2020 at 2:38

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