New syllabus (tba)
“work”, “power” and “energy” are loosely used in colloquial English
What does “work” mean in science?
Work is the displacement of an object by a force.
Work = force x how far the object moves.
W = F•d
Two kinds of work:
Work done against a force
(archer stretches her bowstring, weightlifter raising her barbell)
work done to change an object’s speed
(slowing down a car, speeding up an airplane)
What are the units of work?
Force is measured in Newtons; distance is measured in meters. Therefore:
[unit of work] = Newtons x Meters = N•m
In honor of the English physicist James Prescott Joule (1818–1889), this unit has been named the joule, abbreviated as J
Note that the name of the unit is lowercase, but the abbreviation is uppercase. Because #reasons
unit of work = N•m = J
Soon enough we will see that work can be converted into energy, and energy into work, so both energy and work have the same units.
I feel like I’m doing work, but I’m not ?!
A force with no motion, or a force perpendicular to the motion, does no work
How can we say that no work is done? We feel “work” …
On the left: No matter the push, if the crate doesn’t move, then you’ve done no work on it
W = F•d d = 0 m Therefore W = F•0 = 0 joules
Why then do you feel fatigued?
When exerting a force, muscle fibers inside your body contract and release.
That involves force and motion, so you do accomplish internal work in your body.
The energy shows up as warming in your muscle tissue.
However, there is no net work on the crate.
On the right: the object is already moving at constant velocity, and kept at a constant height. (zero acceleration.)
This means no net force is necessary to keep it in motion.
You are only supplying force to counteract friction.
Since we didn’t change V, and didn’t change energy, we’ve done no net work.
So why does it feel like you did work?
Because friction was doing (negative) work on us, trying to slow us down.
We’re making our muscles do (positive) work to counteract this friction.
So our positive work is constantly being canceled by negative work. The net work is zero.
another way to think about it:
The force exerted by the person is upward, and = to weight of the box.
This force is perpendicular to the motion.
Since there is no motion in the direction of the force, then no work is done by that force.
Since we didn’t change its height, we didn’t change its PE = done no net work
How much work is done?
Hmm, how is it that no net work was done on the box in the last case?
Let’s break it down:
Man applies forward force: That does + work on the box
But friction applies an opposing force: That does – work on the box
Net force is the total of the forward and backward = they cancel out = add up to 0
Still not convinced?
Logically, friction force must be equal (and opposite) to forward force:
If the forward force were bigger then the man & box would pick up speed!
If the forward force were smaller then the man & box would slow down.
Consider an object is being lifted up at constant velocity (i.e. object is not accelerating.)
If the object is being lifted at constant velocity, then it is not accelerating, so the net force on it is zero.
We know that from Newton’s first law of motion.
We pull up on an object when we lift it – giving an upward force on the object. Yet gravity pulls it down (“weight”).
If the net force on the object is zero, then the upward lifting force is canceled by the downward pull of gravity
So when you lift something at constant velocity, the upward pull = the weight of the object.
Over what distance does this force act? The height of the lift.
How much work is done by lifting?
Work = force x displacement.
object’s weight = mass x (acceleration of gravity) = m•g
displacement over which work is done = height of the lift.
Work done lifting an object = (mass) x (the acceleration due to gravity) x (height of the lift).
W = F•d = m•g•h
This work does not end up as kinetic energy, since after the lift the object is not moving.
The object is higher up, though, so energy has been stored in the gravitational field.
This is called gravitational potential energy , or just PE grav .
PE grav = m∙g∙h
Example: What is the gravitational potential energy of a 4 kg object that is lifted 5 m?
PE = mgh = (4 kg)(9.8 m/s/s)(5 m) = 196 joules
Doing “work” means that you change the amount of energy it has .
* If you change its velocity, then you changed its KE (kinetic energ), thus you’ve done work.
* If you lifted it higher, then you’ve increased its potential energy, thus you’ve done work.
* In any case where you don’t change its energy, then you haven’t done work on it.
Question: A waiter carries a tray full of meals across the room – Is work being done?
We need to ask Does (something) do work on (something else)?
* Is the tray starting at rest, and then moved horizontally?
In this case there would be a positive acceleration.
*Are we only considering the motion of the tray while it is being moved at a constant velocity?
In this case its acceleration would be zero.
* Is the tray initially moving, and then comes to a rest?
In this case there would be a negative acceleration
So let’s restate the question. In fact, let’s ask four different questions.
Does the waiter do work on the tray, as he starts from rest and then reaches some final velocity?
Does the waiter do work on the tray, as he slows down to a stop?
Does the waiter do work on the tray, as he moves at constant velocity?
While moving at a constant velocity, does the waiter do internal work, i.e., inside his body?
Does the waiter do work on the tray, as he starts from rest and then reaches some final velocity?
Yes, he applies a force over a distance, changing the tray’s kinetic energy. The tray starts at v = 0 m/s, but ends up at a non-zero velocity, thus it has more KE.. Work is done.
Does the waiter do work on the tray, as he slows down? Yes – he applies a force over a distance, and changes the tray’s kinetic energy. The tray starts at v = (some positive speed), but then is slowed down, so its KE changes. Work is done.
Does the waiter do work on the tray, as he moves at constant velocity? Nope – no net work is done by the waiter on the tray. Why not? .
Since the waiter is not accelerating, the net force equals zero. W = F∙d.
If F = 0 then W = 0.
The tray’s potential energy doesn’t change; its kinetic energy doesn’t change.
Doing “work” on an object means that you changes the amount of energy it has. |
No change in energy means that no work was done.
Motion by the waiter does positive work on the tray, but it is canceled by the negative work done by the friction between the waiter and the floor. See question (4) for more detail.
While moving at a constant velocity, does the waiter do internal work, i.e., work inside his body?
While moving at a constant V, does the waiter do internal work?
Yes, he does do internal work.
Friction between the waiter and the floor constantly retards the motion of the waiter (and thus also the motion of the tray.)
Friction does negative work on the waiter. He uses his muscles to overcome this friction force, to move at a constant speed. The waiter thus does positive work – inside his body.
The + work done on the tray is canceled by the – work done on the tray.
So the net work on the waiter equals zero, and the net work done on the tray must equal zero.
The work-energy principles
joule = unit of work – Newton x meter
PowerPoint (Slides) presentation
2016 Massachusetts Science and Technology/Engineering Curriculum Framework
HS-PS3-1. Use algebraic expressions and the principle of energy conservation to calculate the change in energy of one component of a system when the change in energy of the other component(s) of the system, as well as the total energy of the system including any energy entering or leaving the system, is known
HS-PS3-1. Create a computational model to calculate the change in the energy of one component in a system when the change in energy of the other component(s) and energy flows in and out of the system are known.
[Clarification Statement: Emphasis is on explaining the meaning of mathematical expressions used in the model.]
[Assessment Boundary: Assessment is limited to basic algebraic expressions or computations; to systems of two or three components; and to thermal energy, kinetic energy, and/or the energies in gravitational, magnetic, or electric fields.]
DCI – Energy is a quantitative property of a system that depends on the motion and interactions of matter and radiation within that system. That there is a single quantity called energy is due to the fact that a system’s total energy is conserved, even as, within the system, energy is continually transferred from one object to another and between its various possible forms.
Conservation of energy means that the total change of energy in any system is always equal to the total energy transferred into or out of the system.
Energy cannot be created or destroyed, but it can be transported from one place to another and transferred between systems.
Mathematical expressions, which quantify how the stored energy in a system depends on its configuration (e.g., relative positions of charged particles, compression of a spring) and how kinetic energy depends on mass and speed, allow the concept of conservation of energy to be used to predict and describe system behavior.
The availability of energy limits what can occur in any system.
Massachusetts Science and Technology/Engineering Curriculum Framework
2. Conservation of Energy and Momentum
The laws of conservation of energy and momentum provide alternate approaches to predict and describe the movement of objects.
2.3 Describe both qualitatively and quantitatively how work can be expressed as a change in mechanical energy.
2.4 Describe both qualitatively and quantitatively the concept of power as work done per unit time.
SAT Physics Subject Test Learning Standards
Mechanics: Energy and momentum, such as potential and kinetic energy, work, power, impulse, and conservation laws
Learning Standards: Common Core Math
CCSS.MATH.CONTENT.7.EE.B.4 Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities.
CCSS.MATH.CONTENT.8.EE.C.7 Solve linear equations in one variable
CCSS.MATH.CONTENT.HSA.SSE.B.3 Choose and produce an equivalent form of an expression to reveal and explain properties of the quantity represented by the expression. (including isolating a variable)
CCSS.MATH.CONTENT.HSA.CED.A.4 Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V = IR to highlight resistance R.