Mechanics 203-SN1-RE • CEGEP, Montreal

Revision sheet: work, energy and power (203-SN1-RE)

The work-energy theorem and the conservation of mechanical energy are the two most rewarding tools in 203-SN1-RE (formerly 203-NYA): they give a speed without going through time, and they ignore the shape of the path. This comfort has a price: every term of the balance must be written with the right sign, the right height and the right distance.

This sheet lists what costs marks on this particular chapter, in the order the errors show up: the choice of states, the count of forces, the signs, then the cases where energy alone is not enough and you have to go back to Newton. It is written for CEGEP students in the Science program (Sciences de la nature).

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The thread of the chapter

An energy balance is not computed, it is COMPARED between two states. Choose the two states, fix the reference level once and for all, and never touch it again: almost every error in this chapter comes from having changed one of the three along the way.

This chapter is part of 203-SN1-RE, Mechanics

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 force
Work-energy theorem
Kinetic energy
Gravitational potential energy
Elastic potential energy
Hooke's law
Conservative force
Conservation of mechanical energy
Energy dissipated by friction
Mechanical power
Potential energy curve

The essentials

Work, and what does no work

• $W=Fd\cos\theta$ , with $\theta$ the angle between the FORCE and the DISPLACEMENT, not between the force and some arbitrary axis.
• A force perpendicular to the displacement does no work: the normal force on a plane, the tension of a pendulum, the centripetal force in uniform circular motion.
• A force opposite to the displacement does negative work: this is always the case for kinetic friction, in both directions of motion.
• Variable force: $W=\displaystyle\int F\,dx$ , that is, the SIGNED AREA under the curve $F(x)$ . For a spring, this area is $\tfrac{1}{2}kx^{2}$ .
• Work-energy theorem: $W_{net}=\Delta E_{k}$ . The word NET is essential: it means the sum of the work of ALL the forces.

Same force, same displacement

d

: the work is positive when

F

pulls in the direction of motion, zero when it is perpendicular to it, and negative for friction, which opposes it.

Two methods coexist and are never mixed: either you count the work of the weight in $W_{net}$ , or you write it as a potential energy $mgh$ . Doing both in the same balance counts gravity twice.

The energy balance, in the only form that works

• $E_{k,i}+E_{p,i}+E_{e,i}=E_{k,f}+E_{p,f}+E_{e,f}+|W_{f}|$ , where $|W_{f}|=f_{k}d$ is the energy dissipated.
• $E_{k}=\tfrac{1}{2}mv^{2}$ , always positive, never directed: energy has no sign of direction.
• $E_{p}=mgh$ , where $h$ is the VERTICAL HEIGHT above the reference level, never the distance travelled along a slope.
• $E_{e}=\tfrac{1}{2}kx^{2}$ , where $x$ is the DEFORMATION of the spring from its natural length, never its length.
• $d$ in the dissipated term is the distance ACTUALLY SLID. A round trip over a rough patch $1.2$ m long dissipates $f_{k}\times 2.4$ m.

At the ends,

v=0

: everything is potential. At the lowest point, everything is kinetic. The tension, perpendicular to

\vec{v}

, never does work, and

h

is measured vertically.

The reference level of $E_{p}$ is a free choice, but it applies to both states of the balance. Putting it at the lowest point of the problem cancels a term and removes a sign to get wrong.

Reading a potential energy curve

• The horizontal line $E$ is the total mechanical energy, constant if there is no friction.
• $E_{k}(x)=E-U(x)$ : the VERTICAL distance between the line and the curve, to be read at a glance.
• Where the curve meets the line, $E_{k}=0$ : these are the turning points, where the object stops and turns back.
• A minimum of $U$ is a STABLE equilibrium, a maximum an UNSTABLE one. The force is $F=-\dfrac{dU}{dx}$ , so it always pushes toward the bottom of the curve.
• An object trapped in a well can only get out if $E$ exceeds the top of the barrier bordering it.

Kinetic energy is the vertical gap between the line and the curve: zero at the four marked points, greatest at the bottom of the well. An object released on the right stays trapped there, the hump being too high.

This reading answers in five seconds questions that would take a page of calculation: where does it go, how far, where is it fastest. It shows up on almost every final exam.

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. Counting the weight twice, as work and as potential energy

the whole question, 5 marks, with a speed too large by a factor of

\sqrt{2}

What not to write

“ $W_{net}=W_{weight}+W_{normal}=mgh$ , so $mgh+mgh_{i}=\tfrac{1}{2}mv_{f}^{2}$ .”

What to write

“Either $W_{weight}$ appears in the work-energy theorem, or $mgh$ appears in the mechanical balance. Never both: $mgh_{i}=\tfrac{1}{2}mv_{f}^{2}$ is enough.”

Why: Potential energy IS the work of the weight, with its sign changed. Writing both means making gravity act twice on the same path.

2. Using the length of the slope instead of the vertical height in $mgh$

3 marks, and a factor of

2

on the energy

What not to write

“The block slides $4.0$ m along a slope at $30°$ , so it loses $mg\times 4.0$ of potential energy.”

What to write

“The height lost is $h=4.0\sin 30°=2.0$ m, so the change in potential energy is $mg\times 2.0$ .”

The block slides

L=4.0

m but only drops

h=4.0\sin 30°=2.0

m: it is this vertical measure that goes into

mgh

Why: $E_{p}$ depends only on the VERTICAL position. That is precisely what makes the weight conservative: the length of the path must not enter the calculation, otherwise you would be back to a friction force.

3. Writing the work of friction with a positive sign

2 marks, and a final speed greater than the initial speed

What not to write

“ $E_{m,f}=E_{m,i}+f_{k}d$ , the energy of friction is added to the balance.”

What to write

“ $E_{m,f}=E_{m,i}-f_{k}d$ : friction REMOVES mechanical energy, in both directions of motion.”

Why: Friction always opposes the slipping, so its work is always negative. A plus sign makes a block speed up while it is being braked, which you can spot in the result without doing any calculation.

4. Taking the displacement instead of the distance slid in the dissipated energy

the whole question, 5 marks

What not to write

“The block climbs the rough slope for $1.5$ m then slides back to its starting point: the displacement is zero, so nothing is dissipated.”

What to write

“Friction dissipates $f_{k}\times 3.0$ m: it counts the distance ACTUALLY travelled, there AND back, never the net displacement.”

Why: This is the signature of a non-conservative force: its work depends on the path. If the displacement were enough, friction would be conservative and there would be no reason to tell it apart from the weight.

5. Changing the reference level between the initial and the final state

3 marks, with no visible calculation error

What not to write

“At the start the block is $2.0$ m above the floor, so $E_{p,i}=mg\times 2.0$ ; at the end it is $1.0$ m above the table, so $E_{p,f}=mg\times 1.0$ .”

What to write

“One single reference level for the whole problem, for example the floor: $E_{p,i}=mg\times 2.0$ and $E_{p,f}=mg\times 1.8$ if the table is $0.8$ m high.”

Why: Only the DIFFERENCE in potential energy has physical meaning. It means nothing if the two terms are measured from two different origins, and the error is invisible in the calculation.

6. Confusing the deformation of the spring with its length

the whole question, 4 marks

What not to write

“The spring measures $0.25$ m under the load, so $E_{e}=\tfrac{1}{2}k(0.25)^{2}$ .”

What to write

“Its natural length is $0.40$ m, so the deformation is $x=0.15$ m and $E_{e}=\tfrac{1}{2}k(0.15)^{2}$ .”

x

is the difference from the natural length (dotted), not the length of the spring:

x=0.40-0.25=0.15

Why: $k$ measures the stiffness relative to the undeformed state. The $x$ in Hooke's law is a difference, never a position or a length, and a spring at rest stores nothing.

7. Approaching a vertical loop with the height of its top

the whole question, 6 marks

What not to write

“To go around the loop of radius $R$ , you must at least reach its top, so start from $h=2R$ .”

What to write

“At the top you still need a speed: $v_{top}^{2}=gR$ . The balance then gives $h_{min}=2R+\dfrac{R}{2}=2.5R$ .”

The top of the loop is at

2R

, but the starting height needed is

2.5R

: the extra half radius is exactly the kinetic energy that must be kept at the top to stay on the rail.

Why: Arriving at the top with zero speed means leaving the rail just before. The condition is not geometric but dynamic, and it comes from the previous chapter: $N=0$ gives $v_{min}=\sqrt{gR}$ .

8. Applying the work-energy theorem with the work of only one force

3 marks, and a speed that is always overestimated

What not to write

“The applied force does $120$ J of work over $6.0$ m, so $\tfrac{1}{2}mv_{f}^{2}-\tfrac{1}{2}mv_{i}^{2}=120$ J.”

What to write

“ $W_{net}$ is the sum of ALL the work: $120$ J from the applied force minus $45$ J from friction, that is $75$ J.”

Why: The theorem is about the resultant. Forgetting a force amounts to assuming it is perpendicular to the motion, which is almost never the case for friction.

9. Confusing average power and instantaneous power on a climb

2 marks

What not to write

“The car climbs $200$ m of elevation in $60$ s at constant speed: its instantaneous power halfway up is $\dfrac{mgh}{t}$ for the whole climb.”

What to write

“ $P_{avg}=\dfrac{W_{total}}{\Delta t}$ describes the whole trip; $P_{inst}=Fv$ describes one precise instant, with the force and the speed of THAT instant.”

Why: The two coincide only if the force and the speed are constant over the whole trip. A climb at constant speed but on a varying slope does not meet this condition.

10. Treating a collision with conservation of mechanical energy

the whole question, 8 marks

What not to write

“The ballistic pendulum: the bullet embeds itself in the block, so $\tfrac{1}{2}m v^{2}=(m+M)gh$ .”

What to write

“A collision in which the bodies stick together dissipates energy: you conserve MOMENTUM during the impact, then mechanical energy only during the swing up.”

Why: The impact and the swing are two phases with two different laws. Applying energy to the impact overestimates the initial speed by a factor that can reach several units.

Which method to choose

Energy or Newton's laws

What the question asks, and what the statement provides

If a SPEED is asked at two positions, and time appears nowhere → energy balance

Example: speed at the bottom of a slide of unknown shape
If a TIME or an ACCELERATION is asked → Newton's laws and kinematics, since energy does not contain time

Example: duration of a braking, acceleration on a slope
If a FORCE is asked at a precise point of a curved path → energy to get $v$ at that point, then Newton with $\dfrac{mv^{2}}{r}$

Example: tension at the bottom of a pendulum, normal force at the top of a loop
If the path is complicated but frictionless → energy, since only the initial and final states count

Example: ball on a rail of any shape
If friction acts over a known distance → energy with the term $-f_{k}d$ , using the distance slid

Example: block crossing a rough patch of given length
If two bodies come into contact and leave together or separately → momentum for the impact, energy only before and after

mechanical energy is conserved only in a collision explicitly stated to be elastic

The two methods are not opposed: most exam problems chain them, energy for the speed and Newton for the force. What costs marks is choosing the second where the first was enough, and losing ten minutes.

Which term to write in the balance

The objects mentioned in the statement, one by one

If the statement mentions a spring and a compression → $\tfrac{1}{2}kx^{2}$ , with $x$ measured from the natural length

Example: $k=800$ N/m compressed by $0.15$ m gives $9.0$ J
If the statement mentions a height difference, a height, a slope → $mgh$ with $h$ VERTICAL, reference chosen only once

Example: $4.0$ m along a slope at $30°$ gives $h=2.0$ m
If the statement mentions a rough surface and a length → $-f_{k}d$ with $d$ the distance actually slid

Example: a round trip over $1.2$ m gives $d=2.4$ m
If the statement gives a graph of $F(x)$ → $W$ is the signed area under the curve, counted negative below the axis

Example: a triangle of base $4$ m and height $30$ N gives $60$ J
If the statement mentions a speed at two instants → $\tfrac{1}{2}mv^{2}$ on both sides of the equals sign

kinetic energy never takes a sign, even if the object moves backward

Write one line per term, in a column, rather than one long equation: absent terms are then immediately visible, and it is almost always a missing term, not a wrong one, that lowers the mark.

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 up an energy balance

When to use it: The statement asks for a speed, a height, a compression or a distance, and never mentions time.

1 Name the two states being compared, in one sentence: “state 1, spring compressed and block at rest; state 2, block at the top of the slope”.
2 Choose and WRITE DOWN the reference level of potential energy, preferably the lowest point of the problem.
3 Write the inventory of energies in state 1, term by term, explicitly setting to zero those that are zero.
4 Do the same for state 2, then add the dissipated term $f_{k}d$ on the side of the final state.
5 Set up the equality, isolate the unknown, and give the result with its unit.

Concluding sentence

“Between state 1, spring compressed by $0.15$ m and block motionless, and state 2, block at its highest point, with the floor as reference: $\tfrac{1}{2}kx^{2}=mgh+f_{k}L$ , hence $h=0.25$ m.”

The trap: Not writing the zero terms. A paper that skips $E_{k,i}=0$ loses nothing, but a paper that skips $E_{e,f}=0$ without thinking often forgets that there was a spring on the final side.

Marking: Typically 1 mark for the two states and the reference, 3 marks for the complete inventory, 2 for the solution, 1 for the unit and the plausibility check.

Chaining energy then Newton to get a force

When to use it: The statement asks for a tension, a normal force or a contact force at a point of a curved path.

1 First get the speed at that point with an energy balance between the start and that point.
2 Then draw the free-body diagram AT the point in question, with the positive axis pointing toward the centre of the circle.
3 Write $\Sigma F_{\text{toward the centre}}=\dfrac{mv^{2}}{r}$ with the $v^{2}$ obtained in step 1, without going back through the square root.
4 Isolate the force asked for and check its sign: a negative normal force or tension means contact has already been lost.

Concluding sentence

“The energy balance gives $v^{2}=2gL(1-\cos\theta)$ at the bottom; the free-body diagram there gives $T-mg=\dfrac{mv^{2}}{L}$ , so $T=mg(3-2\cos\theta)$ .”

The trap: Extracting $v$ in step 1 only to square it again in step 3. Keeping $v^{2}$ throughout avoids a root, a rounding and a line of writing.

Check before you hand in

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

The sign of each term in the balance

Reread the friction line: it must REMOVE energy. Reread the potential energy line: it must increase when the object goes up. Two seconds, two errors avoided.

A final speed greater than the initial speed on an uphill run with friction: a sign is reversed.
Unit consistency in joules

Every term must be in joules. Check $\tfrac{1}{2}kx^{2}$ as newtons per metre times square metres, and $mgh$ as kilograms times metres per second squared times metres.

A $k$ left in N/cm gives a result a hundred times too large, invisible without a unit check.
The upper bound on the height reached

The final height can never exceed the one you would get without friction. Computing this bound in your head gives an immediate check.

Without friction, $9.0$ J on $1.5$ kg gives $0.61$ m; an answer of $0.25$ m is therefore plausible, an answer of $0.9$ m is not.
The limiting case of zero friction

Set $\mu_{k}=0$ in the final expression. The dissipated term must disappear and you must recover pure conservation, often a known formula.

An expression that still carries a trace of $\mu_{k}$ after setting it to zero contains a copying error.
Agreement with Newton on a simple case

On a free fall, the energy balance must give back $v=\sqrt{2gh}$ , exactly what kinematics gives. This check validates the structure of the balance, not just the numbers.

A balance that gives $v=\sqrt{gh}$ on a free fall has lost a factor of $2$ in the kinetic energy.

The typical problem, taken apart

From the compressed spring to the top of the slope, by way of a rough patch

A spring of constant $k=800$ N/m is compressed by $0.15$ m by a $1.5$ kg block resting on a horizontal plane. The block is then released.

After leaving the spring, it crosses a rough patch of length $L=1.20$ m where $\mu_{k}=0.30$ , then reaches a perfectly smooth slope inclined at $30°$ .

Find the speed of the block at the foot of the slope, then the maximum height it reaches. Take $g=9.8$ m/s $^{2}$ .

Three zones, three terms: the spring supplies the energy, the hatched zone removes a fixed share, the smooth slope converts the rest into height. The distance between the spring and the hatching does not enter anywhere.

Step 1

State 1: spring compressed by $0.15$ m, block motionless. State 2: block at the foot of the slope. State 3: block at its highest point. Reference of $E_{p}$ : the horizontal plane.

Why

Three named states are better than a long chain of calculations: each balance links only two of them, and at every moment you know what you are talking about.

Step 2

Initial energy: $E_{e}=\tfrac{1}{2}kx^{2}=\tfrac{1}{2}(800)(0.15)^{2}=9.0$ J. The other terms are zero: $E_{k,1}=0$ and $E_{p,1}=0$ .

Why

Write the zero terms rather than omit them: it guarantees that none has been forgotten, and it is often half a mark of rigour.

Step 3

Energy dissipated in the rough patch: $f_{k}=\mu_{k}mg=0.30\times 1.5\times 9.8=4.41$ N, so $W_{f}=f_{k}L=4.41\times 1.20=5.29$ J.

Why

The normal force equals $mg$ here because the plane is horizontal and no force has a vertical component. On the slope it would be $mg\cos 30°$ , but the slope is smooth and the question does not arise.

Step 4

Balance between states 1 and 2: $9.0-5.29=\tfrac{1}{2}(1.5)v^{2}$ , so $v^{2}=\dfrac{2\times 3.71}{1.5}=4.94$ and $v=2.22$ m/s.

Why

The rough patch is treated as a single subtraction, not as a motion to follow. That is the whole point of the energy method: you never calculate what happens between the two states.

Step 5

Balance between states 2 and 3, on a smooth slope: $\tfrac{1}{2}(1.5)v^{2}=mgh$ , so $h=\dfrac{3.71}{1.5\times 9.8}=0.252$ m.

Why

The angle of the slope appears nowhere: the height reached does not depend on how steep the path is, only on the energy available. This is the most frequent trick question of the chapter.

Step 6

Distance travelled along the slope, if asked: $d=\dfrac{h}{\sin 30°}=\dfrac{0.252}{0.5}=0.504$ m.

Why

The height and the distance along the slope are two different answers to two different questions. Reading which one is asked is worth two marks here.

Step 7

Check: without friction, the height would be $\dfrac{9.0}{1.5\times 9.8}=0.612$ m. The result $0.252$ m is indeed well below it.

Why

This bound can be computed in your head and rules out an absurd answer. It checks at the same time the sign of the dissipated term and the order of magnitude of the result.

The conclusion, written out

“The block reaches the foot of the slope at $2.22$ m/s and rises to a height of $0.25$ m, that is $0.50$ m measured along the slope.”

The classic mistake on this problem: Adding the dissipated term instead of subtracting it, which gives $14.3$ J at the foot of the slope and a height of $0.97$ m, greater than the one you would get with no friction at all. The check in the last step is enough to catch it.

Learn by heart

• $W=Fd\cos\theta$ : a force perpendicular to the displacement never does work.
• $W_{net}=\Delta E_{k}$ , with NET meaning all the forces together.
• Either the work of the weight, or $mgh$ , never both in the same balance.
• $h$ is a VERTICAL height; $x$ is a DEFORMATION; $d$ is the distance SLID.
• The reference level of $E_{p}$ is free but single for the whole problem.
• Vertical loop: $h_{min}=2.5R$ , never $2R$ , because you need $v_{top}=\sqrt{gR}$ .
• $P_{avg}=\dfrac{W}{\Delta t}$ and $P_{inst}=Fv$ : they coincide only at constant force and speed.
• On a $U(x)$ curve: $E_{k}$ is the vertical gap to the line $E$ , and $F=-\dfrac{dU}{dx}$ .

Frequently asked questions

When should I use energy rather than Newton's laws?

As soon as the question is about a speed between two positions and time appears neither in the statement nor in the answer. Energy completely ignores the shape of the path, which makes it unbeatable on rails, slides and pendulums. As soon as a time, an acceleration or a force is asked for, you have to go back to Newton.

Why is the work of friction always negative?

Because kinetic friction opposes the slipping, so its direction is always opposite to the displacement. The angle between the two is one hundred and eighty degrees and its cosine is minus one. This is true on the way out and on the way back, which explains why a round trip dissipates twice as much energy as a one-way trip.

How do I choose the reference level of potential energy?

It is entirely free, because only the difference in potential energy has meaning. The profitable choice is the lowest point of the problem: one of the two terms becomes zero and there is no negative height to handle. The one absolute rule is to keep the same level for both states being compared.

What height do I need to start from to complete a vertical loop?

Two and a half times the radius, measured from the bottom of the loop, if the rail is smooth. The top is at two radii, but you must arrive there with a nonzero speed, otherwise the object leaves the rail. That minimum speed corresponds to a zero contact force and equals the square root of the product of gravity and the radius.

Can kinetic energy be negative?

No, never. It equals half the mass times the square of the speed, and a square is always positive. An object moving backward has the same kinetic energy as an object moving forward at the same speed. That is exactly what distinguishes energy from momentum, which changes sign with the direction of motion.

Why does the height reached not depend on the inclination of the slope?

Because potential energy depends only on the vertical height, and on a smooth slope nothing is dissipated. The distance travelled along the slope changes with the angle, but not the height reached. This stops being true as soon as there is friction, since the dissipated term depends on the length of the path.

Practise it

Corrected exercises: Work, energy and power, 203-SN1-RE

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

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-phys1-work-energy-power. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Get in touch for a first session. On energy, we mostly work on setting up the balance: two states, one reference level, one line per term. The calculation that follows is then no problem.

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Revision sheet: work, energy and power (203-SN1-RE)

The essentials

Work, and what does no work

The energy balance, in the only form that works

Reading a potential energy curve

The mistakes that cost marks

1. Counting the weight twice, as work and as potential energy

2. Using the length of the slope instead of the vertical height in mghmghmgh

3. Writing the work of friction with a positive sign

4. Taking the displacement instead of the distance slid in the dissipated energy

5. Changing the reference level between the initial and the final state

6. Confusing the deformation of the spring with its length

7. Approaching a vertical loop with the height of its top

8. Applying the work-energy theorem with the work of only one force

9. Confusing average power and instantaneous power on a climb

10. Treating a collision with conservation of mechanical energy

Which method to choose

Energy or Newton's laws

Which term to write in the balance

How the answer is expected to be written

Writing up an energy balance

Chaining energy then Newton to get a force

Check before you hand in

The typical problem, taken apart

From the compressed spring to the top of the slope, by way of a rough patch

Learn by heart

Frequently asked questions

When should I use energy rather than Newton's laws?

Why is the work of friction always negative?

How do I choose the reference level of potential energy?

What height do I need to start from to complete a vertical loop?

Can kinetic energy be negative?

Why does the height reached not depend on the inclination of the slope?

Corrected exercises: Work, energy and power, 203-SN1-RE

See also

Looking for a 203-SN1-RE tutor in Montreal?

2. Using the length of the slope instead of the vertical height in $mgh$