We hope you don’t recoil when listening to Episode 14. It begins with discussing the common spring scale.
We hope you don’t recoil when listening to Episode 14. It begins with discussing the common spring scale (1:08) before diving into Hook’s law (1:40). Don’t forget that you can use springs vertically and horizontally (4:12). Springs can do work on other objects, or they can possess elastic potential energy if they are defined as part of the system. You can look at springs through a graphical or algebraic lens, and still get the same answer! (5:12)
The Question of the Day asks: (6:22)
Question: If a given spring is deformed by twice as much, then how much more work can it do on an object?
a) half as much
b) twice as much
c) four times as much
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Hi and welcome to the APsolute RecAP: Physics 1 Edition. Today’s episode will focus on energy in springs and Hooke’s Law.
Let’s Zoom out:
Unit 4 – Energy
Topic 4.1-4.3
Big Idea – Force Interactions, Change, and Conservation
A lab cart is attached to a horizontal spring which is secured to a rigid wall and the spring is compressed and then released. The spring returns to its normal length, and the cart moves with a newly gained velocity. Systems that involve springs can feel confusing, especially when masses on a spring are set in motion. Springs are nothing to be afraid of, and we will sort out many of their intricacies in this episode.
Let’s Zoom in:
In many physics labs, you utilize a piece of equipment known as a spring scale. It usually measures the mass or weight of an object. You have probably used it many times already, but, how does it work? As it turns out, we have springs to thank… hence the name, spring scale. As you add more mass, gravity applies a larger gravitational force and yet the object is in force equilibrium (net force is zero). There must be a restorative force keeping the gravitational force in check. That is the spring force.
Imagine you place a 1 N weight on a scale and the spring attached to the scale’s needle stretches by 1 cm. Then you notice that each successive 1 N weight that you add, the needle moves an additional 1 cm. This is exactly what Robert Hooke saw in the 17th century. Hooke saw that a direct linear relationship existed between the force a spring applies to an object and the spring’s deformation. You will notice in our weight on a scale example, that the direction of the deformation, down in our case, is opposite the direction of the force the spring applies to the weight to maintain equilibrium. As a result, Hooke’s law is stated as: the absolute value of the spring force is equal to the product of the spring’s deformation (stretch or compression) and a constant we call the spring stiffness constant (k).
If you try the weight on the scale example, you can graph the spring force which would have to equal each of the weights added versus the deformation distance in meters. The slope of that line would be the spring constant. In our perfect example, that’d be a rise of 1 N for every 0.01 m of deformation yielding a spring constant of... 100 N/m. This value will be true for this spring no matter what. Well, kind of. In physics we love dealing with idyllic scenarios so I should mention that the initial bit of stretch for a spring is actually non-linear, but then it quickly gets linear and works great!
Now, imagine we take that same spring (k = 100 N/m) and we lay it flat on a table and are able to compress the spring 0.5 meters. We attach a 1 kg lab cart with frictionless wheels, and we allow the spring to do work on the car over the 0.5 m distance. The car would speed up wouldn’t it? But, how fast would it be going? Without knowing the applied force, it feels impossible. But, if we look at the force vs deformation distance graph, it would have a 100 N/m slope, and so for 0.5 m of run, it should rise from 0N to 50 N of force. That triangular graph’s area would equal the work done on the cart by the spring and can be set equal to the kinetic energy gained by the cart. That area is ½ the base (0.5m) times the height (50N). You should see that the cart will now be traveling with a speed of 5 m/s.
Another way to look at that problem is to define the system to include the spring, and then evaluate how the elastic potential energy of the cart-spring system is ½ kx2. Or, 12.5 J in our case. No surprises there, that’s what we got when we looked at the work done by the spring on the cart previously. Ultimately, in most cases, it is up to you to define the system. Depending on what is given in the particular situation, then it may be easier to define the system to include the spring, other times maybe not. Personally, I think of questions from more of a graphical mindset so I would choose the work done by the spring on the cart, but many of my students prefer the more algebraic approach. Either way - the physics checks out!
To Recap…
Each spring has a stiffness constant associated with it, that doesn’t change under normal working loads. Springs can do work on other objects, or they can possess elastic potential energy if they are defined as part of the system. That’s it! See, springs are nothing to worry about.
Coming up next on the APsolute RecAP Physics 1 Edition, Mechanical Energy Types and the Conservation of Energy
Today’s Question of the Day focuses on the work done by a spring.
Question: If a given spring is deformed by twice as much, then how much more work can it do on an object?
a) half as much
b) twice as much
c) four times as much