Click to go back to home page Chapter Four: Work and Energy

Work Work is a term that is used in everyday life for a variety of reasons. For example, I could walk up to a boulder and try to push it with all of my might and not accomplish one inch, but I would walk away saying "wow, that was hard work." In physics, if I failed to move the boulder no matter how hard I "worked" at it, I did absolutely NO work on the boulder. I would have to move the boulder before I could say I did any work. Therefore, the work on an object depends on the distance moved and the force used to move it. The equation is given by, W = Fx Now, that is assuming that the force is along the same axis of motion. That is very key when dealing with work. The forces involved on the object must be along the same axis of motion in order for them to do work. What that means is that if the object is moving horizontally, all horizontal forces acting on the object do work. All vertical forces do not. If you consider the above diagram, there is a force that is acting on the box at a certain angle. That force has two components to it. There is Fx which is the component that is acting horizontally, and there is Fy which is the component that is acting vertically. Therefore, Fy is not contributing to the motion of the box if it is moving along the floor. So, Fx is the only force that is doing work because mg and Fno are vertical forces. In order to find Fx, you have to use the same method used in chapter two to find Vx. Here, Fx is one of the legs of the vector triangle that makes up F. To find the Fx component, you have the equation, Fx = Fcos(angle) So, the work equation becomes W = Fcos(angle)(x) For example, if the force is 100 N at 30 degrees AH, and the distance moved is 10 meters, the work equation becomes Fx = 100cos(30)(10) = 866 J Where J is the units of work which is the Joule. Now let's consider friction as shown below. The applied force is to the right and the frictional force is to the left, because friction will always point to the opposite direction of motion. Suppose the kinetic coefficient of friction is .3, the mass is 50 Kg, and the applied force is 500 N. It follows from the kinetic friction equation that kinetic friction equals 147 N. We can now find the work done by the person who applied the 500 N, and we can find the work done by friction, if the crate is moved 20 meters. Work done by person: W = (500)(20) = 10,000 J Work done by friction: W = (-147)(20) = -2940 J The work done by friction is negative because it is working against the person pulling the crate. The negative does not mean direction here because work is a scalar quantity, so it does not have a vector arrow associated with it. However, it is proper to designate what the work is doing. The applied force is doing work to speed the crate up. The frictional force is working to slow the crate down. Those two forces are working agaisnt each other, so it is necessary to show one as positive and the other as negative, because in order to find the total work done, we have to add them together. Therefore, the total work is found by, Total Work = 10.000 - 2940 = 7060 J This means that of the 10,000 J done by the person, 2940 J went into just overcoming the frictional work, which means that 7060 J is left to actually accelerate the crate. It is obvious that the crate is accelerating because the applied force is larger than the kinetic force. Here, we have shown that if there is work left over after overcoming the frictional work, the crate will accelerate. What happens if the work done by the person is exactly equal to the work done by friction? If that is the case then the applied force will have to equal the frictional force which means the net force is zero. Whenever, the net force is zero and the crate is already moving, it will continue to move with a constant speed. Therefore, If the total work is zero, the object will not speed up or slow down. Conceptually speaking, the work done by friction is trying to slow the object down, and the work done by the person is trying to speed the object up. So, if they equal each other, there is no work left over after overcoming the frictional work, so the object will do neither. It will just continue at a constant speed. So, what happens when the work done by friction is greater than the work done by the person? You guessed it. The crate will slow down. The three scenarios can be expressed neatly as: 1. If you outwork friction, the object will accelerate (speed up). 2. If your work equals the frictional work, the object will not accelerate (it will maintain a constant speed). 3. If friction outworks you, the object will accelerate (slow down). Work and Energy As shown above, if the total work on an object is greater than zero, the object will accelerate, which means the object will have a final speed once the work is finished. For example, imagine pushing a toy car that is at rest for a certain distance. Once you let go, the car will continue with a certain speed. So, the question becomes, how does the work done on an object relate to the speed given? At first, you would possibly consider that work is directly proportional to the speed, which is like force and acceleration. If that were true, if you doubled your speed, you would have to double the work. However, it turns out that work is actually directly proportional to the speed squared, which is shown by, In order to make this an equation, we have to multiply the speed squared by a constant. It turns out that the constant is 1/2 times the mass of the object, which yields the equation, A square relationship is very different than a direct relationship. For example, if you want to double the speed, you have to actually do four times the amount of work. The reason is due to the squared speed. Here is an example to illustrate the point. Let's say an object with mass of 1.0 kg starts from rest and accelerates to 5.0 m/s. The work needed to accomplish this is given by, W = 1/2(1)(5^2) = 12.5 J Now, let's see how much work is required to take the same object from rest to 10 m/s (double the original speed). The work is again given by, W = 1/2(1)(10^2) = 50 J You can see that the second case is 4 times the size as the first case. So, to double the speed, you had to quadruple the work. So, how much more work is needed to triple the speed. To solve this, you have to square the three which is nine. So, it takes nine times the work to go from rest to 15 m/s as it does to just go 5 m/s! The above equation 1/2mV^2 is actually the equation for the kinetic energy of a moving object. Kinetic energy is the energy of motion. So, we can now say that doing work on an object gives the object energy. In this case, the energy is kinetic. This relationship with work and energy is represented in the Work-Energy Theorem, which states the total work done on an object is equal to its change in kinetic energy, which can be stated mathematically, Where Vf is the final speed and Vi is the initial speed. This is the more general equation that applies to all situations. When we used the expression in the above examples and introduction of the topic, we assumed the initial speed was zero, which reduces to, It should be noted that like work, kinetic energy is a scalar quantity. You cannot put a direction to kinetic energy even though there is a v in the equation. The v represents speed in this case and not velocity. This will be more apparent with later examples. Let's now do an example that utilizes the work energy theorem. Suppose a baseball pitcher throws a 100 mph fastball. We can calculate the force in pounds required to achieve that speed. We can assume that between the wind up and the release, the distance through which the force was exerted is roughly 60 inches, which is about 1.5 meters. We need to convert 100 mph to meters per second, which equals around 44.7 m/s. The ball's initial velocity is zero, and a baseball's mass is .15-kg. So the equation becomes, F(1.5) = 1/2(.15)(44.7^2)-1/2(.15)(0) Solving for F yields 99.9 Newtons. Dividing by 4.448 to convert to pounds gives 22.5 pounds. This doesn't seem like much, but imagine doing that over the course of two hours. It adds up. Which situation do you think would require the most energy: accelerating a car from 0 to 100 mph, or from 100 to 200 mph. Or, would it be equal amounts of energy for do both? Most people would say equal because both situations entail the same difference of speed. However, appling the equations tells a different story. We have already converted 100 mph to m/s in the last problem. 200 mph is about 89.4 m/s. If we consider the mass of the vehicle to be 1000-kg, we can find the work to accelerate from 0 to 100 mph by, W = 1/2(1000)(44.7^2) - 1/2(1000)(0) = 999045 J Finding the work for 100 to 200 mph, W = 1/2(1000)(89.4^2) - 1/2(1000)(44.7^2) = 2997135 J The last example required almost three million Joules! That is exactly three times the energy of the first amount. The reason depends on the square of the velocity. Remember when the velocity is doubled, the work is quadrupled. Because of this exponential growth, the energy needed to accelerate a car greatly increases as the car's speed gets bigger. This is why it is important to maintain an even speed on the highway. Constantly speeding up and slowing down by even 10 mph at high speeds consumes more gas as compared to doing the same thing from start to stop. Returning to the relationship between work and speed, there is a flip side to that. If it takes a certain amount of work to accelerate to 30 mph, that work is converted to kinetic energy. However, something has to happen to that energy when the car stops. That energy is converted back into work, either over a long distance such as breaking, or it could be over a small distance such as hitting a tree. Either way, the car does work equal to the kinetic energy it attained. If the work is over a long distance, the force will be small. If it is over a small distance, the force will be large. As long as the product of the force and the distance equals the amount of work needed to stop, you can have any number of forces. This concept is summed up nicely in what I call concept equations. This illustrates how even though the work is the same, if the distance is large, the force is small and vice versa. The point of all this is to show the danger of high speed crashes. If it takes a certain amount of work to accelerate to 30 mph, and the car crashes into a tree, the work done to stop the car can be correlated to the amount of damage sustained to the car. So, if the car is traveling at twice the speed, 60 mph, the damage sustained is not twice as much. It is four times. At 90 mph, the damage would be nine times as much! This of course is assuming the car hits a rigid tree that does not move. Nevertheless, you get the idea. This actually gets worse. Remember relative velocity. Say one car is traveling on a road at 30 mph and another car is traveling on the same road towards the other car with the same speed. The relative velocity between the cars is 60 mph. Therefore, if they crash head on, the damage would be four times more as compared to justing hitting the tree at 30 mph. This gets scary when talking about the highway. Two cars traveling at 60 mph in a head on collision will sustain not 9 but 16 times the damage! Needless to say, getting behind the wheel is not to be taken lightly. There are more example problems that use this theorem in the questions archive page under chapter four. Work Done Agaisnt Gravity Of the four ramps pictured below, which one would give the ball the greatest speed at the bottom? Assume the heights of the ramps are equal.

It turns out that the four ramps would give the exact same speed! The reason lies in the fact that they all have the same height. To understand this, we need to analyze this in terms of energy.

Work must be done in order to get the ball onto the ramp. However, this time the object is being moved vertically, which means the work is done against the gravitational force that is pulling the object down. Therefore, the force used to move the ball is equal to the ball's weight, which is mg. If we use h, which stands for height, for the distance moved, the work equation becomes,

W = mgh

This equation is a result of substituting mg for F and h for x. Recall from the previous section that work gives an object kinetic energy. In the case agaisnt gravity, the work used to put the ball on top of the ramp did not result in kinetic energy, because the ball remains at rest after the work was finished. In this case, work still gives an object energy, only this time the energy is not a matter of velocity, it is a matter of position with respect to the surface it will hit when let go. This energy of position is called Potential Energy . You can think of work against gravity as giving an object the potential to move. In the case of the ramps, the ball has the potential to roll to the bottom. Since the work done equals the potential energy, the equation becomes,

PE = mgh

In summary, the potential energy depends on mass and how high an object is with respect to the surface it will hit when let go. This means that a book could be on the second floor of a building and have zero potential with respect to the table it is on, and have a certain potential with respect to the sidewalk at the same time. It all depends on the reference frame.


Back to the ramps. What happens to the potential energy of the ball when it rolls down to the bottom? Since the height is getting smaller as the ball rolls, the potential energy is getting smaller as well. What else is the ball doing as it rolls? It is gaining speed, which means it is gaining kinetic energy. So, the potential energy is actually converting to kinetic until the ball reaches the bottom. At the bottom, the height is zero, so the potential energy is zero. At this point, all of the potential is converted to kinetic, which means the ball has attained its maximum speed.


Potential energy is a vector quantity in the sense that it is positional. It is possible to have a negative height if the ball ends up below the zero position. This will be illustrated further in a later example.

The Conservation of Energy
This transfering of energy from potential to kinetic is based on one of the most important laws in all physics: The Conservation of Energy. It states:

The initial energy will equal the final energy in a closed system.
Essentially, the conservation law is saying energy cannot be created or lost. The closed system refers to the parameters set on the system. In the case of our example, the ball, ramp and earth encompass the system. In the case of planetary orbits, the Solar System would be the system. It all depends on the scope of the energy.

The profound nature of this law is really quite impressive. It says that you can account for every Joule of energy in any system. If the ball is traveling slower then it should at the bottom of the ramp due to friction, the energy "lost" due to friction would be exactly equal to the work done by the friction. So, the energy is not really lost. It is just converted to another form, namely heat between the ramp and ball. Scientists have never found a violation of this law in nature. If it were to ever be violated, it would cause a major revolution because so much of physics is built around this law.

The conservation law can be written in equation form as:


The ho and vo is the initial height and velocity, and hf and vf is the final height and velocity. It is very important to note that the velocities in this equation are vertical velocities only! They cannot include horizontal velocities because this depends on gravity, which does not affect horizontal velocities. This will make more sense when we do an example.

For the first example, we are going to revisit a problem from chapter one. Recall the picture below.


When we dealt with this picture in chapter one, we used the kinematic equations to determine the final velocity and displacement. We are now going to use the energy conservation equation to solve for the velocities and distances to show that it finds the same answers as before. So, to begin we will find the maximum height the ball attains when the initial speed is 29.4 m/s. The key to dealing with this equation is to start from the left and work your way to the right. Ask yourself what is the starting height, and is there an initial speed? This will fill in the first two terms on the initial side. You then decide if there is a final height and a final velocity to fill in the final side of the equation. Therefore, for this example we know the initial height is zero, the initial speed is 29.4 m/s, the final height is the question, and the final speed is zero. The conservation equation becomes,

mg(0) + 1/2m(29.4^2) = mghf + 1/2m(0)

Notice since we started from zero height and the final velocity at the top of the path is zero, we set ho and Vf equal to zero. So, the equation reduces down nicely to

1/2m(29.4^2) = mghf

Notice now that both terms have mass so they cancel. Again, mass does not matter. Plugging in 9.8 for g and solving for hf gives the answer of 44.1 meters, which is the same answer found in chapter one. Let's now find the final velocity after the ball returns. Again, ho will be zero, as well as hf this time because the ball is returning to the original position. Therefore,

mg(0) + 1/2m(29.4^2) = mg(0) + 1/2mVf^2



1/2m(29.4^2) = 1/2mVf^2

Again, the masses cancel, and taking the square root to solve for Vf gives 29.4 m/s! Notice that the conservation equation does not give the actual direction. For example, the answer found in chapter one was a negative 29.4 m/s. The conservation equation deals with speed, not velocity, again because kinetic energy is a scalar quantity. However, notice we used zero for both ho and hf. That is because potential energy deals with displacement rather than distance.

I want to now take a look at the picture above solely in terms of energy to further illustrate the conservation law. We will isolate the first four positions after the ball was thrown up and calculate both the kinetic and potential energies at each position. Assume the ball's mass is 1.0-kg.


Using the data from chapter one's calculations, this picture illustrates that the total energy throughout the ball's trip remains constant. In this case, it starts with all kinetic energy and converts to potential on the way up until it reaches its maximum height where the energy is now all potential.

to wrap this problem up, let's consider what would happen if the man was standing on a cliff and the ball dropped passed the original starting height of zero. This is where h can be negative. Using the kinematic equations, we can determine what the displacement would be one second after it passed the man on the cliff. The initial velocity is -29.4 m/s, yo would be zero, and time would be one second. The kinematic free fall equation for distance would become,

yf = 0 + (-29.4)(1) - 4.9(1^2) = -34.3 m

This means the ball would be 34.3 meters below the man after one second of fall. Also, since the ball was falling for one second, the final speed would be increased by 9.8 m/s, so it would be -39.2 m/s.


Now, we can use the conservation equation to solve for the final speed starting from when the ball was thrown up in the air. This means Vo will equal 29.4 and h0 is zero.

mg(0) + 1/2m(29.4^2) = mg(-34.3) + 1/2mvf^2

The masses cancel and since the final potential energy is negative, it can be moved to the other side and made positive.

1/2(29.4^2) + g(34.3) = 1/2vf^2

Solving for Vf indeed gives the answer of 39.2 m/s!

What about the total energy at this point? It would seem that the energy would be greater, but you have to keep in mind the negative potential. It turns out, the total energy is still constant!


The negative energies seems a bit contradicting in this example. You have to remember that these energies are all with respect to the starting position at the man's hand. If you made the man's hand not the zero height but the height that his hand was from the actual ground, and therefore, made the ground zero, all of the energies would be positive.



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