A spring with a constant of 120 newtons per metre is 0.35 metres long when unloaded. A force is exerted on the end of the spring. And the spring’s length changes to 0.45 metres. What is the magnitude of the force exerted on the end of the spring?
Now to answer this question, the first thing to do is to draw a diagram of the spring when it’s unloaded and when a force is exerted on the end of the spring. So first of all, here’s the spring when it’s unloaded.
We know that the length of this spring is 0.35 metres. And we know its spring constant is 120 newtons per metre. Okay, so when a force is exerted on the end of the spring, so let’s say we exert a force on the right end of the spring, the spring ends up extending. And we know that it extends to a length of 0.45 metres now.
Now what we’re trying to work out is the force that we exert on the spring. We’ll call this force 𝐹. As well as this, we’ll label the spring constant of the spring. And we’ll call it 𝑘. The reason for this is that we can use something known as Hooke’s law.
Hooke’s law tells us that the force applied to a spring is equal to the constant of the spring multiplied by the extension of the spring. And it’s really important that 𝑥 represents the extension, in other words, how much the spring has stretched or compressed by. In this case, the spring is stretching. So the value of 𝑥 is this distance here, labelled in blue, because that’s how much longer the spring has become.
Now happily, we can actually calculate the value of 𝑥 because we have the length of the spring before it was stretched and the length of the spring after it was stretched. So the value of 𝑥 is simply this length, 0.45 metres, minus this length, 0.35 metres. So we write this down here. And the right-hand side of the equation evaluates to 0.10 metres.
Now at this point, we’ve just worked out the extension of the spring. And we’ve already been given the force constant of the spring. So we can work out the value of 𝐹, the force exerted on the spring. To do this, we just have to substitute in values to the equation. We say that 𝐹, which is what we’re trying to find out, is equal to 𝑘, 120 newtons per metre, multiplied by 𝑥, which is 0.10 metres.
And, importantly, we’re working in standard units here. The value of 𝑘 that we have is 120 newtons per metre. And newtons per metre is indeed the standard unit of the force constant. As well as this, we found the extension of the spring in metres, which is the standard unit of distance. So we’ve got both of these quantities in their standard units, which means that the force is also going to be in its standard unit, which is the newton, at which point we can evaluate the right-hand side of the equation to give us 12 newtons. And so our final answer is that the magnitude of the force exerted on the end of the spring is 12 newtons.