Question Video: Elastic Potential Energy Physics • 9th Grade

A spring has 50 J of energy stored in it when it is extended by 2.5 m. What is the spring’s constant?


Video Transcript

A spring has 50 joules of energy stored in it when it is extended by 2.5 meters. What is the spring’s constant?

Okay, so to answer this question, we need to recall the relationship between the energy stored in the spring, the extension of the spring, and the spring constant. The relationship that we’re looking for is this one. It tells us that the energy stored in the spring which is elastic potential energy is equal to half multiplied by the spring constant 𝑘 multiplied by the extension of the spring 𝑥 squared. Now in this equation, we already know the value of the potential energy. It’s 50 joules; that’s the energy stored in the spring. And we know that this energy is stored in the spring when the extension is 2.5 meters.

So based on this, we can calculate the value of 𝑘, the spring constant. To do this, we first need to rearrange this equation. We can multiply both sides of the equation by two, so that on the right hand side this factor of half cancels with the two. And then we can divide both sides of the equation by the extension 𝑥 squared because this way it cancels on the right-hand side. And so all we’re left with on the right is the spring constant 𝑘. At this point then, all that’s left for us to do is to plug in some values on the left-hand side.

We can say that two multiplied by the potential energy stored, that’s 50 joules, divide by the extension 2.5 meters whole squared is equal to the spring constant 𝑘. Now before we do anything, let’s deal with the squared term by making sure that we square everything inside the parentheses. In other words, we need to square 2.5 and the unit meters. And when we do this, we find of that the denominator is 6.25 meters squared. So now, we can start thinking about units. We know that, in the numerator, the only unit that we’ve got is joules; that’s the unit of energy. And in the denominator, we’ve got meter squared. And so our expression on the left-hand side is going to be in units of joules per meter squared. And we can give our answer in these units if we wish. But most commonly, the spring constant of a spring is given in units of newtons per metre.

So we need to make sure that joules per metre squared and newtons per metre are equivalent. And to do this, we can recall that work done, which we’ll call 𝑤, has units of joules. So when we use these square braces it means that we’re talking about the units of the thing inside the square braces. The units of work are jewels. And we can also recall that work done is defined as the force applied to an object multiplied by the distance moved by that object. Now, a force has a unit of newtons and a distance as a unit of meters. So the unit of work done is the same as the unit of force multiplied by the unit of distance or newtons times meters. And so now we see that newton times meters is the same thing as joules.

Hence, we can say that joules per meter squared is the same thing as newton multiplied by meters divided by meters squared. Then one power of meters in the numerator cancels with one power of meters in the denominator. And we’re left with simply newtons per meter which is exactly what we wanted. So at this point, all we have to do is to evaluate numerically what the left-hand side of this equation is. And we know that our units are going to be in newtons per meter. So when we evaluate numerically the left-hand side of our equation, we find that the spring constant is 16 newtons per meter. And hence, that is the answer to our question.

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