Hey, guys. So we've already seen one type of potential energy, which was called gravitational. And in this video, I want to introduce you to the other type of which is called elastic or spring potential energy here. Alright? So let's take a look here. The idea is the same. Potential energy, remember, is just stored energy. So just like you store energy when you lift something, springs store energy when you compress or stretch them. So this energy is called elastic potential energy. So let's take a look here. Right? So if you're at the ground, for gravitational potential, your gravitational potential energy is 0. But if you raise it to some height of y, then your U g = m g y here. Well, it's the same idea elastics for springs. Basically, the ground is like the equilibrium position. When springs are relaxed, they have no stored energy so their elastic potential energy is 0. But what happens is that when you push against them and you deform them by compressing or stretching them, then you have some applied forces and some spring forces. And the spring force here depends on your deformation, which is this x here. Alright. So remember that we said the relationship between work and gravitational potential energy was that W g = - δ U g . It's the same idea for springs. The work that's done by springs is going to be the change in spring or elastic potential energy, this - δ U e . So what we saw here is that if this - δ U g = - m g δ y , then the equation for gravitational potential was just U g = m g y . It's the same exact thing we can do for springs. We can basically cancel out these negative signs here, and we can say that the elastic potential energy is really equal to 1 2 k x 2 . So this is the equation that we're going to use in our energy conservation equations. Now how does this change our energy conservation equation? It actually really doesn't. We're still going to write K+U+work done=K+U. The only thing is that up until now, we've only been focused on gravitational potential energies. But now we're actually just going to include elastic potential energies because these things are the same type. They're both potential energies, so we can just combine them. So our potential energy is going to be U g + U e . So all you have to do now is just keep track or keep on the lookout for any springs in our problems. Let's go ahead and take a look here. We have a block that's attached to a horizontal surface. We have the spring constant k and we're going to push the block with a force of 100 newtons. So I've got my applied force. The magnitude is 100, which means that the magnitude of the spring force that pushes back is also 100. So what we want to calculate is the compression distance, how far we've actually compressed the spring. So that's actually this distance right here. This is x . So how do we solve this? Do we use energy? Do we use something else? Basically, the idea here is that whenever you have spring problems in which objects are stationary like we have in this first part here, we're still just holding the block up against the compressed spring. Then we're going to solve this by using forces. And the idea here is if we want to solve the compression distance, remember, which is just x , we can solve this by using Hooke's law, which says that the compression, sorry, the absolute value of the spring force equals k x . So we actually have the spring force and the applied force. They're both 100. And we have the spring constant. So we can figure out what our x is by just rearranging for this. Let's go ahead and do that. So x is really just going to be equal to the magnitude of your applied force divided by k which is just 100 divided by 500

# Springs & Elastic Potential Energy - Online Tutor, Practice Problems & Exam Prep

### Energy in Horizontal Springs

#### Video transcript

### Springs in Rough Surfaces

#### Video transcript

Hey, guys. Hopefully, you got a chance to work this one out on your own. Let's go ahead and check it out together. So we have a 4 kilogram box that is moving to the right with an initial speed of 20, and it's going to collide with the spring. The force constant of the spring is 600. What happens is the spring is going to compress. So this box is pushing up against the spring as it collides with it, and then it's going to compress. So the box comes to a stop due to maximum compression when the spring is coiled up a little more.

Now, we want to write our energy conservation equation. We can't solve this using forces because the force that occurs as you're compressing the spring isn't constant; it varies. We have to use energy conservation. Our equation is: K i + U i + W nonconservative = K f + U f . The initial kinetic energy is significant as the box is moving with some speed. There's no initial gravitational potential energy, and since the spring isn't compressed initially, there's no elastic potential energy. There's no work done by non-conservative forces, and the final kinetic energy is zero, so all its kinetic energy has transformed into elastic potential energy.

We write our expressions: 1 2 m v initial 2 = 1 2 k x final 2 . We move k to the other side to get m v initial 2 k = x final 2 . Take the square root to find: x final = m v initial 2 k . Plug in the numbers to get a compression distance of 1.63 meters.

So, after the block has stopped, it has transferred all of its energy and compressed the spring by 1.63 meters. That's the answer. Let me know if you guys have any questions, and I'll see you in the next one.

A 4-kg block moving on a flat surface strikes a massless, horizontal spring of force constant 600 N/m with a 20 m/s. The block-surface coefficient of friction is 0.5. Calculate the maximum compression that the spring will experience.

## Do you want more practice?

More sets### Your Physics tutor

- A spring of negligible mass has force constant k = 1600 N/m. (b) You place the spring vertically with one end ...
- A spring of negligible mass has force constant k = 800 N/m. (a) How far must the spring be compressed for 1.20...
- A spring of negligible mass has force constant k = 1600 N/m. (a) How far must the spring be compressed for 3.2...
- (I) A spring has a spring constant k of 78.0 N/m. How much must this spring be compressed to store 45.0 J of p...
- If you stand on a bathroom scale, the spring inside the scale compresses 0.60 mm, and it tells you your weight...
- You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car...
- As a 15,000 kg jet plane lands on an aircraft carrier, its tail hook snags a cable to slow it down. The cable ...
- A horizontal spring with spring constant 100 N/m is compressed 20 cm and used to launch a 2.5 kg box across a ...
- The elastic energy stored in your tendons can contribute up to 35% of your energy needs when running. Sports s...
- CALC A 10 kg box slides 4.0 m down the frictionless ramp shown in FIGURE CP10.73, then collides with a spring ...
- A freight company uses a compressed spring to shoot 2.0 kg packages up a 1.0-m-high frictionless ramp into a t...
- A 50 g ice cube can slide up and down a frictionless 30° slope. At the bottom, a spring with spring constant 2...
- The ice cube is replaced by a 50 g plastic cube whose coefficient of kinetic friction is 0.20. How far will th...
- A stretched spring stores 2.0 J of energy. How much energy will be stored if the spring is stretched three tim...
- In a physics lab experiment, a compressed spring launches a 20 g metal ball at a 30° angle. Compressing the sp...
- The spring in FIGURE EX10.21a is compressed by 10 cm. It launches a block across a frictionless surface at 0.5...
- (II) Chris jumps off a bridge with a 15-m-long bungee cord (a heavy stretchable cord) tied around his ankle, F...
- (III) An engineer is designing a spring to be placed at the bottom of an elevator shaft. If the elevator cable...
- (III) A spring ( k = 75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.5...
- (III) A spring ( k = 75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.5...
- (III) A spring ( k = 75 N/m) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.5...
- (II) A vertical spring (ignore its mass), whose spring constant is 875 N/m, is attached to a table and is comp...
- (II) A 1400-kg car moving on a horizontal surface has speed v = 85 km/h when it strikes a horizontal coiled s...
- (II) The spring of Problem 71 has a stiffness constant k = 160 N/m . The mass m = 6.0 kg is released from rest...