Chapter 7 Part 1 Kinetic Energy, Potential Energy, Gravity, Springs

 Prof. Clements Notes/Hints for Physics Courses

Chapter 7 Part 1 Open Stax College Physics or Most Introductory Physics Courses

Work, Kinetic Energy, Potential Energy

Terms you should know at the end of this unit: energy, work, kinetic energy, potential energy, conservative force, force constant

Work

Physics has a specific definition for work. For the work value to be non-zero the force that acts on the body must have a component in the direction of the displacement. You should read the text explanation and examples.

The work done on a body is NOT simply calculated by multiplying the net force acting on a body by the distance that the object travels. In some cases the net force will not be parallel to the displacement. In those cases you must use the component of the force that is parallel to the displacement when you calculate work.

Work= (Force component that is parallel to the displacement vector) * displacement

W = F d cos(θ) or W = F cos(θ) d; θ is the angle between the force vector and the displacement vector

The metric unit of work is the Joule. 1 Newton of force acting parallel to a displacement of 1 meter does 1 Joule of work on the object.

Kinetic Energy (KE)

Energy also has a physics definition. Energy is not a substance. You will use energy calculations in order to solve problems related to the position or motion of objects. Energy can be defined as the ability of a body or system of bodies to perform work.

When work is done on an object, energy is transferred to an object or energy is taken away from an object. The work and energy concepts provide a powerful tool for solving problems that cannot easily be solved with the methods you have studied before this chapter.

Work-Energy Theorem

(½) mV² is a form of energy called Translational Kinetic Energy (KE). m is the mass in kg of the object that is moving and V is its speed in meters/second. You will use a slightly different calculation to compute rotational kinetic energy in a future chapter.

Your textbook probably derives Work = ½ mV² – ½ m V_o² or, Work = change in KE, using the kinematic equations that rely on constant acceleration. Using the tools of calculus the same relationship can be derived for situations in which the acceleration changes. This relationship assumes that the object is moving in a horizontal direction (not closer or farther from the center of the Earth). This relationship assumes that the object is not compressing or extending any springs, etc. The reason for these assumptions will be explained as you read the chapter in your textbook.

You may notice that the kinetic energy has no direction associated with it. Kinetic energy is a scalar. You won’t have to use components of vectors as much in this chapter compared to the chapters on Force.

Suppose objects A and B have the same mass. Object A is traveling at a velocity of 4 m/s North. Object B is traveling at a velocity of 12 m/s East. Object B does not have three times as much translational kinetic energy (kinetic energy) as object A. Object B will have 9 times more kinetic energy. The velocity is squared when the kinetic energy is calculated.

Notice that the direction of the velocity has no effect on the amount of kinetic energy.

Applications and Concepts

Stopping Distance for a Car

The stopping distance of a car depends on the square of the car’s speed.

The brakes perform work to take away kinetic energy from the car and reduce its speed. The work done by the brakes is F_friction * d. To stop the car, F_friction * d = (1/2)mV_o². Given that the force of friction of the brakes and the mass of the car are constant, distance is proportional to the square of the velocity (speed). The distance the brake pads “move” across the metal plate in the brake is related to the distance the car moves as it stops.

Work Done on Orbitting Satellite

Suppose a satellite is moving in uniform circular motion around the earth. The force of gravity between the earth and the satellite is not equal to zero but does the force of gravity do any work on the satellite?

To answer this question first identify the value of the angle between the direction of the force of gravity on the satellite and the direction of the instantaneous velocity of the satellite. This angle is 90 degrees (for a circular orbit). The force of gravity thus has no component in the direction of motion. The value of the work done by the force of gravity is zero * distance traveled. The work done by the force of gravity of the earth on the satellite is equal to zero.

Acceleration Not Constant

Suppose that in a certain situation the force acting on an object is not constant. You cannot use the four equations of motion V = V_o + at, etc., to calculate the final velocity and final kinetic energy because the acceleration is not constant. But, if you are told the amount of work done on the object you can calculate the change in kinetic energy for the object and find the final velocity. This will be a common type of problem in this chapter.

e.g. A certain object has 20 Joules of kinetic energy. It is moving on a horizontal, frictionless surface. If its speed is doubled by some unknown acceleration over an unknown time, what is the new value of its kinetic energy?

Answer: The kinetic energy is proportional to the square of the velocity. The velocity has increased by a factor of 2. The kinetic energy increases by a factor of 2² . The new KE is 4 * 20 Joules = 80 Joules.

Gravitational Potential Energy (PE)

A system has “potential” energy due to its position (e.g. height above the floor) The system can gain potential energy when work is done on the system. The system can release potential energy into the form of kinetic energy at some later time.

Work is required to lift an object in a vertical direction. To move the object upward with no acceleration, the force applied must match the weight (mg) of the object.

Work = F d cos(θ) . Consider the application of a force on an object off mass m in an upward, vertical direction. You can replace F with mg. The angle is 0 degrees and the value of cos(0 degrees) is 1. The symbol “h” is the traditional symbol for a vertical displacement. When you make these substitutions you find that the work done in lifting an object, at constant speed, by a height “h” near the surface of the Earth is

Work = m * g * h

Note that this calculation is only allowed if "g" and m are constant. If the vertical distance is "too large" "g" will change and the work value is different than mgh. The value of "g" due to Earth's gravity is reduced by 1% if h has a value of about 32 km (20 miles). When h is larger than 30 km you should use a different method to calculate the work done by, or against, gravity. The more accurate equation will be discussed later in this unit.

If the force is directed upward the work has given the object gravitational potential energy. There will be an elastic potential energy when work is done on a spring. During this process you should assume that the value of the kinetic energy of the object did not change (i.e. started at rest, ends at rest at the new height). mgh actually calculates the change in the PE of the system due to the change in location. Only changes in PE are important in solving problems. The specific value of the PE at one location does not help you solve a problem (unless you know the PE is zero at the other point in the problem).

e,g. A student has left a 2 kg book on the floor. Suppose you do work on the book and lift it a distance of 1.5 meters such that you have given the book 29.4 Joules of potential energy. If the book slips out of your hand and falls back to the floor what is the speed just before it hits the floor? Ignore air resistance.

Answer: The potential energy the book has before it slips out of your hand will become kinetic energy as the book moves towards the floor. Just at the instant the book hits the floor all of the potential energy you gave to the book becomes kinetic energy for the book. The book has 29.4 J of KE. 29.4 J = ½ m V² . 29.4 J = ½ (2 kg) V² .

Solving for V you should have the result that V = 5.42 m/s.

(This is a preview of the next unit in chapter 7.)

The location where the potential energy is zero (the reference level) is not always at ground level. You can set the zero point for PE at any location. Typically, PE will be zero at the start or end of the motion.

Which situation has the greatest change in gravitational potential energy?

a) A 5 kg ball is lifted from the ground to a point 2 meters above the ground.

b) A 5 kg ball is moved from a table that is 0.5 meters above the ground to a shelf that is 2.5 meters above the ground.

Answer: The change in the PE is the same for both situations. Only the height difference is important. “h” in mgh is the change in the vertical location. I will assume that “g” is constant near the surface of the Earth.

Conservative Forces and Potential Energy

For conservative forces the work done in moving between two locations only depends on the start and end positions. The work does not depend on the particular path (route) taken. Gravity and spring forces are conservative. The force of friction is not conservative as the work done by friction will be greater if the distance traveled is greater. This is because the force of friction vector always changes to be opposite the direction of motion. The gravitational force vector is always pointed towards the center of the Earth and does not change direction as the object has a mixture of horizontal and vertical motion on its path.

Conservative forces have an associated PE. Non-conservative forces do not have an associated PE. There is a PE for gravity and for springs. There is no PE associated with the force of friction.

Hooke’s Law F = - k X

This notation emphasizes (using the minus sign) that the direction of the force due to the spring at the end of the spring is in the opposite direction to the stretch or compression of the spring.

When a spring is relaxed its stretch or compression value, X, is zero. In order to stretch or compress a spring you must apply a force to the spring. There is a linear relationship between the force applied and the stretch or compression of the spring.

Recall that “k” represents the force constant. k = -F/X

e.g. If a 10 Newton force is required to stretch a spring by 0.2 meters, the force constant is

- (-10)/0.2 or 50 Newtons/meter. “k” is a constant value for one particular spring. The force required to stretch this spring by 0.1 meters will be F = 50 N/m * 0.1 meters or 5 Newtons.

Springs that have smaller k values are easier to stretch.

The value of “k” is ALWAYS positive. If you calculate "k" and have a negative value you have not used the + or - sign correctly for the force or the value of X.

e.g. Calculate k for a car spring if adding 900 kg of cargo to the car makes the body of the car sink toward the ground by 0.05 meters. Treat the four springs at the corners of the car as a single spring.

Answer: 900 kg creates a downward force of 900 * (- 9.8) or -8820 Newtons. The spring pushes back with the force of +8820 Newtons. k = F/x or -(8820 N) / (-0.05 meters)

k = 1.76 x 10⁵ N/m

Note that the force of the spring is not constant as a spring is stretched. This leads to non-constant acceleration for spring motion problems. This is why springs were not discussed in the earlier chapters. By using energy methods (KE, PE) you will be able to study the motion of springs.

Potential Energy for Spring

The elastic potential energy is PE = ½ k X² . Note that the PE is always positive since X is squared. Work is done on a spring when it is compressed or extended from its relaxed position.

e.g. A spring has a force constant of 300 N/m. How much work is done when this spring is stretched by 6 centimeters? Assume that the spring is at rest before and after it is stretched. You do not know the value of the mass attached to the end of the spring. Ignore friction.

Answer: The work done is equal to the PE now stored in the spring. PE = ½ k X²

PE = ½ (300 N/m) (0.06 m)²

(Note the units N/m times m² yield Newtons * meters, the units of work.)

In this problem W = PE. W = 0.54 Joules

You should perform your own calculations of gravitational and elastic potential energy. Ask your instructor when you have questions on this material. The next unit will also give your more practice in performing these calculations.

YouTube videos of my (usually short) lectures and example problems are indexed at http://www.physics.gpclements.com/ . There are also a few videos that give review of basic math tools that are used in an introductory physics course.

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