Work Page: 420
Big Ideas Details Unit: Energy, Work & Power
Work
Unit: Energy, Work & Power
NGSS Standards/MA Curriculum Frameworks (2016): HS-PS3-1
AP® Physics 1 Learning Objectives/Essential Knowledge (2024): 3.2.A, 3.2.A.1,
3.2.A.1.i, 3.2.A.1.ii, 3.2.A.1.iii, 3.2.A.1.iv, 3.2.A.1.v, 3.2.A.2, 3.2.A.3, 3.2.A.3.i,
3.2.A.3.ii, 3.2.A.4, 3.2.A.4.i, 3.2.A.4.i, 3.2.A.4.ii, 3.2.A.4.iii, 3.2.A.5
Mastery Objective(s): (Students will be able to…)
• Calculate the work done when a force displaces an object .
Success Criteria:
• Variables are correctly identified and substituted correctly into equation(s).
• Algebra is correct and rounding to appropriate number of significant figures is
reasonable.
Language Objectives:
• Explain why a longer lever arm is more effective.
Tier 2 Vocabulary: work, energy
Notes:
In high school physics, there are two ways that we will study of transferring energy
into or out of a system:
work (W): mechanical energy transferred into or out of a system by a net force
acting over a distance.
heat (Q): thermal energy transferred into or out of a system. Heat is covered in
Physics 2.
If you lift a heavy object off the ground, you are giving the object gravitational
potential energy (in the object-Earth system). The Earth’s gravitational field can
now cause the object to fall, turning the potential energy into kinetic energy.
Therefore, we would say that you are doing work against the force of gravity.
Work is the amount of energy that was added to the object (W = ΔE)*. (In this case,
because the work was turned into potential energy, we would say that W = ΔU.)
* Many texts start with work as the application of force over a distance, and then discuss energy. Those
texts then derive the work-energy theorem, which states that the two quantities are equivalent. In
these notes, we instead started with energy, and then defined work as the change in energy. This
presentation makes the concept of work more intuitive, especially when studying other energy-
related topics such as thermodynamics.
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
Mathematically, work is also the effect of a force applied over a distance:
E = W = Fd
Remember that if the force is not in the same direction as the (instantaneous)
displacement, you will need to use trigonometry to find the component of the force
that is in the same direction as the displacement:
F = F cos and therefore W = F cos = Fd cos
Work is measured in joules (J) or newton-meters (N∙m), which are equivalent.
1N m 1 J 1 kgsm2
2
Positive vs. Negative Work
Recall that in physics, we use positive and negative numbers to indicate direction.
So far, we have used positive and negative numbers for one-dimensional vector
quantities (e.g., velocity, acceleration, force) to indicate the direction of the vector.
We can also use positive and negative numbers to indicate the direction for energy
(and other scalar quantities), to indicate whether the energy is being transferred
into or out of a system.
• If the energy of an object or system increases because of work (energy is
transferred into the object or system), then the work is positive with respect
to that object or system.
• If the energy of an object or system decreases because of work (energy is
transferred out of the object or system), then the work is negative with
respect to that object or system.
However, we often discuss work using the prepositions on (into) and by (out of).
• If energy is transferred into an object or system, then we can say that work
was done on (into) the object or system, or that work was done by (out of) the
surroundings.
• If energy was transferred out of an object or system, we can say that work was
done by (out of) the object, or we can say that work was done on (into) the
surroundings.
Example:
A truck pushes a 1000 kg car up a 50 m hill. The car gained
Ug = mgh = (1000)(10)(50) = 500 000 J of potential energy. We could say that:
• 500 000 J of work was done on the car (by the truck).
• 500 000 J of work was done by the truck (on the car).
• −500 000 J of work was done on the truck (by the car).
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
A simple way to tell if a force does positive or negative work on an object is to use
the vector form of the equation, W = F • d . If the force and the displacement are in
the same direction, then the work done by the force is positive. If the force and
displacement are in opposite directions, then the work done by the force is
negative.
Example:
Suppose a force of 750 N is used to push a cart against 250 N of friction for a
distance of 20 m. The work done by the force is W = F||d = (750)(20) = 15 000 J . The
work done by friction is W = F d = (−250)(20) = −5 000 J (negative because friction is
in the negative direction). The total (net) work done on the cart is
15 000 + (−5 000) = 10 000 J .
We could also figure out the net work done on the cart directly by using the net
force: Wnet = Fnet , d = (750 − 250)(20) = (500)(20) = 10 000 J
Notes:
• If the displacement is zero, no work is done by the force. E.g., if you hold a heavy
box without moving it, you are exerting a force (counteracting the force of gravity)
but you are not doing work.
• If the net force is zero, no work is done by the displacement (change in location)
of the object. E.g., if a cart is sliding across a frictionless air track at a constant
velocity, the net force on the cart is zero, which means no work is being done.
• If the displacement is perpendicular to the direction of the applied force (θ = 90°,
which means cos θ = 0), no work is done by the force. E.g., you can slide a very
heavy object along a roller conveyor, because the force of gravity is acting
vertically and the object’s displacement is horizontal, which means gravity and the
normal force cancel, and you therefore do not have to do any work against
gravity.
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
Work Done by Conservative vs. Nonconservative Forces
conservative force: a force for which the amount of work done does not depend on
the path.
nonconservative force: a force for which the amount of work done depends on the
path.
For example, consider a situation in which one person lifts a block, and another
person slides a block of the same weight up a frictionless ramp.
Both blocks gained the same amount of gravitational potential energy, so the
amount of work done on the blocks (against gravity) is the same. Work that is
converted to potential energy is therefore done by a conservative force.
However, consider instead a block sliding on the ground, with friction:
In this case, the amount of work done against friction depends on the distance
traveled; Path #1 requires more work than Path #2. Friction is therefore a
nonconservative force.
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
Work Done “Against” a Force
When an object is moved in the presence of an opposing force, questions often ask
about the work done “against” that force. This means “calculate the work done as if
the specified force were the only force acting on the object”.
Consider the previous example. Suppose that both blocks have a mass of 2 kg.
The work that either person does against gravity is the change in gravitational
potential energy. W = U = mgh = (2)(10)(3) = 60 J .
Now, suppose that the coëfficient of friction between the block and the ramp is
k = 0.4 . The normal force is FN = Fg cos = (20) ( 45 ) = 16 N , which means the force
of friction is Ff = k FN = (0.4)(16) = 6.4 N . The work that the woman does against
friction is therefore W = Fd = (6.4)(5) = 32 J .
How Can You Tell If Work is Done?
1. Look for a change in mechanical energy.
• Kinetic energy: K = 12 mv 2 . Mass is almost certainly constant, so look for a
change in velocity. If the change in velocity was caused by a force, then
work was done.
• Potential energy: Ug = mgh . Mass and the strength of the gravitational
field are almost certainly constant, so look for a change in height. If the
change in potential energy was caused by a force, then work was done.
2. Look for a force applied over a distance.
• Work: W = Fd . If a force is applied over a distance, look for a resulting
change in kinetic energy (velocity) or potential energy (height). If either of
those is the case, then work was done.
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
Force vs. Distance Graphs
Recall that on a graph, the area “under the graph” (between the graph and the
x-axis)* represents what you get when you multiply the quantities on the x and y-axes
by each other.
Because W = F||d , if we plot force vs. distance, the area “under the graph” is
therefore the work:
In the above example, (3N)(3m) = 9N m = 9 J of work was done on the object in the
interval from 0–3 s, 2.25 J of work was done on the object in the interval from
3–4.5 s, and −2.25 J of work was done on the object in the interval from 4.5–6 s.
(Note that the work from 4.5–6 s is negative, because the force was applied in the
negative direction during that interval.) The total work is therefore
9 + 2.25 + (−2.25) = +9 J .
* In most physics and calculus textbooks, the term “area under the graph” is used. This term always
means the area between the graph and the x-axis.
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
Sample Problems:
Q: How much work does it take to lift a 60. kg box 1.5 m off the ground at a
constant velocity over a period of 3.0 s?
A: The box is being lifted, which means the work is done against the force of gravity.
W = F|| d = Fg d
W = Fg d = [mg]d = [(60)(10)](1.5) = [600](1.5) = 900 J
Note that the amount of time it took to lift the box has nothing to do with the
amount of work done.
It may be tempting to try to use the time to calculate velocity and acceleration
in order to calculate the force. However, because the box is lifted at a constant
velocity, the only force needed to lift the box is enough to overcome the weight
of the box (Fg).
In general, if work is done to move an object vertically, the work is done against
gravity, and you need to use a = g = 10 sm2 for the acceleration when you
calculate F = ma.
Similarly, if work is done to move an object horizontally, the work is not against
gravity and either you need to know the force applied or you need to find it
from the acceleration of the object using F = ma.
Q: In the picture to the right, the adult is pulling on
the handle of the wagon with a force of 150. N
at an angle of 60.0°.
If the adult pulls the wagon a distance of
500. m, how much work does he do?
A: W = F||d
W = [F cos ]d = [(150. )cos60.0](500. ) = [(150. )(0.500)](500.) = 37500 J
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
Homework Problems
1. (S) How much work is done against gravity by a weightlifter lifting a 30. kg
barbell 1.5 m upwards at a constant speed?
Answer: 450 J
2. (M) A 3000. kg car is moving across level ground at 5.0 ms when it begins an
acceleration that ends with the car moving at 15.0 ms . Is work done in this
situation? How do you know?
3. (S) A 60. kg man climbs a 3.0 m tall flight of stairs. How much work was
done by the man against the force of gravity?
Answer: 1 800 J
Physics 1 In Plain English Jeff Bigler
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Big Ideas Details Unit: Energy, Work & Power
4. (M) The following graph shows the force on a 2.0 kg object vs. its position
on a level surface.
The object has a velocity of +4.0 m
s
at time t = 0.
a. (M) What is the kinetic energy of the object at time t = 0?
Answer: 16 J (Note: this is the starting kinetic energy for parts (b) & (c).)
b. (M) How much work was done on the object by the force during the
interval from 0–2 m? What are the kinetic energy and velocity of
the object at position x = 2 m?
Answer: W = 8 J; K = 24 J; v = +4.9 ms
c. (M) How much work was done on the object by the force during the
interval from 0–9 m? What are the kinetic energy and velocity of
the object at position x = 9 m?
Answer: W = 50 J; K = 66 J; v = +8.1 ms
(Note: 66 J is the starting kinetic energy for part (d).)
Physics 1 In Plain English Jeff Bigler
Work Page: 429
Big Ideas Details Unit: Energy, Work & Power
d. (M) How much work was done on the block during the interval from
9–20 m? What are the kinetic energy and velocity of the block at
position x = 20 m?
Answer: W = −40 J; K = 26 J; v = +5.1 ms
5. (M) A dog pulls a sled using a 500. N force across a 10. m wide street. The
force of friction on the 90. kg sled is 200. N. How much work is done by the
dog? How much work is done by friction? How much work is done on the
sled? How much work is done by gravity?
honors & AP® 6. (M – honors & AP®; A – CP1) Find
the work done when a 100. N
force at an angle of 25° pushes a
cart 10. m to the right, as shown
in the diagram to the right.
Answer: 906 J
Physics 1 In Plain English Jeff Bigler
