Spring Force (Hooke's Law) - Online Tutor, Practice Problems & Exam Prep

When a force is applied to a spring, it compresses or stretches, demonstrating Hooke's Law, which states that the spring force (F_s) is equal to the negative of the applied force (F_a), or $F=-kx$, where k is the spring constant and x is the deformation. The spring force acts as a restoring force, always opposing the deformation. Understanding these principles is crucial for analyzing mass-spring systems and their motion.

Spring Force (Hooke's Law)

So whenever you're pushing or pulling against a spring with some applied force, that spring pushes back on you in the opposite direction, and it forms an action-reaction pair. So, for instance, in this example, I'm going to take this spring and push against it with some applied force. That's the action. And then the spring pushes back on me with an equal but opposite direction force. That's the reaction force. And so we say that the spring force is equal to the negative of your applied force, and that's equal to negative k times x, where x represents the object's deformation here, and k is just a constant. It's just a force constant. The only thing you need to know about this negative sign here is that it just means that that force points in the opposite direction. And because that negative sign often just goes away and we're only just asked for the magnitude of these forces, we're going to use this more powerful equation. The absolute value of all these forces are equal, and that's just equal to k times x. The negative sign just goes away.

So to see how all this stuff works, let's just go ahead and take a look at an example. So I've got, in this first example here, I'm pushing the spring to the left, so that means that my applied force is equal to 120. And so that means that the spring force is in the opposite direction also equal to 120. Now I'm told that the k constant, which is that force constant, is equal to 20. And that k constant really just means it's a measure of how stiff the spring is, how hard it is to push or pull it. And so what I'm asked for is how much it compresses by. So I've got these two forces and I'm pushing on the spring and I've compressed it by some distance x, and that is the object's deformation. That deformation is always measured relative to this dotted line here, which is basically the relaxed position. By the way, the other word that you might hear for that often is the equilibrium position, where x is equal to 0.

So let's go ahead and just set up my equation here. So I've got the magnitude of these forces, right, are equal to each other, and that's just equal to k times x. I'm looking for x and I've got k, right, and I've also got the force. So my applied force, which is the spring's force, was equal to 120. My k is equal to 20, and then I've got x, which is my variable here. So you go ahead and solve for that, and you get a deformation of 6 meters. That's how much I compressed it by.

So let's take a look at a different example here because here I was pushing against the spring. Now what I'm going to do is I'm going to pull against it. So now what happens is I'm taking this spring, pulling it out so my applied force is here, which means that the spring force is equal and opposite to that. Now I'm looking for how much force I need to pull it out to some distance here. So let's just write out my formula. F_s=F_a=k⋅x. Now I'm looking for the actual forces here. I know what the k is. Now I just need to know what my deformation is. I've got 10 meters, 16 meters. Which one of those things represents my distance, my deformation? Well, before, we had a compression distance that was given to us. But here, all we're told is that the spring was originally 10 meters, and then once you've pulled on it, the spring has now become 16 meters long. So, the amount that you've pulled on it or deformed it is 6 meters. That's equal to 6 meters.

So now I've got my x, that's just equal to 6 meters. Now just go ahead and solve the problem. So I've got F_s=k⋅x, and I've got k equal to 40, and x is 6 here. So I've got the spring force equal to 240 newtons. In this first example, I had a deformation of 6, a k constant of 20, and then the force was equal to 120. Well, in this example, I've got the same exact deformation, 6. I've got the k constant, which is 40 (double that), and the spring force was equal to 240. So what happens is I doubled the spring constant, the force constant, and then I doubled the force. This shows that when the x is the same, and k gets doubled, then the force doubles as well. This clearly demonstrates that k represents how hard it is to deform a spring. You have to use more force to deform it the same amount. The units for k are in newtons per meter, which measures how many newtons you must use per meter of displacement. Now you should remember that, but if you ever forget, it's just F_s=k⋅x. You can get that. The units for F_s are in newtons, k is in newtons per meter, and the units for x are in meters.

So in both of these examples here, whether I've pushed or pulled against the spring, the force that acts in the opposite direction always wants to pull this system back to the center. Therefore, F_s is known as a restoring force, and it always opposes the deformation, whether you push or pull. That's everything we need to know about Hooke's law. Let's go ahead and use some examples.

A 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring's force constant (b) How much force is needed to compress it to 0.7 m?

• A. 250 N/m; 175 N
• B. 50 N/m; 15 N
• C. 250 N/m; 75 N
• D. 50 N/m; 35 N

Acceleration of Mass-Spring Systems

Attach a mass to a spring, it becomes a mass-spring system. So now you're not pushing up against the spring itself. You're going to take some block and push it up against the spring with some applied force. Now we know that the spring is going to push back against you in the opposite direction with some F_S. And we know that those two forces are equal to each other, except for this minus sign here. So we say that F_S is equal to negative F_A, and that's equal to negative k times x. So all that means is you're pushing up against something and you're compressing it some distance here.

And so now if we consider all the forces that are acting on this object, we can use f=ma to figure out what's going on. The first thing is that this m always refers to the mass of the object itself, and so we're always going to assume that the mass of the spring is equal to 0. So now we've read all the forces, we've got negative F_A, and then we've got the positive spring force, and those two things are equal and opposite, so they're going to cancel out. That means ma is equal to 0. And so therefore, if you're just pushing up against this thing and keeping it there, there's no acceleration.

So now what happens if I release that applied force? If I remove my hand, now the only thing that's pushing up against the spring or this object here is that spring force. The applied force goes away. And so that's equal to negative kx. So now the spring force is the force that's going to want to push or pull it back to the equilibrium. So now if we consider all these forces here, we've got F_S=ma. We know that's kx, so we have −kx=ma. This is a powerful formula.

Now if we want to solve and calculate for the acceleration, we can just go ahead and divide over the mass, and we get acceleration is equal to negative k over m times x, where again, this negative sign reminds you that it's in the opposite direction of wherever you're pushing or pulling it.

So, let's check out an example. In this example, we've got a 0.60 kilogram block that's attached to some spring. The k constant is equal to 15. And we're told that this thing is stretched 0.2 meters to the right beyond its equilibrium point. So we've got this deformation x is equal to 0.2 meters. So now, given those two components, we're supposed to figure out the force that's acting on this object and also its acceleration.

For the force on the block, that's going to be the spring force. So let's write the whole equation now. We've got −kx and that's equal to ma. So if I want to figure out the spring force here, all I need is the compression distance or the stretching distance and the force constant, and I have both of those. So I've got the spring force is equal to negative 3 N. The reason we got a negative sign is that we're taking the right direction to be positive, so this negative sign just means it points to the left. That makes sense because once you release it, that force is going to be acting in that direction.

So now we're supposed to find the acceleration. Let's just go ahead and use our formula: A=−k/mx. We've got all of those numbers. So a is just equal to −15/0.6⋅0.2, and we get an acceleration equal to −5m/s². Again, this negative sign just means that it points to the left. That makes sense because this is the only force that's acting on this thing. That means the acceleration must be towards the left. If you ever forget this formula, you can always get back to this just by using the spring force is equal to mass times acceleration. Those two things are equal to each other.

Alright, guys. That's it for this one. Let's keep going!

You push a 3-kg mass against a spring and release it from rest. Its maximum acceleration is 10m/s² when pushed back 0.5m. What is the (a)spring constant and (b) restoring force at this point?

• A. 60 N/m; 30 N
• B. 120 N/m; 45 N
• C. 100 N/m; 35 N
• D. 30 N/m; 15 N

Here’s what students ask on this topic:

What is Hooke's Law and how is it mathematically represented?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement or deformation of the spring from its equilibrium position. Mathematically, it is represented as:

$F=-kx$

where $F$ is the spring force, $k$ is the spring constant, and $x$ is the deformation. The negative sign indicates that the force exerted by the spring is in the opposite direction of the applied force.

How do you calculate the spring constant (k) in Hooke's Law?

The spring constant $k$ can be calculated using Hooke's Law formula:

$F=-kx$

Rearranging the formula to solve for $k$ gives:

$k=\frac{F}{x}$

Here, $F$ is the force applied to the spring, and $x$ is the deformation of the spring. The units of $k$ are Newtons per meter (N/m).

What is the significance of the negative sign in Hooke's Law?

The negative sign in Hooke's Law, $F=-kx$, indicates that the spring force is a restoring force. This means that the force exerted by the spring is always in the opposite direction of the applied force. If you compress the spring, it pushes back; if you stretch the spring, it pulls back. The negative sign ensures that the spring force always acts to return the spring to its equilibrium position.

How do you determine the deformation of a spring when a force is applied?

To determine the deformation $x$ of a spring when a force $F$ is applied, you can use Hooke's Law:

$F=-kx$

Rearranging the formula to solve for $x$ gives:

$x=\frac{F}{-}$

Here, $F$ is the applied force, and $k$ is the spring constant. The deformation $x$ is measured in meters (m).

What is a mass-spring system and how does it relate to Hooke's Law?

A mass-spring system consists of a mass attached to a spring. When the mass is displaced from its equilibrium position, the spring exerts a restoring force described by Hooke's Law:

$F=-kx$

In this system, the mass experiences an acceleration due to the spring force, which can be described by Newton's second law:

$F=ma$

Combining these equations gives:

$-kx=ma$

This relationship helps analyze the motion of the mass-spring system, including oscillations and vibrations.