Video: Calculating a Mass Attached to a Spring from Spring Constant and Spring Expansion

The spring shown in the diagram has a constant of 50 N/m. What is the mass of the object attached to it? Answer to the nearest gram.

04:25

Video Transcript

The spring shown in the diagram has a constant of 50 newtons per meter. What is the mass of the object attached to it? Answer to the nearest gram.

Taking a look at our diagram, we see in the first case our unstretched spring, the spring with no mass attached to it. This spring has an equilibrium length of 15 centimeters. But then, when we attach a mass to the end of our spring and we can call that mass 𝑚, we see the spring stretches out and that stretch is given as a distance of 2.5 centimeters. From this information, we want to solve for the mass of the object attached to the spring.

As we start doing this, it’s helpful to realize that once the mass is attached to the spring and the spring is stretched out, then this mass-spring system is in equilibrium. This means that the force upward on the mass is equal to the force downward on it. And to see just what those forces are, let’s sketch out a free body diagram of this mass. Just like any mass of object, this mass is subject to the force of gravity. And we know that that force is equal to the mass of our object multiplied by the acceleration due to gravity.

Now, if this were the only vertical force acting on our mass, then it would be accelerating downward. But it’s not the only force, there’s the restoring force of the spring that acts on the mass as well. And to know what this force is, it will be helpful to recall Hook’s law. This law says that the restoring force provided by a stretched or compressed spring is equal to the spring constant 𝑘 multiplied by the displacement of the spring from equilibrium.

So, when it comes to the upward force on our mass, that’s equal to the force provided by the spring. It’s equal to the spring constant times 𝑥, the displacement of the spring from its natural length. And just like we said, this mass is stationary as it hangs off the end of the spring. That means these two forces are equal in magnitude to one another. In other words, we can write that 𝑘 times 𝑥 is equal to 𝑚 times 𝑔. If this weren’t true, if these two forces weren’t equal in magnitude to each other, then the mass would be accelerating. But we know that it’s not. So, 𝑘 times 𝑥 is equal to 𝑚 times 𝑔. And it’s 𝑚, the mass of our object, that we want to solve for.

To do that, let’s divide both sides of the equation by the acceleration due to gravity 𝑔, causing that term to cancel out on the right. And what we find is that the object’s mass is equal to 𝑘 times 𝑥 divided by 𝑔. At this point, we can recall that the acceleration due to gravity is approximately 9.8 meters per second squared. And then, looking up at our problem statement, we see that the spring constant 𝑘 of our spring is 50 newtons per meter. The last thing we want to solve for and then plug into our equation is 𝑥.

And at this point, we’ll need to be a bit careful. Notice that over in our diagram, we’re given this 15-centimeter distance. That’s this distance here, the length of our unstretched spring. But the 𝑥 in Hook’s law, and the 𝑥 we’re using in our equation, is not the natural length of a spring. Rather, it’s the stretch or compression of that spring from equilibrium. In other words, is the displacement of the spring. And that is given as a distance of 2.5 centimeters. So, the 𝑥 in our equation is 2.5 centimeters. 𝑘 is 50 newtons per meter. And 𝑔 is 9.8 meters per second squared. At this point, with all our numbers plugged in, we’re almost ready to calculate the mass 𝑚.

There are two things we should keep in mind though. First, our displacement is in units of centimeters, whereas the distances in the other terms in our expression are in units of meters. This means that the terms in our expression aren’t all on equal footing. And we’ll want to make sure they are before we combine them. To do that, let’s convert 2.5 centimeters to that same distance in meters. Since 100 centimeters is equal to one meter, that tells us if we shift the decimal place two spots to the left in this distance, we’ll get that same distance in meters. It’s 0.025 meters. And now, we’re ready to multiply all these values together.

When we do, we get this answer 0.12755 and so on kilograms. But we know that we want to give our answer to the nearest gram. Since there are 1000 grams in one kilogram, that means to convert this answer to some number of grams, we’ll want to shift the decimal place three spots to the right. Doing so, we get 127.55 dot dot dot grams. And lastly, we want to round this answer to the nearest gram. That means we look at this value right to the right of the decimal point, and since it’s equal to or greater than five, that indicates that to round this answer to the nearest gram, we’ll round this number up by one. In other words, our answer is 128 grams. That’s the mass of the object attached to the spring to the nearest gram.

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