The Physics Philes, lesson 17: Oh, Snap…
In which springs are stretched, springs are relaxed, and gliders glide.
You’ve been very patient with me, reader. I should have attempted to do elastic potential energy questions last week, but I spent a lot of time goofing off. But your patience will be rewarded this week. Let’s snap to it! (Ba dum bum.) Let’s start simple.
A glider with a mass of 0.200 kg is on a horizontal track, which we will assume is frictionless. This glider is connected to a spring with a force constant of 5.00 N/m. I pull the spring 0.100 m and release it with no initial velocity. The glider begins to move back toward its equilibrium position. What is the x-velocity at 0.80 m?
You might be wondering if this equation could be solved by the constant acceleration equations. It can’t because the spring force varies with position. So we have to use elastic potential energy equations. Gravitational energy isn’t a factor in this problem because the spring stays at the same level.
The only force acting on the glider is the spring force, so we don’t need to factor in any other work. We will use the equation
Use only when elastic force does work.
Let’s call the point where the glider is released as point 1 and x = 0.080 m as point 2. The velocity at point 1 is 0 (because there was no initial velocity). Our target variable is the velocity is the velocity at point 2. Now that we’ve identified our target variable, all we need to do is plug in values. Oh wait! We need to find those values. As you’ll remember, the equation for kinetic energy is:
Kinetic energy equation.
And the equation for elastic energy is:
Elastic potential energy equation.
We’ll use those equations to find the kinetic and potential energy values we need.
Kinetic energy at point 1.
Elastic potential energy at point 1.
Kinetic energy at point 2. This equation holds our target variable.
Elastic potential energy at point 2.
Now that we have three of the four needed variables, we can solve for K sub 2:
We’ve solved for the kinetic energy at point 2!
All that is left to do is rearrange the K sub 2 equation to solve for the velocity at point 2.
Because the glider is moving in the negative x-direction, the answer is – 0.30 m/s. Eventually, the spring will pop back and push the glider in the positive x-direction. That’s why one 0.30 m/s is a potential answer.
What if this system was acted upon by other forces? Let’s say the glider starts out at rest at x = 0. The spring is unstretched. This time I apply a 0.610 N force to the glider in the positive x-direction. What is the glider’s velocity when it is moved 0.100 m?
Even though the force I apply on the glider is constant, the spring force isn’t. Also, the total mechanical energy isn’t conserved because of the additional force I exert on the glider. So for this problem we’ll use the generalized energy relationship, which is:
Generalized energy relationship.
Point 1 is at x = 0 and velocity at that point is 0. Point 2 is x = 0.100 m. The target variable is the velocity point 2. As in the prior question, we need to find the values of kinetic and potential energy and the work done by the other force involved:
The value of kinetic energy at point 1.
The value of elastic potential energy at point 1.
This equation for kinetic potential energy at point 2 includes our target variable, the velocity at point 2.
The value of elastic potential energy at point 2.
The value of the work done by the other force exerted.
To find the value of the other work we just multiplied the magnitude of the force by the displacement. As the equations show, there is no mechanical energy at first. But the force I exerted on the glider increased the total mechanical energy to 0.0610 J. Elastic potential energy accounts for 0.0250 J of the total mechanical energy.
Now that we have four of the five variables needed to plug into the generalized energy relationship equation, we can find the kinetic energy at point 2. We just need to rearrange the equation like this:
The rearranged equation.
We just need to plug in the values to find K sub 2.
All that is left to do is rearrange the kinetic energy equation to find the velocity at point 2, our target variable.
The answer is positive because the glider is moving in the positive x-direction.
So that’s it! That’s elastic potential energy. Huzzah!
I’m thinking about doing something a bit different next week. I’m thinking about taking a few of weeks to just brush up on my math. You know, remind myself of all the trigonometry I know and expose myself to a little calculus. I think it would really enhance my ability to explain some of this physics stuff. So, barring a commenter’s uprising, next week you can expect a slight detour on my quest to understand the physical world. It should be terrifying.
Let me know in the comments whether there are any questions or if I’ve made a huge mistake. I always find it helpful.
Featured image credit: anniehp
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excellent fun, once again.
This is just a test
$ K_{1}+U_{el,1}=K_2+U_{el,2} K={1 \over 2} m v^2 $
I DON’T HAVE ANY EQUATIONS ON MY FLASH CARDS WITH DOLLAR SIGNS!
NO LaTeX for you!
$latex It is done like this $
Do’h! I thought it would work like in Freethoughtblogs. ^_^