Answer key · Unit 5

Energy.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 25 minutes

Work

A force that moves a body does work on it: the force times the distance moved along it. Forces at right angles to the motion do no work, friction does negative work, and the net work is the kinetic energy the body gains.

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The 6 steps and their answers

Step 1 of 6
Open this step
A force over a distance
Practice · Calculate
Question
The fan pushes the cart with a force of 0.5 N, and in the first two seconds the cart moves 0.98 m along the track. How much work does the push do on the cart in that time?

Correct answer
0.49 J ± 3%
Setup, choices and feedback
The Scene at this step
Fan push
0.5 Nfixed
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
Offdefault
Other settings
Probe time/nodes/cart/props/probeTime = 2
Accepted within
± 3%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1 J ± 3%
  Misconception: Multiplies the force by the time instead of the distance: W = F·t.
  Feedback from the catalogue
  Work is force times distance, not times time: $W = F d$ ⁠. A force that pushes on a wall for an hour, while the wall does not move, does no work at all. Time comes in only when you want to know how fast the work is done, and that is power.
Feedback for a right answer
$W = F d = 0.5 \cdot 0.98 \approx 0.49\ \mathrm{J}$ ⁠. The push and the motion point the same way, so all of the force does work.

Hint
Work is force times the distance along the force: $W = F d$ ⁠.

Worked solution
1. The givens: the force $F = 0.5\ \mathrm{N}$ ⁠, along the motion, and the distance $d = 0.98\ \mathrm{m}$ ⁠.
2. The formula: $W = F d$ ⁠.
3. Substitute: $W = 0.5 \cdot 0.98 = 0.49$ ⁠.
4. The result: 0.49 J, that is 0.49 N·m.
If you got 1 J, you multiplied the force by the time, two seconds, instead of the distance. There is no time in work: what counts is how far the cart moves.
Step 2 of 6
Open this step
Forces that do no work
Practice · Choose
Question
The cart’s weight is 9.8 N. How much work does it do on the cart while the cart travels 0.98 m along the level track?

Correct answer
- Correct 0 J
Setup, choices and feedback
The Scene at this step
Fan push
0.5 Nfixed
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
On
Choices
Shown in a random order.
- Correct 0 J
- Wrong 9.6 J
  Misconception: Thinks a force at right angles to the motion, such as the weight on a level track, does work.
  Feedback from the catalogue
  Only the part of a force along the motion does work: $W = F d \cos\theta$ ⁠. The weight pulls down and the cart moves sideways, so $\theta = 90^\circ$ ⁠ and $\cos 90^\circ = 0$ ⁠. The weight neither speeds the cart up nor slows it down along the track, so it does no work on it.
- Wrong −9.6 J
  Misconception: Thinks a force at right angles to the motion, such as the weight on a level track, does work.
  Feedback from the catalogue
  Only the part of a force along the motion does work: $W = F d \cos\theta$ ⁠. The weight pulls down and the cart moves sideways, so $\theta = 90^\circ$ ⁠ and $\cos 90^\circ = 0$ ⁠. The weight neither speeds the cart up nor slows it down along the track, so it does no work on it.
- Wrong 0.49 J, the same as the push
  Feedback
  Each force does its own work, set by its own size and direction. The air’s push acts along the track; the weight is at right angles to it.
Feedback for a right answer
The weight pulls down and the cart moves sideways. A force at right angles to the motion does no work, even when it is twenty times the push.

Hint
At what angle to the motion does the weight act?

Worked solution
1. The weight acts straight down and the cart moves horizontally: the angle between the force and the motion is $\theta = 90^\circ$ ⁠.
2. The formula: $W = F d \cos\theta$ ⁠.
3. Substitute: $W = 9.8 \cdot 0.98 \cdot \cos 90^\circ = 9.8 \cdot 0.98 \cdot 0 = 0$ ⁠.
4. The result: the weight does no work. The same goes for the normal force, which pushes up.
If you chose 9.6 J, you multiplied the whole weight by the distance. But the weight neither carries the cart along the track nor holds it back, so it gives it no energy and takes none away.
Step 3 of 6
Open this step
A force at an angle
Practice · Calculate
Question
You pull a suitcase along a level floor with a force of 40 N on its handle, and the handle is at $60^\circ$ ⁠ above the horizontal. The suitcase moves 25 m. How much work does the pull do?

Correct answer
500 J ± 2%
Setup, choices and feedback
The Scene at this step
Fan push
0.5 N
Cart mass
1 kg
Labels
offdefault
Show the forces
Offdefault
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1000 J ± 2%
  Misconception: Uses W = F·d for a force at an angle to the motion, leaving out cos θ.
  Feedback from the catalogue
  When the force pulls at an angle, only its component along the motion, $F\cos\theta$ ⁠, does work. The perpendicular component lifts the suitcase a little but does not carry it forward. So $W = F d \cos\theta$ ⁠, and at $60^\circ$ ⁠ that is half of $F d$ ⁠.
- 866 J ± 2%
  Misconception: Uses sin θ instead of cos θ for the component along the motion.
  Feedback from the catalogue
  The angle is measured between the force and the direction of motion, and the component next to that angle is $F\cos\theta$ ⁠: the adjacent side. $F\sin\theta$ ⁠ is the component at right angles to the motion, the one that does no work. Check the extremes: at $\theta = 0^\circ$ ⁠ all the force does work, and $\cos 0^\circ = 1$ ⁠.
Feedback for a right answer
$W = F d \cos\theta = 40 \cdot 25 \cdot \cos 60^\circ = 500\ \mathrm{J}$ ⁠. The vertical part of the pull only lightens the suitcase; it does not move it along.

Hint
Only the component of the force along the floor does work: $F\cos\theta$ ⁠.

Worked solution
1. The givens: $F = 40\ \mathrm{N}$ ⁠, $d = 25\ \mathrm{m}$ ⁠, and the angle between the force and the motion $\theta = 60^\circ$ ⁠.
2. The component along the motion: $F\cos 60^\circ = 40 \cdot 0.5 = 20\ \mathrm{N}$ ⁠.
3. The work: $W = F d \cos\theta = 20 \cdot 25 = 500$ ⁠.
4. The result: 500 J.
If you got 1000 J, you multiplied the whole force by the distance and left out the angle. If you got 866 J, you used $\sin 60^\circ$ ⁠: that is the vertical component, which is at right angles to the motion and does no work.
Step 4 of 6
Open this step
Negative work
Practice · Calculate
Question
On the rougher track friction pulls the cart back with a force of 0.49 N. In the first two seconds the cart moves 1.02 m forward. How much work does friction do on the cart in that time?

Correct answer
-0.5 J ± 0.02 J
Setup, choices and feedback
The Scene at this step
Fan push
1 Nfixed
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0.05
Other settings
Probe time/nodes/cart/props/probeTime = 2
Accepted within
± 0.02 J
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.5 J ± 5%
  Misconception: Gives the work of friction as positive, though friction acts against the motion.
  Feedback from the catalogue
  Friction pulls against the motion: the angle between it and the motion is $180^\circ$ ⁠, and $\cos 180^\circ = -1$ ⁠. So its work is negative. Negative work means the force takes kinetic energy from the body instead of giving it.
Feedback for a right answer
$W = f d \cos 180^\circ = 0.49 \cdot 1.02 \cdot (-1) \approx -0.50\ \mathrm{J}$ ⁠. Friction takes half a joule from the cart, which becomes heat.

Hint
Friction acts against the motion. What is $\cos 180^\circ$ ⁠?

Worked solution
1. The givens: the friction force $f = 0.49\ \mathrm{N}$ ⁠, backwards, and the distance $d = 1.02\ \mathrm{m}$ ⁠, forwards. The angle between them is $\theta = 180^\circ$ ⁠.
2. The formula: $W = f d \cos\theta$ ⁠.
3. Substitute: $W = 0.49 \cdot 1.02 \cdot (-1) \approx -0.50$ ⁠.
4. The result: about −0.50 J.
In the same time the fan did 1 N times 1.02 m, about 1.02 J, so the net work is about 0.52 J. If you wrote 0.5 J with no minus sign, you forgot that friction acts against the motion: the work of such a force is always negative.
Step 5 of 6
Open this step
Net work and speed
Practice · Calculate
Question
The fan pushes a 0.5 kg cart with a force of 1 N, and friction is negligible. The cart starts from rest. What is its speed after the first metre?

Correct answer
1.995 m/s ± 3%
Setup, choices and feedback
The Scene at this step
Fan push
1 Nfixed
Cart mass
0.5 kgfixed
Labels
offdefault
Show the forces
Offdefault
Other settings
Probe time/nodes/cart/props/probeTime = 1.0025
Accepted within
± 3%
Computed from the Scene at this step’s setting
Speed at probeTime (flat, ramp)/nodes/cart/outputs/velocityAt
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 1.41 m/s ± 3%
  Misconception: Leaves the ½ out of the kinetic energy: E = m·v².
  Feedback from the catalogue
  Kinetic energy is $E_k = \tfrac12 m v^2$ ⁠, with a half. The half comes from how the speed builds up: a steady force takes the body from zero to $v$ ⁠, so the average speed is $\tfrac v2$ ⁠. Without the half the speed comes out $\sqrt2$ ⁠ times too small.
- 4 m/s ± 3%
  Misconception: Solves ½·m·v² = E for v² and gives that as the speed.
  Feedback from the catalogue
  From $\tfrac12 m v^2 = E$ ⁠ you first get $v^2 = \dfrac{2E}{m}$ ⁠, and then there is one more step: $v = \sqrt{\dfrac{2E}{m}}$ ⁠. Check the units: $\dfrac{\mathrm{J}}{\mathrm{kg}} = \mathrm{m^2/s^2}$ ⁠, a speed squared.
Feedback for a right answer
$\tfrac12 \cdot 0.5 \cdot v^2 = 1 \cdot 1$ ⁠, so $v^2 = 4$ ⁠ and $v = 2\ \mathrm{m/s}$ ⁠. The Scene shows about 1.99 m/s because of a very small friction.

Hint
The net work is $F d$ ⁠, and it equals the kinetic energy $\tfrac12 m v^2$ ⁠. Solve for $v$ ⁠.

Worked solution
1. The givens: $F = 1\ \mathrm{N}$ ⁠, $d = 1\ \mathrm{m}$ ⁠, $m = 0.5\ \mathrm{kg}$ ⁠, from rest.
2. The net work: $W = F d = 1 \cdot 1 = 1\ \mathrm{J}$ ⁠.
3. The work–energy theorem: $\tfrac12 m v^2 = W$ ⁠, that is $\tfrac12 \cdot 0.5 \cdot v^2 = 1$ ⁠, so $v^2 = \dfrac{2 \cdot 1}{0.5} = 4$ ⁠.
4. The result: $v = \sqrt{4} = 2\ \mathrm{m/s}$ ⁠.
If you got about 1.41 m/s, you dropped the half from the kinetic energy. If you got 4 m/s, you stopped at $v^2$ ⁠ and did not take the square root.
Step 6 of 6
Open this step
In summary
No practice question in this step.
The Scene at this step
Fan push
0.5 Ndefault
Cart mass
1 kgdefault
Labels
offdefault
Show the forces
Offdefault

Lesson 2 of 4 · About 25 minutes

Kinetic and potential energy

A roller-coaster car without friction: its height is potential energy, its speed kinetic energy, and their sum is conserved. How fast is it at the bottom and at the top of the loop, and from what height does it make it round?

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The 7 steps and their answers

Step 1 of 7
Open this step
Energy stored in height
Practice · Calculate
Question
The car, of mass 500 kg, waits at rest 15 m above the bottom of the track. How much gravitational potential energy does it have? Use $g = 9.81\ \mathrm{m/s^2}$ ⁠.

Correct answer
73.58 kJ ± 1%
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kgfixed
Labels
offdefault
Ride / Energy
Ridedefault
Accepted within
± 1%
Computed from the Scene at this step’s setting
Potential energy m·g·h at the release, the car's whole energy (coaster)/nodes/coaster/outputs/startEnergy
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 7.5 kJ ± 2%
  Misconception: Leaves g out: uses the mass in kilograms where the weight in newtons belongs (m·h, μ·m·d, m·h/t).
  Feedback from the catalogue
  Mass in kilograms is not a force. The force pulling a body down is its weight, $mg$ ⁠, in newtons, so the potential energy is $mgh$ ⁠ and the friction is $\mu m g$ ⁠. Without $g$ ⁠ the answer comes out $9.81$ ⁠ times too small, and not in a unit of energy.
Feedback for a right answer
$E_p = m g h = 500 \cdot 9.81 \cdot 15 \approx 73{,}600\ \mathrm{J}$ ⁠, about 73.6 kJ: the same as the total column at Energy.

Hint
$E_p = m g h$ ⁠. You can answer in joules or kilojoules.

Worked solution
1. The givens: $m = 500\ \mathrm{kg}$ ⁠, $h = 15\ \mathrm{m}$ ⁠, $g = 9.81\ \mathrm{m/s^2}$ ⁠.
2. The formula: $E_p = m g h$ ⁠.
3. Substitute: $E_p = 500 \cdot 9.81 \cdot 15 = 73{,}575$ ⁠.
4. The result: about 73,600 J, which is 73.6 kJ.
If you got 7,500 J, you multiplied the mass by the height without $g$ ⁠. The force it takes to lift the car is its weight, $mg$ ⁠, not its mass.
Step 2 of 7
Open this step
Height becomes speed
Practice · Calculate
Question
The car is released from rest 15 m up, with no friction. What is its speed at the bottom of the loop, at height zero? Use $g = 9.81\ \mathrm{m/s^2}$ ⁠.

Correct answer
17.16 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kgfixed
Labels
offdefault
Ride / Energy
Energy
Accepted within
± 2%
Computed from the Scene at this step’s setting
Speed at the foot of the loop (coaster)/nodes/coaster/outputs/bottomSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 12.13 m/s ± 2%
  Misconception: Leaves the ½ out of the kinetic energy: E = m·v².
  Feedback from the catalogue
  Kinetic energy is $E_k = \tfrac12 m v^2$ ⁠, with a half. The half comes from how the speed builds up: a steady force takes the body from zero to $v$ ⁠, so the average speed is $\tfrac v2$ ⁠. Without the half the speed comes out $\sqrt2$ ⁠ times too small.
- 294.3 m/s ± 2%
  Misconception: Solves ½·m·v² = E for v² and gives that as the speed.
  Feedback from the catalogue
  From $\tfrac12 m v^2 = E$ ⁠ you first get $v^2 = \dfrac{2E}{m}$ ⁠, and then there is one more step: $v = \sqrt{\dfrac{2E}{m}}$ ⁠. Check the units: $\dfrac{\mathrm{J}}{\mathrm{kg}} = \mathrm{m^2/s^2}$ ⁠, a speed squared.
Feedback for a right answer
$v = \sqrt{2 g h} = \sqrt{2 \cdot 9.81 \cdot 15} \approx 17.2\ \mathrm{m/s}$ ⁠, more than 60 km/h.

Hint
All the potential energy becomes kinetic: $m g h = \tfrac12 m v^2$ ⁠.

Worked solution
1. At the start: height $h = 15\ \mathrm{m}$ ⁠ and zero speed. At the bottom: zero height and speed $v$ ⁠.
2. Conservation of energy: $m g h = \tfrac12 m v^2$ ⁠. The mass cancels, leaving $v = \sqrt{2 g h}$ ⁠.
3. Substitute: $v = \sqrt{2 \cdot 9.81 \cdot 15} = \sqrt{294.3}$ ⁠.
4. The result: $v \approx 17.2\ \mathrm{m/s}$ ⁠.
If you got about 12.1 m/s, you dropped the half: $g h = v^2$ ⁠. If you got about 294, that is $v^2$ ⁠, and it still needs a square root.
Step 3 of 7
Open this step
Does the mass matter?
Practice · Predict, then look
Question
You double the car’s mass from 500 kg to 1000 kg and release it from the same height, 15 m. What will its speed be at the bottom of the loop?

Correct answer
- Correct The same speed, with twice the kinetic energy
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kg
Labels
offdefault
Ride / Energy
Energy
Choices
- Correct The same speed, with twice the kinetic energy
- Wrong A higher speed
  Misconception: Thinks a heavier car reaches a higher speed down the same drop.
  Feedback from the catalogue
  A heavier car has more potential energy, $mgh$ ⁠, but it also needs more kinetic energy for every metre per second, $\tfrac12 m v^2$ ⁠. The mass is on both sides and cancels: $v = \sqrt{2gh}$ ⁠. Just as bodies fall with the same acceleration, they reach the same speed.
- Wrong A lower speed, because a heavy car is harder to speed up
  Feedback
  A heavy car is harder to speed up, but its weight, pulling it downhill, is bigger in the same ratio. The two cancel.
Once the prediction is locked in, the Scene changes
Car mass
1000 kg
The Scene then shows: 17.16 m/s · Speed at the foot of the loop (coaster)/nodes/coaster/outputs/bottomSpeed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The energy columns doubled, and the speed stayed about 17.2 m/s. Each kilogram of the car gets the same energy, $g h$ ⁠, so it reaches the same speed.

Worked solution
1. In both cases $m g h = \tfrac12 m v^2$ ⁠.
2. Divide both sides by $m$ ⁠: $g h = \tfrac12 v^2$ ⁠. The mass has left the equation.
3. So $v = \sqrt{2 g h} = \sqrt{2 \cdot 9.81 \cdot 15} \approx 17.2\ \mathrm{m/s}$ ⁠ for any mass.
4. The energy itself does depend on the mass: $m g h$ ⁠ for 1000 kg is about 147 kJ, twice 73.6 kJ.
The idea that heavier is faster comes from air: in everyday life it slows a feather more than a stone. Without friction and air the mass does not change the speed.
Step 4 of 7
Open this step
At the top of the loop
Practice · Calculate
Question
The car is released from rest 15 m up, with no friction. The top of the loop is 10 m up. What is the car’s speed at the top of the loop?

Correct answer
9.905 m/s ± 2%
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kgfixed
Labels
offdefault
Ride / Energy
Energy
Accepted within
± 2%
Computed from the Scene at this step’s setting
Speed at the top of the loop; NaN when the car rolls back or falls first (coaster)/nodes/coaster/outputs/topSpeed
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 17.16 m/s ± 2%
  Misconception: Uses the wrong height for the speed at a point: the start height alone, or the point's own height, instead of the drop between them.
  Feedback from the catalogue
  The speed at a point is set by how far the car has dropped from the start to there: $\tfrac12 m v^2 = m g (h_1 - h_2)$ ⁠. At the top of the loop, 10 m up, a car that started at 15 m has dropped only 5 m. The rest of its energy is still potential.
- 14 m/s ± 2%
  Misconception: Uses the wrong height for the speed at a point: the start height alone, or the point's own height, instead of the drop between them.
  Feedback from the catalogue
  The speed at a point is set by how far the car has dropped from the start to there: $\tfrac12 m v^2 = m g (h_1 - h_2)$ ⁠. At the top of the loop, 10 m up, a car that started at 15 m has dropped only 5 m. The rest of its energy is still potential.
Feedback for a right answer
The car has dropped only 5 m, so $v = \sqrt{2 \cdot 9.81 \cdot 5} \approx 9.9\ \mathrm{m/s}$ ⁠. Two thirds of its energy is still potential.

Hint
At the top of the loop the car is still high. How far has it dropped from the start?

Worked solution
1. At the start: $h_1 = 15\ \mathrm{m}$ ⁠, at rest. At the top of the loop: $h_2 = 10\ \mathrm{m}$ ⁠, speed $v$ ⁠.
2. Conservation of energy: $m g h_1 = m g h_2 + \tfrac12 m v^2$ ⁠, so $v = \sqrt{2 g (h_1 - h_2)}$ ⁠.
3. Substitute: $v = \sqrt{2 \cdot 9.81 \cdot (15 - 10)} = \sqrt{98.1}$ ⁠.
4. The result: $v \approx 9.9\ \mathrm{m/s}$ ⁠.
If you got about 17.2 m/s, that is the speed at the bottom: you forgot that at the top of the loop the car is high again. If you got about 14 m/s, you used the height of the top, 10 m, instead of the drop, 5 m.
Step 5 of 7
Open this step
Is the loop's height enough?
Practice · Predict, then look
Question
The top of the loop is 10 m up. You release the car from rest at exactly 10 m, with no friction. What will happen?

Correct answer
- Correct The car will leave the rail and fall before it reaches the top
Setup, choices and feedback
The Scene at this step
Start height
15 m
Car mass
500 kgfixed
Labels
offdefault
Ride / Energy
Ridedefault
Choices
- Wrong The car will reach the top of the loop at zero speed and just get over it
  Misconception: Thinks a car released from the height of the top of the loop gets just round it.
  Feedback from the catalogue
  A car starting at the height of the top of the loop would get there with zero speed, if it got there at all. But at the top it is upside down, and the rail can only push it, not pull it. With no speed, gravity pulls it inwards more sharply than the rail curves, and it leaves the rail and falls before it reaches the top. It has to arrive at the top with some speed, so it must start higher: without friction, at least $2.5$ ⁠ times the radius.
- Correct The car will leave the rail and fall before it reaches the top
- Wrong The car will not even get halfway up the loop, and will roll back
  Feedback
  Halfway up the loop is 5 m, and the car starts at 10 m. It gets there with speed to spare.
- Wrong The car will go round the loop easily
  Misconception: Thinks a car released from the height of the top of the loop gets just round it.
  Feedback from the catalogue
  A car starting at the height of the top of the loop would get there with zero speed, if it got there at all. But at the top it is upside down, and the rail can only push it, not pull it. With no speed, gravity pulls it inwards more sharply than the rail curves, and it leaves the rail and falls before it reaches the top. It has to arrive at the top with some speed, so it must start higher: without friction, at least $2.5$ ⁠ times the radius.
Once the prediction is locked in, the Scene changes
Start height
10 m
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The car got past halfway up the loop, but near the top it was too slow: the rail would have had to pull it, and it can only push. It left the rail and fell.

Worked solution
1. Conservation of energy says a car starting at 10 m can reach 10 m only at zero speed.
2. But at the top of the loop the car is upside down, and what bends its path is gravity and the rail pushing down on it. The rail cannot pull it upwards.
3. A car that is too slow falls inwards more sharply than the rail curves: the rail no longer touches it, and it flies off in an arc and falls.
4. For this car that happens about 8.3 m up, before the top.
The common slip is to think it is enough to reach the height. In a loop the car has to reach the top with some speed, so it must start higher.
Step 6 of 7
Open this step
The lowest start
Practice · Reach the target
Question
Find the lowest Start height, in steps of half a metre, from which the car makes it round the loop, and press Check.

Goal
- 2 ± 0.1
  At the start of the step: 2
- 12.5 m ± 0.1 m
  At the start of the step: 19.5 m
Setup, choices and feedback
The Scene at this step
Start height
19.5 m
Car mass
500 kgfixed
Labels
offdefault
Ride / Energy
Ridedefault
Read from the Scene at
0 the car rolls back, 1 it leaves the rail and falls, 2 it goes round the loop (coaster)/nodes/coaster/outputs/loopOutcome
Read from the Scene at
Start height/nodes/coaster/props/startHeight
One setting that reaches it
Start height
12.5 m
Labels
off
Feedback for a right answer
12.5 m: two and a half times the loop’s radius. From here the car reaches the top at about 7 m/s, just enough for the rail to stay pressed against it.

Feedback for a wrong answer
Not yet. If the car falls, go higher. If it makes the loop, try half a metre lower.

Hint
From 10 m the car falls. Go up a little at a time and watch what happens at the top of the loop.

Worked solution
1. Change the Start height: from 10 m the car falls, and from 12 m it still falls, near the top.
2. At 12.5 m it makes the loop, and at 13 m of course too. So the answer is 12.5 m.
3. The reckoning behind it: at the top of the loop, with the rail not pushing, gravity alone has to bend the path. That happens when the car’s speed there is $v^2 = g R$ ⁠, here $9.81 \cdot 5$ ⁠, about 7 m/s.
4. Conservation of energy between the start and the top, at height $2R$ ⁠: $g h = g \cdot 2R + \tfrac12 g R$ ⁠, so $h = 2.5R = 12.5\ \mathrm{m}$ ⁠.
The condition $v^2 = g R$ ⁠ comes in the circular-motion Unit; here it is enough to see that the car has to reach the top with some speed.
Step 7 of 7
Open this step
In summary
No practice question in this step.
The Scene at this step
Start height
15 mdefault
Car mass
500 kgdefault
Labels
offdefault
Ride / Energy
Ridedefault

Lesson 3 of 4 · About 25 minutes

Heat and power

On a real track rolling resistance turns part of the energy into heat, but the total is conserved. How much heat is made, how much energy is left for motion, and what is the power of a motor that lifts the car?

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
The total
Practice · Predict, then look
Question
There is rolling resistance, and it warms the wheels and the rails. What will happen to the total column while the car rides?

Correct answer
- Correct It will stay at the same height
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kgfixed
Labels
offdefault
Show the energy
Off
Choices
- Correct It will stay at the same height
- Wrong It will slowly shrink, because friction uses the energy up
  Misconception: Thinks energy gets used up and the total goes down when there is friction.
  Feedback from the catalogue
  The energy does not run out, it changes form. Friction turns kinetic energy into heat: the wheels and the rails warm up. Count the heat too, and the total stays exactly what it was at the start. What shrinks is only the energy that is useful for motion.
- Wrong It will grow as the car speeds up and shrink as it slows down
  Feedback
  As the car speeds up, its kinetic energy grows exactly as its potential energy shrinks. The sum does not follow the speed.
Once the prediction is locked in, the Scene changes
Show the energy
On
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The red heat column grows slowly and then all at once at the brakes, but in the total column it simply takes the place that kinetic and potential energy held before. The sum stays 73.6 kJ for the whole ride.

Worked solution
1. At the start all the energy is potential: $m g h = 500 \cdot 9.81 \cdot 15 \approx 73.6\ \mathrm{kJ}$ ⁠.
2. During the ride it is shared among three forms: potential, kinetic and heat. Friction does not delete energy; it moves it from motion into heat.
3. So the sum $E_p + E_k + Q$ ⁠ stays 73.6 kJ, and even after the brakes, with the car at rest, all of it is in the heat column.
If you think friction uses energy up, you are counting only what can still be used for motion. The energy itself does not vanish.
Step 2 of 7
Open this step
How much heat
Practice · Calculate
Question
Rolling resistance pulls the 500 kg car back with a force of $0.015$ ⁠ times its weight. From the release, 15 m up, to the bottom of the loop the car covers 26.9 m of track. How much energy has become heat by then? Use $g = 9.81\ \mathrm{m/s^2}$ ⁠.

Correct answer
1.982 kJ ± 2%
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kgfixed
Labels
offdefault
Show the energy
On
Accepted within
± 2%
Computed from the Scene at this step’s setting
Energy turned to heat by rolling resistance by the foot of the loop (coaster)/nodes/coaster/outputs/bottomHeat
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.202 kJ ± 3%
  Misconception: Leaves g out: uses the mass in kilograms where the weight in newtons belongs (m·h, μ·m·d, m·h/t).
  Feedback from the catalogue
  Mass in kilograms is not a force. The force pulling a body down is its weight, $mg$ ⁠, in newtons, so the potential energy is $mgh$ ⁠ and the friction is $\mu m g$ ⁠. Without $g$ ⁠ the answer comes out $9.81$ ⁠ times too small, and not in a unit of energy.
Feedback for a right answer
$f = 0.015 \cdot 500 \cdot 9.81 \approx 73.6\ \mathrm{N}$ ⁠, and $Q = f d \approx 73.6 \cdot 26.9 \approx 1{,}980\ \mathrm{J}$ ⁠: about 2.0 kJ, under $3\%$ ⁠ of the energy.

Hint
First the force, $f = \mu m g$ ⁠, then its work along the track, $f d$ ⁠.

Worked solution
1. The givens: $\mu = 0.015$ ⁠, $m = 500\ \mathrm{kg}$ ⁠, $d = 26.9\ \mathrm{m}$ ⁠ of track.
2. The resisting force: $f = \mu m g = 0.015 \cdot 500 \cdot 9.81 \approx 73.6\ \mathrm{N}$ ⁠.
3. The heat equals the size of its work: $Q = f d \approx 73.6 \cdot 26.9 \approx 1{,}980\ \mathrm{J}$ ⁠.
4. The result: about 1.98 kJ.
If you got about 202 J, you multiplied $0.015$ ⁠ by the mass rather than the weight: $g$ ⁠ is missing. And note that the distance is the length of track, 26.9 m, not the height, 15 m.
Step 3 of 7
Open this step
What is left for motion
Practice · Calculate
Question
The car started with 73.6 kJ of potential energy, and by the bottom of the loop 1.98 kJ has become heat. How much kinetic energy does it have at the bottom of the loop?

Correct answer
71.59 kJ ± 1%
Setup, choices and feedback
The Scene at this step
Start height
15 mfixed
Car mass
500 kgfixed
Labels
offdefault
Show the energy
On
Accepted within
± 1%
Computed from the Scene at this step’s setting
Kinetic energy at the foot of the loop (coaster)/nodes/coaster/outputs/bottomKinetic
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 73.6 kJ ± 0.5%
  Misconception: Leaves out the energy turned to heat: takes all the starting energy as still kinetic.
  Feedback from the catalogue
  With friction, the energy at the bottom is shared: $m g h = \tfrac12 m v^2 + Q$ ⁠. Part of the starting energy has already become heat, and only the rest is kinetic. So subtract the heat from the starting energy.
Feedback for a right answer
$E_k = 73.6 - 1.98 \approx 71.6\ \mathrm{kJ}$ ⁠. At the bottom the car moves at about 16.9 m/s, a little less than the 17.2 m/s without friction.

Hint
$m g h = E_k + Q$ ⁠, and at the bottom the height is zero.

Worked solution
1. Conservation of energy with friction: $m g h = m g h_2 + E_k + Q$ ⁠.
2. At the bottom $h_2 = 0$ ⁠, so $E_k = m g h - Q$ ⁠.
3. Substitute: $E_k = 73.6 - 1.98 = 71.6$ ⁠.
4. The result: about 71.6 kJ. The speed follows: $v = \sqrt{\dfrac{2 \cdot 71{,}600}{500}} \approx 16.9\ \mathrm{m/s}$ ⁠.
If you answered 73.6 kJ, you took all the starting energy to be still kinetic. With friction, part of it is already heat.
Step 4 of 7
Open this step
A higher start
Practice · Choose
Question
Without friction the car makes the loop from 12.5 m. Here, with rolling resistance, it needs 13.5 m. Why?

Correct answer
- Correct Part of the energy becomes heat on the way, so the car has less kinetic energy left at the top of the loop
Setup, choices and feedback
The Scene at this step
Start height
13.5 m
Car mass
500 kg
Labels
offdefault
Show the energy
On
Choices
Shown in a random order.
- Correct Part of the energy becomes heat on the way, so the car has less kinetic energy left at the top of the loop
- Wrong With friction energy is not conserved, so energy goes missing
  Misconception: Thinks energy gets used up and the total goes down when there is friction.
  Feedback from the catalogue
  The energy does not run out, it changes form. Friction turns kinetic energy into heat: the wheels and the rails warm up. Count the heat too, and the total stays exactly what it was at the start. What shrinks is only the energy that is useful for motion.
- Wrong Friction pulls the car off the rail at the top of the loop
  Feedback
  Friction acts along the rail, against the motion; it does not pull the car away from it. It only slows it down.
Feedback for a right answer
The car has to reach the top of the loop with the same speed as before. To pay for the heat made on the way as well, it has to start with more energy, that is, higher up.

Hint
What does friction do to the car’s energy along the way?

Worked solution
1. To stay on the rail, the car needs the same speed at the top of the loop, with friction or without.
2. With friction: $m g h = m g \cdot 2R + \tfrac12 m v^2 + Q$ ⁠. The term $Q$ ⁠ is extra that has to be paid for.
3. Up to the top of the loop the car covers about 40 m of track, and each kilogram loses about $0.015 \cdot 9.81 \cdot 40 \approx 6\ \mathrm{J}$ ⁠, like dropping about another 0.6 m.
4. So the lowest start rises from 12.5 m to about 13.1 m, and in half-metre steps to 13.5 m.
Energy is conserved here too: it is just shared with heat as well.
Step 5 of 7
Open this step
Power
Practice · Calculate
Question
The chain hauls the 500 kg car at a steady speed to the top of the hill, 20 m above the bottom, in 40 s. Ignore friction. What is the chain’s power? Use $g = 9.81\ \mathrm{m/s^2}$ ⁠.

Correct answer
2.45 kW ± 2%
Setup, choices and feedback
The Scene at this step
Start height
19.5 m
Car mass
500 kgfixed
Labels
offdefault
Show the energy
Offdefault
Accepted within
± 2%
A fixed value in the question
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.25 kW ± 3%
  Misconception: Leaves g out: uses the mass in kilograms where the weight in newtons belongs (m·h, μ·m·d, m·h/t).
  Feedback from the catalogue
  Mass in kilograms is not a force. The force pulling a body down is its weight, $mg$ ⁠, in newtons, so the potential energy is $mgh$ ⁠ and the friction is $\mu m g$ ⁠. Without $g$ ⁠ the answer comes out $9.81$ ⁠ times too small, and not in a unit of energy.
- 3924 kW ± 3%
  Misconception: Multiplies the work by the time to get power, instead of dividing.
  Feedback from the catalogue
  Power is work per second: $P = \dfrac{W}{t}$ ⁠. A motor that does the same work over a longer time is weaker, not stronger. Check the units: $\dfrac{\mathrm{J}}{\mathrm{s}} = \mathrm{W}$ ⁠.
Feedback for a right answer
$W = 500 \cdot 9.81 \cdot 20 = 98{,}100\ \mathrm{J}$ ⁠, and $P = \dfrac{98{,}100}{40} \approx 2{,}450\ \mathrm{W}$ ⁠, about 2.45 kW.

Hint
First the work, $W = m g h$ ⁠, then the power, $P = \dfrac{W}{t}$ ⁠.

Worked solution
1. The givens: $m = 500\ \mathrm{kg}$ ⁠, $h = 20\ \mathrm{m}$ ⁠, $t = 40\ \mathrm{s}$ ⁠. At a steady speed the chain pulls with a force equal to the weight, so its work is the potential energy the car gains.
2. The work: $W = m g h = 500 \cdot 9.81 \cdot 20 = 98{,}100\ \mathrm{J}$ ⁠.
3. The power: $P = \dfrac{W}{t} = \dfrac{98{,}100}{40} \approx 2{,}450\ \mathrm{W}$ ⁠.
4. The result: about 2.45 kW.
If you got 250 W, you left out $g$ ⁠. If you got a huge number, about 3,900 kW, you multiplied the work by the time instead of dividing.
Step 6 of 7
Open this step
Faster, the same work
Practice · Choose
Question
Another chain hauls the same car to the same height in 20 s instead of 40 s. Compared with the first chain:

Correct answer
- Correct It does the same work, at twice the power
Setup, choices and feedback
The Scene at this step
Start height
19.5 m
Car mass
500 kg
Labels
offdefault
Show the energy
Offdefault
Choices
Shown in a random order.
- Correct It does the same work, at twice the power
- Wrong It does twice the work, at twice the power
  Misconception: Confuses power with work: thinks a faster lift does more work.
  Feedback from the catalogue
  Work depends only on the force and the distance: lifting the same car to the same height is the same work, $mgh$ ⁠, fast or slow. What changes is the rate: the same work in half the time is twice the power, $P = \dfrac{W}{t}$ ⁠.
- Wrong It does the same work, at the same power
  Feedback
  The work is indeed the same, but it is done in half the time. Twice the work is done each second.
- Wrong It does the same work, at half the power
  Misconception: Multiplies the work by the time to get power, instead of dividing.
  Feedback from the catalogue
  Power is work per second: $P = \dfrac{W}{t}$ ⁠. A motor that does the same work over a longer time is weaker, not stronger. Check the units: $\dfrac{\mathrm{J}}{\mathrm{s}} = \mathrm{W}$ ⁠.
Feedback for a right answer
The same car to the same height: the same work, $mgh \approx 98\ \mathrm{kJ}$ ⁠. In half the time the power is doubled, about 4.9 kW.

Hint
Work depends on the force and the distance. Power is work divided by time.

Worked solution
1. The chain’s work is $W = m g h$ ⁠. The mass and the height have not changed, so neither has the work: about 98 kJ.
2. Power is $P = \dfrac{W}{t}$ ⁠. The same work in half the time: $\dfrac{98{,}100}{20} \approx 4{,}900\ \mathrm{W}$ ⁠.
3. So the faster chain does the same work at twice the power.
Thinking the faster chain does more work mixes up work, how much energy is moved, with power, how fast.
Step 7 of 7
Open this step
In summary
No practice question in this step.
The Scene at this step
Start height
15 mdefault
Car mass
500 kgdefault
Labels
offdefault
Show the energy
Offdefault

Lesson 4 of 4 · About 25 minutes

Simple machines

A lever, pulleys and gears lift a heavy load with a small force. The mechanical advantage, the distance the effort moves, the work that does not change, and what efficiency is.

Open the lesson

The 7 steps and their answers

Step 1 of 7
Open this step
Less force
Practice · Calculate
Question
The block and tackle hangs a 40 N load on four runs of rope. How hard do you have to pull the free end of the rope to lift the load? Work it out before you look at the spring balance.

Correct answer
10 N ± 0.2 N
Setup, choices and feedback
The Scene at this step
Machine
Pulleysfixed
Mechanical advantage
4 ×fixed
Load
40 Nfixed
Show the forces
Machine
Accepted within
± 0.2 N
Computed from the Scene at this step’s setting
The effort the spring balance reads, load ÷ advantage/nodes/machine/outputs/effort
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 160 N ± 2%
  Misconception: Multiplies the load by the mechanical advantage to get the effort, instead of dividing.
  Feedback from the catalogue
  The mechanical advantage is the load divided by the effort, so the effort is the load divided by the advantage. A machine with a mechanical advantage of $4$ ⁠ makes the force you need four times smaller. If you had to pull harder than the load, there would be no point in using it.
Feedback for a right answer
$\dfrac{40}{4} = 10\ \mathrm{N}$ ⁠: each run holds a quarter of the load, and your hand holds one run. That is also what the spring balance reads.

Hint
Each of the four runs carries an equal share of the load.

Worked solution
1. The givens: a load $F_1 = 40\ \mathrm{N}$ ⁠, held by $4$ ⁠ runs of rope.
2. The mechanical advantage is the number of runs, $4$ ⁠, so $F_2 = \dfrac{F_1}{4}$ ⁠.
3. Substitute: $F_2 = \dfrac{40}{4} = 10$ ⁠.
4. The result: 10 N.
If you got 160 N, you multiplied by the mechanical advantage instead of dividing. The machine makes the force you need smaller, not bigger.
Step 2 of 7
Open this step
The price: distance
Practice · Predict, then look
Question
The load rises 9 cm. How far do you have to pull the free end of the rope?

Correct answer
36 cm ± 1 cm
A number prediction counts as right within ± 1 cm
Setup, choices and feedback
The Scene at this step
Machine
Pulleysfixed
Mechanical advantage
4 ×fixed
Load
40 Nfixed
Show the forces
Machine
Once the prediction is locked in, the Scene changes
Show the forces
Forces
The Scene then shows: 36 cm · How far the effort moves in one lift/nodes/machine/outputs/effortDistance
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
36 cm: each of the four runs gets 9 cm shorter, and all that rope passes through your hand. Four times less force, four times the distance.

Worked solution
1. When the load rises $9\ \mathrm{cm}$ ⁠, each of the $4$ ⁠ runs of rope holding it gets $9\ \mathrm{cm}$ ⁠ shorter.
2. The rope that is freed comes out at the free end: $4 \cdot 9 = 36$ ⁠.
3. The result: 36 cm, the mechanical advantage times the load’s rise.
That is the price of the small force: your hand moves four times as far as the load.
Step 3 of 7
Open this step
The work does not change
Practice · Calculate
Question
You pull the rope with a force of 10 N through 36 cm, and the 40 N load rises 9 cm. How much work do you do?

Correct answer
3.6 J ± 2%
Setup, choices and feedback
The Scene at this step
Machine
Pulleysfixed
Mechanical advantage
4 ×fixed
Load
40 Nfixed
Show the forces
Forces
Accepted within
± 2%
Computed from the Scene at this step’s setting
Work done by the effort in one lift, effort × its distance/nodes/machine/outputs/workIn
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.9 J ± 3%
  Misconception: Thinks a machine reduces the work needed, multiplying the smaller effort by the load's distance.
  Feedback from the catalogue
  A machine shrinks the force, but your hand moves farther in the same ratio. The effort’s work is the effort times its own distance, not times the load’s. In an ideal machine the work in equals the work out exactly.
- 360 J ± 3%
  Misconception: Multiplies newtons by centimetres and calls the result joules.
  Feedback from the catalogue
  A joule is a newton times a metre. Before you multiply, change centimetres to metres: 36 cm = 0.36 m. A newton times a centimetre is a hundredth of a joule.
Feedback for a right answer
$10 \cdot 0.36 = 3.6\ \mathrm{J}$ ⁠, exactly the work done on the load, $40 \cdot 0.09 = 3.6\ \mathrm{J}$ ⁠. The machine saved no work.

Hint
Your work is your force times the distance your hand moves, in metres.

Worked solution
1. The givens: the effort $F_2 = 10\ \mathrm{N}$ ⁠ and its distance $d_2 = 36\ \mathrm{cm} = 0.36\ \mathrm{m}$ ⁠.
2. The work: $W = F_2 d_2 = 10 \cdot 0.36 = 3.6$ ⁠.
3. The result: 3.6 J.
4. Check: the work on the load is $F_1 d_1 = 40 \cdot 0.09 = 3.6\ \mathrm{J}$ ⁠, the same.
If you got 0.9 J, you multiplied your force by the load’s distance: as if the machine saved work. If you got 360, you multiplied by centimetres, not metres.
Step 4 of 7
Open this step
The lever
Practice · Calculate
Question
On the lever a 60 N load sits 20 cm from the fulcrum, and the effort pulls 60 cm from it, on the other side. What effort balances the load?

Correct answer
20 N ± 0.2 N
Setup, choices and feedback
The Scene at this step
Machine
Leverfixed
Mechanical advantage
3 ×fixed
Load
60 Nfixed
Show the forces
Machinefixed
Accepted within
± 0.2 N
Computed from the Scene at this step’s setting
The effort the spring balance reads, load ÷ advantage/nodes/machine/outputs/effort
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 180 N ± 2%
  Misconception: Inverts the lever's arm ratio: puts the effort's longer arm on the wrong side, so the effort comes out larger than the load.
  Feedback from the catalogue
  The lever balances when the moments are equal: $F_1 d_1 = F_2 d_2$ ⁠. The effort acts on the long arm, so it is smaller than the load: $F_2 = F_1 \dfrac{d_1}{d_2}$ ⁠. A force far from the fulcrum turns harder, so you need less of it.
Feedback for a right answer
$60 \cdot 20 = F_2 \cdot 60$ ⁠, so $F_2 = 20\ \mathrm{N}$ ⁠: the arm is three times as long, and the force three times smaller.

Hint
The lever balances when the moments are equal: $F_1 d_1 = F_2 d_2$ ⁠.

Worked solution
1. The givens: $F_1 = 60\ \mathrm{N}$ ⁠ on an arm $d_1 = 20\ \mathrm{cm}$ ⁠, and the effort on an arm $d_2 = 60\ \mathrm{cm}$ ⁠.
2. Equal moments: $F_1 d_1 = F_2 d_2$ ⁠, so $F_2 = \dfrac{F_1 d_1}{d_2}$ ⁠.
3. Substitute: $F_2 = \dfrac{60 \cdot 20}{60} = 20$ ⁠. The units of distance cancel, so there is no need to change to metres.
4. The result: 20 N.
If you got 180 N, you inverted the ratio: you put the long arm on the load’s side. The force acting far from the fulcrum is the smaller one.
Step 5 of 7
Open this step
Choosing a machine
Practice · Reach the target
Question
The load is 80 N, and you can apply only 20 N. Choose a Machine and a Mechanical advantage so that the spring balance reads exactly 20 N, and press Check.

Goal
- 20 N ± 0.01 N
  At the start of the step: 80 N
Setup, choices and feedback
The Scene at this step
Machine
Lever
Mechanical advantage
1 ×
Load
80 Nfixed
Show the forces
Forces
Read from the Scene at
The effort the spring balance reads, load ÷ advantage/nodes/machine/outputs/effort
One setting that reaches it
Machine
Lever
Mechanical advantage
4 ×
Feedback for a right answer
A mechanical advantage of $4$ ⁠: $\dfrac{80}{20} = 4$ ⁠. Any of the three machines does it, and in all of them your hand moves four times as far as the load.

Feedback for a wrong answer
Not 20 N yet. If the spring balance reads more, raise the Mechanical advantage.

Hint
What mechanical advantage turns 80 N into 20 N?

Worked solution
1. The givens: a load $F_1 = 80\ \mathrm{N}$ ⁠, and the effort you want, $F_2 = 20\ \mathrm{N}$ ⁠.
2. The mechanical advantage needed: $\dfrac{F_1}{F_2} = \dfrac{80}{20} = 4$ ⁠.
3. Set Mechanical advantage to $4$ ⁠, on any of the machines: a lever whose effort arm is four times as long, four runs of rope, or a gear with four times the teeth.
4. The spring balance reads 20 N.
If you tried an advantage of $2$ ⁠, the effort came out 40 N: the advantage has to be the ratio of the load to the effort, not the difference.
Step 6 of 7
Open this step
Efficiency
Practice · Calculate
Question
In a real block and tackle with four runs of rope, lifting a 40 N load by 9 cm takes 36 cm of rope pulled with 12.5 N, not 10 N. What is the system’s efficiency, in per cent?

Correct answer
80 ± 1
Setup, choices and feedback
The Scene at this step
Machine
Pulleysfixed
Mechanical advantage
4 ×fixed
Load
40 Nfixed
Show the forces
Forces
Accepted within
± 1
A fixed value in the question
Wrong answers it recognises
- 125 ± 1
  Misconception: Divides the work in by the work out, getting an efficiency above 100%.
  Feedback from the catalogue
  Efficiency is the share of the work put in that reaches the load: the work out divided by the work in. It cannot be more than $100\%$ ⁠, because a machine does not make energy. If you got more than $100\%$ ⁠, you swapped the top and the bottom.
Feedback for a right answer
$\eta = \dfrac{40 \cdot 0.09}{12.5 \cdot 0.36} = \dfrac{3.6}{4.5} = 0.8$ ⁠, that is $80\%$ ⁠. The remaining 0.9 J becomes heat in the axles or lifts the pulleys themselves.

Hint
Work out the work out, on the load, and the work in, by your hand. The efficiency is their ratio.

Worked solution
1. The work out, on the load: $W_1 = 40 \cdot 0.09 = 3.6\ \mathrm{J}$ ⁠.
2. The work in, by your hand: $W_2 = 12.5 \cdot 0.36 = 4.5\ \mathrm{J}$ ⁠.
3. The efficiency: $\eta = \dfrac{W_1}{W_2} = \dfrac{3.6}{4.5} = 0.8$ ⁠.
4. The result: $80\%$ ⁠.
If you got $125\%$ ⁠, you divided the work in by the work out. An efficiency over $100\%$ ⁠ would mean the machine was making energy.
Step 7 of 7
Open this step
In summary
No practice question in this step.
The Scene at this step
Machine
Leverdefault
Mechanical advantage
2 ×default
Load
40 Ndefault
Show the forces
Machinedefault

Across the unit · 20 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Work

A force over a distance

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Forces that do no work

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

A force at an angle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Negative work

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Net work and speed

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

In summary

Kinetic and potential energy

Energy stored in height

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Height becomes speed

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Does the mass matter?

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

At the top of the loop

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Is the loop's height enough?

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The lowest start

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution