9.1: Conservation of Energy for Particles

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The concepts of work and energy provide the basis for solving a variety of kinetics problems. Generally, this method is called the Energy Method or the Conservation of Energy, and it can be boiled down to the idea that the work done to a body will be equal to the change in energy of that body. Dividing energy into kinetic and potential energy pieces as we often do in dynamics problems, we arrive at the following base equation for the conservation of energy.

\[ W = \Delta KE + \Delta PE \]

It is important to notice that unlike Newton's Second Law, the above equation is not a vector equation. It does not need to be broken down into components which can simplify the process. However, we only have a single equation and therefore can only solve for a single unknown, which can limit the method.

Work:

To understand how to use the energy method we first need to understand the concepts of work and energy. Work in general is a force exerted over a distance. If we imagine a single, constant force pushing a body in a single direction over some distance, the work done by that force would be equal to the magnitude of that force times the distance the body traveled. If we have a force that is opposing the travel (such as friction), it would be negative work.

A box on a flat, horizontal surface experiences a pushing force that moves it to the right, as well as a smaller magnitude of frictional force that points to the left. The box's original position on the left side of the image is indicated by a dotted outline, and the distance between that original position and its current position on the right side of the image is labeled as d. — Figure \(\PageIndex{1}\): In instances with a constant force and a constant direction, the work done to a body will be equal to the magnitude of that force times the distance the body travels. For forces opposing the motion, the work will be negative.

\[ W_{push} = F_{push} * d \]

\[ W_{friction} = - F_{friction} * d \]

For instances where forces and the direction of travel do not match, the component of the force in the direction of travel is the only piece of the force that will do work. Following through with this logic, forces that are perpendicular to the direction of travel for a body will exert no work on a body because there is no component of the force in the direction of travel.

A box on a flat, horizontal surface experiences a pushing force that moves it to the right. The push force vector is angled, making an angle of theta with the horizontal as it points down and to the right. The box's original position on the left side of the image is indicated by a dotted outline, and the distance between that original position and its current position on the right side of the image is labeled as d. The box also experiences an upwards normal force from the surface it sits on. — Figure \(\PageIndex{2}\): Only the components of a force in the direction of travel exert work on a body. Forces perpendicular to the direction of travel will exert no work on the body.

\[ W_{push} = F_{push} \cos (\theta) * d \]

\[ W_{normal} = 0 \]

In the case of a force that does not remain constant, we will need to account for the changing force. To do this we will simply integrate the force function over the distance traveled by the body. Just as before, only the component of the force in the direction of travel will count towards the work done, and forces opposing travel will be negative work.

\[ W = \int\limits_{x_1}^{x_2} F(x) \, dx \]

Energy:

When discussing energy in engineering dynamics, we will usually break energy down into kinetic energy and potential energy. Kinetic energy is the the energy mass in motion, while potential energy represents the energy that is stored up due to the position or stresses in a body.

In its equation form, the kinetic energy of a particle is represented by one half of the mass of the body times its velocity squared. If we wish to determine the change in kinetic energy, we would simply take the final kinetic energy minus the initial kinetic energy.

\[ KE = \frac{1}{2} m v^2 \]

\[ \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]

As a note, a body that is rotating will also have rotational kinetic energy, but we will save that for our discussion of work and energy with rigid bodies.

Potential energy, unlike kinetic energy, is not really energy at all. Instead, it represents the work that a given force will potentially do between two instants in time. Potential energy can come in many forms, but the two we will discuss here are gravitational potential energy and elastic potential energy. These represent the work that the gravitational force and a spring force will do, respectively. We often use these potential energy terms in place of the work done by gravity or springs. When including these potential energy terms, it's important to not additionally include the work done by gravity or spring forces.

The change in gravitational potential energy for any system is represented by the product of the mass of the body, the value \(g\) (9.81 m/s² or 32.2 ft/s² on the earth's surface), and the vertical change in height between the start position and the end position. In equation form, this is as follows.

\[ \Delta PE_{gravity} = m * g * \Delta h \]

Two identical boxes are located at the top of the image, horizontally in line with each other. A dotted outline shows that the boxes were originally at the bottom of the page, and traveled upwards by the same distance (Delta h) to reach their current locations although the paths they took were different. The box on the right moved directly upwards, while the box on the left took a curving path. — Figure \(\PageIndex{3}\): When finding the change in gravitational potential energy, we multiply the mass by \(g\) (giving us the weight of the object) and then multiply that by the change in the height of the object, regardless of the path taken.

To find the change in elastic potential energy, we will need to know the stiffness of the spring (represented by \(k\), in units of force per distance) as well as the distance the spring has been stretched or compressed from its natural resting length (represented by \(x\), in units of distance). Once we have those values, the elastic potential energy can be calculated by multiplying one half of the stiffness by the square of the distance \(x\). To find the change in elastic potential energy, we simply take the final elastic potential energy and subtract the initial elastic potential energy.

\[ \Delta PE _{spring} = \frac{1}{2} k x_f^2 - \frac{1}{2} k x_i^2 \]

Going back to our original conservation of energy equation, we simply plug the appropriate terms on each side (work on the left and energies on the right) and balance the two sides to solve for any unknowns. Terms that do not exist or do not change, such as elastic potential energy in a problem with no springs or \(\Delta KE\) in a problem where there is no change in the speed of the body, can be set to zero.

Video lecture covering this section, delivered by Dr. Jacob Moore. YouTube source: https://youtu.be/FMkHrOjKlXs.

Example \(\PageIndex{1}\)

A 16-pound crate slides down a ramp as shown below. The crate is released from a height of 10 feet above the ground.

What is the work done by gravity?
What is the change in gravitational potential energy?

A box is on a ramp with an incline of 60°, held at the point 10 feet above the ground. Then it is released and allowed to slide down the incline. — Figure \(\PageIndex{4}\): problem diagram for Example \(\PageIndex{1}\). A 16-lb crate initially 10 feet above the ground slides down a ramp of incline 60°.

Solution: Video \(\PageIndex{2}\): Worked solution to example problem \(\PageIndex{1}\). YouTube source: https://youtu.be/ZkopO1aTj54.

Example \(\PageIndex{2}\)

A spring with an unstretched length of 40 cm and a \(k\) value of 120 N/cm is used to lift a 5-kg box from a height of 20 cm to a height of 30 cm. If the box starts at rest, what would you expect the final velocity to be?

On the left side of the image, a vertically oriented spring supports a block on its upper end; the block is currently 20 centimeters above some reference point. On the right side of the image, the same spring has now been stretched upwards, so the block is now 30 centimeters above that reference point. — Figure \(\PageIndex{5}\): problem diagram for Example \(\PageIndex{2}\). A vertically oriented spring supporting a box on its upper end is stretched to lift the load.

Solution: Video \(\PageIndex{3}\): Worked solution to example problem \(\PageIndex{2}\). YouTube source: https://youtu.be/T_5JT4XFQN8.

Example \(\PageIndex{3}\)

A 2,000-pound wrecking ball hangs from the end of a 40-foot cable. If the wrecking ball is released from an angle of 40 degrees from vertical, what would the expected maximum velocity at the bottom point of the travel path be?

A cable with one end fixed in place and the other end attached to a wrecking ball. Its final position, hanging straight down from the attachment point, is drawn in dotted lines. Its current position, slanting down and to the right with the cable making a 40° angle with the dotted-line cable, is drawn in solid lines. Another dotted line shows the curved path the ball will take to get from its current to its final position. — Figure \(\PageIndex{6}\): problem diagram for Example \(\PageIndex{3}\). A wrecking ball on a cable is raised slightly above its resting position, with the cable kept taut, then released.

Solution: Video \(\PageIndex{4}\): Worked solution to example problem \(\PageIndex{3}\). YouTube source: https://youtu.be/UkcpT1lfIDY.

Example \(\PageIndex{4}\)

A 24,000-kilogram aircraft is launched from an aircraft carrier using a hydraulic catapult. If the force the catapult exerts over the 90-meter runway is shown in the graph below:

What is the work done by the catapult?
What is the speed of the plane at the end of the runway?

Cartesian-coordinate graph of the force exerted by a launching catapult on an aircraft vs distance traveled on the runway. The force is described with a linear function: when x = 0 meters, F = 1240 kN and when x = 90 meters, F = 205 kN. — Figure \(\PageIndex{7}\): problem diagram for Example \(\PageIndex{4}\). Graph of the force exerted by the catapult vs distance traveled over the runway, showing a linear relationship between these quantities.

Solution: Video \(\PageIndex{5}\): Worked solution to example problem \(\PageIndex{4}\). YouTube source: https://youtu.be/DfSMeUm1SG8.