Potential and Kinetic Energy

Potential and Kinetic Energy

Hi, and welcome to this video on kinetic and potential energy. Energy is a very important concept in physics that helps us describe how much work an object can do. While the word energy is often used in everyday language, the meaning in physics is very specific and may not exactly coincide with the meanings you are used to hearing. Today, we’ll be looking at the different types of energy and how they affect the world around us every day.

There are two main categories of energy: kinetic energy and potential energy. Kinetic energy applies to objects in motion and potential energy is the stored energy of an object and depends on the object’s position. Different types of energy may express energy values in different combinations of units, but the standard unit for energy is the Joule (J).

First, let’s talk about kinetic energy. Since kinetic energy applies to moving objects, it is dependent on the object’s velocity. The equation for kinetic energy is: Kinetic energy equals one half times the mass times the velocity squared.

\(KE=\frac{1}{2}mv^{2}\)

Here, it is apparent that when v equals zero, the kinetic energy is also equal to zero. This makes sense, because if the object is at rest, it won’t have any kinetic energy. Also, kinetic energy is directly proportional to the mass of the object and the velocity squared, so the faster an object is moving and the heavier it is, the higher the kinetic energy will be.

Kinetic energy is closely related to work. In physics, work is a measure of the energy it takes to move something a given distance. A force must act on the object in order to move it and cause work to be done. You must take all of the forces acting on the object into consideration when thinking about the total amount of work that was done on the object. So, the equation for work in terms of force and distance is: the net work done on an object, W-net, equals the net force that acted on the object, F-net, times the distance over which the work was done, or D.

\(W_{net}=F_{net} d\)

However, it is sometimes more useful to think of work in terms of energy. The Work-Energy Theorem states that work is equal to the change in the kinetic energy of an object. So, work is equal to the final value of the kinetic energy minus the initial value of the kinetic energy.

\(W=∆KE-KE_{f}-KE_{i}\)

It is also important to note that work can be either positive or negative. For example, if you are pushing a box across a table, the direction you are pushing the box would be considered the positive direction. The friction between the box and the table is a force in the direction opposite to the motion, so the work being done by friction is negative.

Basically, negative work makes it more difficult to get positive work done.

Let’s think about an example using the Work-Energy Theorem. Imagine you are standing at the top of a building and you are holding a ball over the edge.

While you are holding the ball over the edge, it is at rest. If you let it go, it will be in free fall until it reaches the ground and is at rest again.

There are several questions we need to ask here. Are there any forces at work here? If so, what are they? Is work being done when you drop the ball? Does the amount of work change as the ball falls to the ground?

To answer the first question – yes, there is a force at work here. The force that is acting here is gravity. Specifically, the force of gravity equals mass times the object’s acceleration.

\(F_{G}=mg\)

Gravity is pulling the ball to the ground when you drop it and the acceleration due to gravity, g, is making the ball go faster and faster as it gets closer to the ground. The positive direction here is downward, in the direction of the movement. Since there is kinetic energy when the ball is moving (remember, it has velocity), then work is being done on the ball by gravity. So, what is the total amount of work being done here from when the ball is at rest to the time it hits the ground? Well, the velocity is zero at both of those times, so the total work done is zero after the ball hits the ground. This is because the normal force of the ground pushes upward on the ball and causes negative work to be done. The work from gravity is then canceled out at that point.

However, if we look at the amount of work done when the ball is only halfway to the ground, we will have a non-zero value of kinetic energy,

\(\frac{1}{2}mv_{f}^{2}\)

since the ball has a non-zero velocity. When we plug this into the Work-Energy Theorem with one half times the mass times the initial velocity squared equals zero,

\(\frac{1}{2}mv_{i}^{2} = 0\)

we will get a non-zero value for work. So, the value for the amount of work done will change depending on where the ball is on its way to the ground.

So, where does potential energy come in? Potential energy depends on an object’s position. Unlike kinetic energy, an object can have a non-zero potential energy when it is at rest or when it is moving. The example where we are dropping the ball from the building involves a form of potential energy called gravitational potential energy. As you might imagine, the name comes from the fact that we are dealing with the gravitational force. The equation that we use for gravitational potential energy is: Gravitational potential energy equals mass times acceleration times the height of the object from the ground.

\(PE_{grav} = mgh\)

So, in our previous example, the potential energy is at its highest when you are holding the ball at the top of the building because h is at its maximum value. The mass of the ball is not changing here and g is always nine point eight 9.8 meters per second squared near the surface of the Earth, so the potential energy is entirely dependent on how far away the ball is from the ground in this problem. When the ball hits the ground, it has no more potential energy.

It’s important to understand that during the fall from the top of the building to the ground, the ball has both kinetic and potential energy and both are constantly changing.

If you look again at the equation for gravitational potential energy, you might notice that the “mg” is actually the gravitational force acting on the object and when we multiply that by the distance the object travels, we get the work done against gravity or by gravity, depending on the direction. So, if you are lifting the object up from the ground, you are doing work against gravity and when the object is in free-fall, gravity is doing the work.

Work must be computed over some distance, or some change in “h.” It doesn’t make sense to ask what the work is at one point, say 3 meters above the ground. For work, you might ask something like ‘what is the work done when you lift the ball from a height of 2 meters to a height of 3 meters?’ So, the work done by gravity is actually equal to the change in potential energy.

\(W_{g} = ∆PE = mg∆h\)

Another type of potential energy is elastic potential energy. Elastic potential energy involves any type of object that can be compressed, stretched, or otherwise deformed in such a way that it “wants” to move back into an equilibrium position. Common examples are springs or rubber-bands. Elastic materials all have a spring constant, k, that is dependent on the material and describes how elastic it is. The force of a compressed or stretched spring is described by Hooke’s Law,

\(F_{s} = k∆x\)

where k is the spring constant and delta x is the distance that the spring has been displaced. When the spring is in a stretched or compressed state, the potential energy associated with this state is: Potential energy of the spring equals one half times k times delta x squared.

\(PE_{s} = \frac{1}{2}k(∆x)^{2}\)

So, let’s think about an example of a typical spring problem. Imagine a spring attached to a wall. When the spring is not compressed or stretched, the end of the spring rests at the spot we will call x equals 0. At this point, it has no stored energy. If you compress the spring by pushing it to the left, you’ve moved it by a distance delta x, and it now has stored or potential energy. The force of the spring is pressing in the opposite direction of the direction that you’ve compressed it. Conversely, if you stretch the spring by delta x, the force now points in the opposite direction. The elastic potential energy is the same here if delta x is the same.

Now that we’ve learned all about kinetic energy, potential energy, and work, let’s test it out with some problems!

Let’s say you have a bowling ball of mass M on a ramp that is height h from the ground. What is the potential energy when it is at rest at the top of the ramp and when it is at a height of one third times h?

Is it A) zero, zero 0, 0
B) zero, one third Mgh 0, 1/3Mgh
C) Mgh, one third Mgh 1/3Mgh
or D) Mgh, impossible to tell without knowing how long the ramp is

The correct answer is C! The bowling ball will have potential energy as long as it is on the ramp, so it won’t be zero until it reaches the ground. At the top it is a height of h from the ground and when it is at a height of one third h, that is how far it is from the ground. The path it takes doesn’t matter!

In this same example, how much total work is done from the time the bowling ball is going a speed of V to the time that it hits the ground (assuming it stops when it hits the ground)?

Is it
A) negative one half mv squared \(-\frac{1}{2}mv^{2}\)
B) one half mv squared \(\frac{1}{2}mv^{2}\) or
C) zero 0

The correct answer is A! Here the initial velocity equals V and the final velocity equals zero, because the bowling ball is stopped by the ground. Plugging these into the work-energy theorem, we get negative one half mv squared-12mv2. The reason work is negative here is that we have to take into account the work done by the ground to stop the ball. This value is the net work of the ball-ground system.

Imagine you have a spring with spring constant K hanging from the ceiling. In its equilibrium position, it is a height of two 2 meters off of the ground. What is the elastic potential energy of the spring when you pull it down until it is only one point five 1.5 meters off of the ground?

Is it
A) two K
B) one eighth \(\frac{1}{8}K\)
C) K
D) \(\frac{9}{8}K\)

The correct answer is B! You’ve displaced the spring by two minus one point five equals zero point five 2 – 1.5 = 0.5 or one half ½ meters. Plugging this into potential energy of the spring equation , you get one eighth \(\frac{1}{8}K\).

I hope this review was helpful! Thanks for watching, and happy studying!

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by Mometrix Test Preparation | Last Updated: August 6, 2020