PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: work, energy and power (PHYS 101)

This sheet is not a summary of the energy chapter: you already have the course notes. It answers one question only, what makes students lose marks on work, energy and power in PHYS 101 at McGill University, and which precise gesture avoids each loss.

Everything below is algebra, proportions and right-triangle trigonometry. The work of a force that changes with position is read as an AREA cut into triangles and rectangles, because PHYS 101 is the algebra-based course and a solution that starts by differentiating is not a solution here, even when its final number is right.

The thread of the chapter

Energy is silent about time and silent about direction, and that silence is what makes it win: name the initial state, name the final state, write the reference level in words, then write ONE line, initial energy plus the work of the non-conservative forces equals final energy.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

What is on this sheet

1. The essentials
2. The mistakes that cost marks
3. Which method to choose
4. How the answer is expected to be written
5. Check before you hand in
6. The typical problem, taken apart
7. Learn by heart
8. Frequently asked questions

Topics covered

Work of a constant force
Work-energy theorem
Kinetic energy
Gravitational potential energy
Reference level
Hooke's law
Elastic potential energy
Conservative forces
Mechanical energy conservation
Energy dissipated by friction
Average power
Efficiency

The essentials

The three energies and the one line that uses them

• Work of a constant force: $W = Fd\cos\theta$ , with $\theta$ the angle between the FORCE and the DISPLACEMENT, not the angle with anything else on the diagram.
• Kinetic energy $E_{k} = \frac{1}{2}mv^{2}$ , always positive, and proportional to $v^{2}$ : double the speed, quadruple the energy.
• Gravitational potential energy $U = mg\Delta y$ with $g = 9.8$ m/s $^{2}$ , where $\Delta y$ is the VERTICAL height above a level you choose and write down. Elastic: $U = \frac{1}{2}kx^{2}$ .
• The only equation you need: $E_{k,i} + U_{i} + W_{\text{nc}} = E_{k,f} + U_{f}$ , where $W_{\text{nc}}$ is the work of friction and of any applied force, negative for friction.
• Power $P = \dfrac{W}{t} = Fv$ in watts, with $t$ in SECONDS. Efficiency $= \dfrac{\text{useful output}}{\text{total input}}$ , never above 1.

Everything in this chapter is one of two questions: what is the energy HERE, and what happened to it between here and there. If your page has more than one line of algebra per state, you are solving it as a forces problem in disguise.

Work: the angle, the sign, and the forces that do none

• A force at less than $90^{\circ}$ to the motion does POSITIVE work and adds energy. At exactly $90^{\circ}$ it does NO work. Beyond $90^{\circ}$ it does NEGATIVE work and removes energy.
• The three forces that almost never do work: the normal force on any surface, the tension in a pendulum string, and any centripetal force on a circular path. All three are perpendicular to the velocity at every instant.
• Friction on a sliding object is always at $180^{\circ}$ to the motion: $W_{f} = -fd$ , with the minus sign written, not remembered.
• A force that varies with position has no single value to put in $Fd$ : its work is the AREA between the graph of $F$ against position and the horizontal axis.

Same displacement three times: the left force has a component along

d

and adds energy, the middle one is perpendicular and transfers nothing, the right one leans backwards and removes energy.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Using the length of a slope, a ramp or a staircase as the height in mgh

the whole question, and every speed computed after it

What not to write

“ $U = mgL = 45 \times 9.8 \times 5.0 = 2205$ J for a 5.0 m ramp.”

What to write

“The vertical drop is $h = L\sin\theta = 5.0 \times \sin 30^{\circ} = 2.5$ m, so $U = 45 \times 9.8 \times 2.5 = 1102$ J.”

L = 5.0

m is what the block travels;

h = 2.5

m is what gravity counts. Only the vertical side of the triangle goes into

mgh

Why: Gravity is a conservative force: its work counts only the VERTICAL displacement. A ramp 5.0 m long that drops 2.5 m stores the same energy as a 2.5 m vertical fall, and the wrong version is exactly twice too large, which turns a 7.0 m/s answer into 9.9 m/s. The same slip appears as $L\cos\theta$ on a pendulum and as the total length of a staircase instead of its rise.

2. Dropping the cosine, or using the angle with the wrong reference line

2 marks, and the net work in the part that follows

What not to write

“A 60 N pull over 12.0 m does $W = 60 \times 12.0 = 720$ J.”

What to write

“The pull is at $25^{\circ}$ to the displacement, so $W = 60 \times 12.0 \times \cos 25^{\circ} = 652.5$ J.”

Why: Only the component of the force ALONG the displacement transfers energy. The second half of the trap is subtler: $\theta$ is the angle between the force and the displacement, so a rope at $25^{\circ}$ above the horizontal on a horizontal floor gives $\cos 25^{\circ}$ , but the same rope on a slope gives the angle measured from the SLOPE. Draw the displacement vector first and measure from it.

3. Adding the work of friction instead of subtracting it

2 marks, plus an answer that is physically impossible

What not to write

“ $E_{f} = mgh + fd$ , so the block arrives faster than it would without friction.”

What to write

“ $mgh - fd = \frac{1}{2}mv^{2}$ : friction removes energy, so its work is negative and the speed is lower.”

Why: Friction on a sliding object points opposite the displacement, so $\cos 180^{\circ} = -1$ and $W_{f} = -fd$ . The check takes two seconds: a friction term can only ever make the final speed SMALLER than the frictionless value. If your answer with friction beats the ideal one, the sign is wrong and the marker sees it before reading the algebra.

4. Treating kinetic energy as proportional to the speed

the whole question, and a wrong order of magnitude in every part after it

What not to write

“At twice the speed the car has twice the kinetic energy, so it needs twice the stopping distance.”

What to write

“ $E_{k} = \frac{1}{2}mv^{2}$ , so twice the speed is FOUR times the energy and, for the same braking force, four times the distance.”

Why: The square is the reason energy answers road-safety questions at all. A 1200 kg car at 15 m/s carries $135\,000$ J; at 30 m/s it carries $540\,000$ J, not $270\,000$ J. The same reflex covers the spring: doubling the extension multiplies the stored energy by 4, so the second half of a stretch always costs three times the first half.

5. Quoting a potential energy without saying what it is measured from

1 mark for the statement, and the whole balance if two levels get mixed

What not to write

“The ball has $2.94$ J of potential energy.”

What to write

“Taking the table top as the zero of gravitational potential energy, the ball has $1.76$ J; taking the floor, it has $2.94$ J.”

Why: Potential energy is not a property of the object, it is a number relative to a level you pick. Picking is allowed, changing your mind halfway is not: a student who writes $U_{i}$ from the floor and $U_{f}$ from the table invents energy out of nothing. Write the level in WORDS on the first line, before any number, and the rest of the question polices itself.

6. Leaving a time in minutes inside a power calculation

the whole question, and by a factor of exactly 60

What not to write

“ $W = Pt = 95 \times 22 = 2090$ J for 22 minutes of pedalling.”

What to write

“ $t = 22 \times 60 = 1320$ s, so $W = 95 \times 1320 = 125\,400$ J.”

Why: A watt is a joule per SECOND. The conversion is trivial and that is precisely why it is skipped under exam pressure. Build the habit of converting on the line where the time first appears, never later. The same rule catches hours in a kilowatt-hour question and milliseconds in a flash-lamp question.

7. Multiplying by an efficiency where you should divide

2 marks, and an impossible answer

What not to write

“The muscle is 25 per cent efficient, so the chemical energy used is $125\,400 \times 0.25 = 31\,350$ J.”

What to write

“Efficiency is output over input, so the input is $\dfrac{125\,400}{0.25} = 501\,600$ J.”

Why: The input is always LARGER than the useful output, so an answer smaller than the work delivered is wrong on sight, with no algebra needed. Write the definition first, $e = W_{\text{useful}}/E_{\text{input}}$ , and solve it for whatever is missing rather than guessing which way the multiplication goes.

8. Putting the normal force or the string tension into the energy balance

2 marks, and a balance that cannot be solved

What not to write

“The normal force on the slope does work $Nd$ , so it must appear in the balance.”

What to write

“The normal force is perpendicular to the displacement at every instant, so its work is zero and it never enters the balance.”

Why: This is what makes energy methods shorter than force methods: three of the forces on a typical diagram drop out because $\cos 90^{\circ} = 0$ . Normal force, string tension on a pendulum, centripetal force on a circle. The reflex to build is the opposite of the free-body diagram reflex from the previous chapters: there you listed every force, here you list only those with a component along the motion.

Which method to choose

Which tool the statement is actually asking for

Read what the statement GIVES and what it asks, before deciding anything

A force that changes with position has no single value for

Fd

. Its work is the area below the graph, here a triangle of 10 J plus a rectangle of 40 J, total 50 J, with no calculus anywhere.

If a constant force and a distance, and it asks for an energy → $W = Fd\cos\theta$ , with the angle read off the displacement

Example: 60 N at $25^{\circ}$ over 12.0 m gives $652.5$ J
If a force that CHANGES with position, or a graph of force against position → the area under the graph, cut into triangles and rectangles

Example: a spring to $0.40$ m at 24 N gives $\frac{1}{2}(0.40)(24) = 4.8$ J
If two speeds, or a speed and a distance, and NO duration → the work-energy theorem $W_{\text{net}} = \Delta E_{k}$

Example: $25.0$ m/s to rest in $48.0$ m gives $F = 9115$ N
If heights, and the words frictionless or smooth → conservation of mechanical energy, $mgh = \frac{1}{2}mv^{2}$ , the mass cancels

Example: a 28.0 m drop gives $v = \sqrt{2gh} = 23.4$ m/s
If heights AND a measured final speed, or the word friction → the full balance $E_{i} + W_{\text{nc}} = E_{f}$ , and $W_{\text{nc}}$ is what you solve for

Example: $87\,808$ J in, $70\,560$ J at the bottom, so $17\,248$ J lost
If a duration, a rate, or the words how fast can it be done → power, $P = W/t$ , or $P = Fv$ when the speed is constant

Example: $2799$ J in $6.5$ s gives $431$ W

Notice what never appears on the left of this tree: the name of the object. A roller coaster, a pendulum, a spring gun and a patient on an exercise bike all land on the same branch if they give you the same kind of data. Sort by the DATA, never by the story.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing an energy balance that a marker can follow

When to use it: Any question that mentions heights, speeds, springs or friction and does not mention a duration

1 Name the two states in words, state 1 and state 2, and say where each one is. A sketch with the two positions marked earns the first mark on its own.
2 Write the reference level in words before any number: taking the bottom of the slide as the zero of gravitational potential energy.
3 List the energies present in each state and put a zero in front of the ones that are absent: $E_{k,i} = 0$ because the object starts from rest.
4 Write the single balance line $E_{k,i} + U_{i} + W_{\text{nc}} = E_{k,f} + U_{f}$ , substitute, and only then solve. Substituting into a rearranged formula is where signs get lost.
5 Answer in a sentence with a unit, and check the sign of any work you solved for: a friction term that comes out positive is a mistake, not a discovery.

Concluding sentence

“Taking the bottom of the slide as the zero of gravitational potential energy, and with the child starting from rest: $mgh + W_{f} = \frac{1}{2}mv^{2}$ .”

The trap: Jumping straight to a memorised formula such as v equals the square root of 2gh. It is only true with no friction and no initial speed, and the marker cannot give method marks for a line that hides both assumptions.

Marking: Typically 1 mark for the two states and the reference level, 1 for the balance line written before substitution, 2 for the algebra, and 1 for the final sentence with its unit. The balance line alone is worth more than a correct final number with no working.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The unit test, five seconds

A joule is a newton metre, that is kg m $^{2}$ /s $^{2}$ . If a work came out in newtons you lost a distance; if an energy came out in watts you divided by a time you did not need.

$W = Fd\cos\theta$ in N $\times$ m gives J; $P = Fv$ in N $\times$ m/s gives W.
The sign of the net work

Speeding up means positive net work, slowing down means negative, constant speed means zero. Compare that against your arithmetic before writing the final sentence.

$652.5 - 264 = +388.5$ J and the cart accelerates: consistent.
Friction can only ever take energy away

Compute the frictionless answer first, in your head if possible. The real answer must be smaller, never larger, and the difference is the energy dissipated.

ideal $23.4$ m/s against a measured $21.0$ m/s: plausible. A measured $25$ m/s would be impossible.
The order of magnitude of a fall

One metre of fall is about 10 J per kilogram, and gives about $4.4$ m/s. Anything ten times off that scale is an arithmetic slip, usually a forgotten square root or a height read in centimetres.

$0.90$ m gives $v = \sqrt{2 \times 9.8 \times 0.90} = 4.2$ m/s, and $635$ J for a 72 kg person.
Efficiency between 0 and 1

An efficiency above 100 per cent means the input and the output were swapped. An input smaller than the useful output means a multiplication that should have been a division.

$\dfrac{29\,400}{35\,000} = 0.84$ , so 84 per cent, and the missing $5600$ W heats the motor.

The typical problem, taken apart

A slide with friction, solved in one line of accounting

A 45 kg child starts from rest at the top of a straight slide 6.0 m long, whose top is 2.4 m above the bottom. She arrives at the bottom moving at $5.0$ m/s.

Find the energy dissipated by friction, the average friction force, and the fraction of the initial energy that actually reached the bottom as motion.

The 6.0 m is what the child travels; the 2.4 m is what gravity counts. The friction force acts along the 6.0 m, so the two lengths both appear, in two different places.

Step 1

State 1, top, at rest. Taking the BOTTOM of the slide as the zero: $U_{i} = mgh = 45 \times 9.8 \times 2.4 = 1058.4$ J and $E_{k,i} = 0$ .

Why

The reference level is written before any number, so the marker knows which level every later term refers to and the two states cannot drift apart. Starting from rest is worth stating explicitly: it is what allows the kinetic term to be set to zero.

Step 2

State 2, bottom: $U_{f} = 0$ by the choice just made, and $E_{k,f} = \frac{1}{2} \times 45 \times 5.0^{2} = 562.5$ J.

Why

The square goes in before the multiplication. Writing $\frac{1}{2} \times 45 \times 5.0 = 112.5$ J is the standard slip here, and it makes the friction loss look five times worse than it is.

Step 3

One balance line: $1058.4 + W_{\text{nc}} = 562.5$ , so $W_{\text{nc}} = -495.9$ J.

Why

This is the whole chapter in one line. The unknown is isolated by subtraction, not by rearranging a memorised formula, and the negative sign arrives on its own rather than being remembered, which is exactly what you want under exam pressure.

Step 4

The friction force acts along the 6.0 m of SLOPE: $|W_{\text{nc}}| = f L$ gives $f = \dfrac{495.9}{6.0} = 82.65$ N.

Why

Here the slope length is the right length, and the height would be wrong, the exact mirror image of the first trap on this sheet. Friction acts along the path travelled; gravity counts only the vertical drop. Two lengths, two roles, the same problem.

Step 5

Frictionless comparison: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 2.4} = 6.86$ m/s, and the fraction that survived is $\dfrac{562.5}{1058.4} = 0.53$ .

Why

The comparison is the verification. The measured 5.0 m/s must be smaller than the ideal 6.86 m/s, and it is; about half the energy became heat, which is a believable figure for a plastic slide and a pair of jeans.

The conclusion, written out

“Friction dissipated $496$ J, an average force of $83$ N over the 6.0 m of slope, and only 53 per cent of the initial potential energy arrived at the bottom as kinetic energy.”

The classic mistake on this problem: Using the 6.0 m as the height: $U_{i} = 45 \times 9.8 \times 6.0 = 2646$ J, which makes the friction loss $2084$ J and the friction force $347$ N, more than three quarters of the child's weight. A slide like that would not slide.

Learn by heart

• $W = Fd\cos\theta$ , and $\theta$ is measured between the force and the DISPLACEMENT.
• $\cos 90^{\circ} = 0$ : normal force, pendulum tension and centripetal force do no work, ever.
• $E_{k} = \frac{1}{2}mv^{2}$ , $U = mg\Delta y$ with $g = 9.8$ m/s $^{2}$ , $U_{\text{spring}} = \frac{1}{2}kx^{2}$ .
• The one line: $E_{k,i} + U_{i} + W_{\text{nc}} = E_{k,f} + U_{f}$ , with $W_{\text{nc}} = -fd$ for friction.
• $v = \sqrt{2gh}$ only from rest and only with no friction. Say both out loud before using it.
• Work of a variable force $=$ area under the graph of $F$ against position, in triangles and rectangles.
• $P = \dfrac{W}{t} = Fv$ , time in seconds. Efficiency $= \dfrac{\text{useful}}{\text{input}} \le 1$ . One food Calorie $= 4184$ J.
• Double the speed or the extension: FOUR times the energy. Never twice.

Frequently asked questions

When should I use energy instead of the kinematics equations?

As soon as the statement gives you speeds and distances but never mentions a duration, and as soon as the path is curved or the force is not constant. Energy is blind to time and to direction, so a roller coaster hill, a pendulum swing and a straight ramp are all the same problem to it. Keep kinematics for questions that actually ask how long something takes.

Does the work done by gravity depend on the path taken?

No. Gravity is a conservative force, which means its work depends only on the change in height between the start and the end. A hiker who climbs eight hundred metres by a short steep trail or by a long gentle one does exactly the same work against gravity. The trail changes the friction, the time and the power required, but not the work of gravity.

Where should I put the zero of gravitational potential energy?

Wherever it makes the arithmetic easiest, usually the lowest point of the motion. The choice is genuinely free, because only differences in potential energy ever appear in an equation. What is not free is changing your mind halfway: write the level in words on the first line and use the same one for both states, otherwise you create energy out of nothing.

Is mechanical energy still conserved when there is friction?

No, and saying so is worth a mark. Mechanical energy, the sum of kinetic and potential, decreases by exactly the amount friction dissipated. Total energy is still conserved: the missing joules are now thermal energy in the surfaces and the air. The correct sentence is that friction converts mechanical energy into heat, never that energy is lost or destroyed.

Why does PHYS 101 do work and energy without calculus?

Because it is the algebra-based mechanics course, taken by students who have not done calculus-based physics before. Every result you need comes from areas and proportions instead: the work of a variable force is the area under a force against position graph, cut into triangles and rectangles. Most online solutions are written for calculus courses, which is why they look unusable.

Practise it

Corrected exercises: Work, energy and power, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Friction, circular motion and gravitation Next sheet Momentum, impulse and collisions

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Revision sheet: work, energy and power (PHYS 101)

The essentials

The three energies and the one line that uses them

Work: the angle, the sign, and the forces that do none

The mistakes that cost marks

1. Using the length of a slope, a ramp or a staircase as the height in mgh

2. Dropping the cosine, or using the angle with the wrong reference line

3. Adding the work of friction instead of subtracting it

4. Treating kinetic energy as proportional to the speed

5. Quoting a potential energy without saying what it is measured from

6. Leaving a time in minutes inside a power calculation

7. Multiplying by an efficiency where you should divide

8. Putting the normal force or the string tension into the energy balance

Which method to choose

Which tool the statement is actually asking for

How the answer is expected to be written

Writing an energy balance that a marker can follow

Check before you hand in

The typical problem, taken apart

A slide with friction, solved in one line of accounting

Learn by heart

Frequently asked questions

When should I use energy instead of the kinematics equations?

Does the work done by gravity depend on the path taken?

Where should I put the zero of gravitational potential energy?

Is mechanical energy still conserved when there is friction?

Why does PHYS 101 do work and energy without calculus?

Corrected exercises: Work, energy and power, PHYS 101 at McGill

See also

Looking for a PHYS 101 tutor in Montreal?