Answer key · Unit 4

Forces.

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 5

Lesson 1 of 5 · About 20 minutes

The first law: force and motion

Does a body need a force to keep moving? A fan cart on a track with no friction, what happens when the fan switches off, and forces that balance each other.

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The 5 steps and their answers

Step 1 of 5
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A steady push
Practice · Predict, then look
Question
The fan will switch on and push the cart with a constant force of 0.5 N, and it will not switch off. There is no friction. What will happen to the cart’s speed?

Correct answer
- Correct It will keep growing, by the same amount every second
Setup, choices and feedback
The Scene at this step
Fan push
0 Nfixed
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
Off
Other settings
Friction/nodes/cart/props/friction = 0
Choices
- Correct It will keep growing, by the same amount every second
- Wrong It will grow to a certain speed and stay there
  Misconception: Thinks a constant force gives a constant speed.
  Feedback from the catalogue
  A constant force gives a constant acceleration, not a constant speed. As long as the fan pushes and the net force is not zero, the speed keeps growing by the same amount every second: $v = a t$ ⁠. A constant speed comes precisely when the net force is zero.
- Wrong It will be steady from the start: a constant force, a constant speed
  Misconception: Thinks a constant force gives a constant speed.
  Feedback from the catalogue
  A constant force gives a constant acceleration, not a constant speed. As long as the fan pushes and the net force is not zero, the speed keeps growing by the same amount every second: $v = a t$ ⁠. A constant speed comes precisely when the net force is zero.
Once the prediction is locked in, the Scene changes
Fan push
0.5 Nfixed
Show the forces
On
The Scene then shows: 0.5 m/s² · Constant acceleration along the track (flat, ramp)/nodes/cart/outputs/acceleration
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The marks spread apart: in each half second the cart covers more ground than in the half second before. The green arrow shows a net force of 0.5 N and an acceleration of 0.5 m/s²: the speed grows by 0.5 m/s every second, as long as the fan pushes.

Worked solution
1. A constant net force acts on the cart: the 0.5 N push, with no friction to balance it.
2. A net force that is not zero changes the velocity, so the speed cannot stay constant while it acts.
3. The force is constant, so the change each second is constant too: a constant acceleration of 0.5 m/s², that is $v = 0.5\, t$ ⁠.
4. After one second 0.5 m/s, after two 1 m/s, and so on to the end of the track.
If you chose a speed that levels off, you were thinking of a car or a bicycle: there air resistance grows with speed until it balances the push. Here nothing balances it.
Step 2 of 5
Open this step
When the fan switches off
Practice · Choose
Question
The cart is moving, and the fan switches off. There is no friction on the track. What will the cart do from now on?

Correct answer
- Correct Carry on at the speed it had when the fan switched off
Setup, choices and feedback
The Scene at this step
Fan push
0.5 N
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
Off
Other settings
Friction/nodes/cart/props/friction = 0
Choices
Shown in a random order.
- Correct Carry on at the speed it had when the fan switched off
- Wrong Slow down gradually, as the push it was given runs out
  Misconception: Thinks a push is stored in the body and runs out, so it slows down by itself.
  Feedback from the catalogue
  The push is not stored in the cart and does not “run out”. A force acts only as long as something exerts it: when the fan switches off, there is no more push. What remains is the velocity, and it does not change without a net force. A cart slows down only when a force, such as friction, acts against its motion.
- Wrong Stop at once, because no force moves it any more
  Misconception: A moving body needs a force to keep moving; without one it stops.
  Feedback from the catalogue
  A body does not need a force to keep moving. It needs a force to change its motion: to speed up, slow down or turn. With no net force it carries on at the same speed in the same direction, and that is Newton’s first law. In everyday life things stop because a force always acts to slow them down, friction or air resistance.
- Wrong Keep speeding up, but more slowly
  Feedback
  Acceleration needs a net force. With the fan off and no friction, no force acts on the cart along the track, so it has no acceleration at all.
Feedback for a right answer
After the fan stops, the gaps between the marks are equal: the cart covers the same distance every half second. There is no force along the track, so nothing changes its speed.

Hint
Which forces act on the cart along the track after the fan switches off?

Worked solution
1. After the fan switches off, no force acts on the cart along the track: the fan is off and there is no friction.
2. The net force along the track is zero, so the acceleration is too: $a = 0$ ⁠.
3. Without acceleration the velocity does not change. The cart carries on at the speed it had when the fan stopped, to the end of the track.
If you chose that it slows down, you imagined the push staying inside the cart and slowly running out. A force is not stored in a body: what remains is the velocity.
Step 3 of 5
Open this step
Forces that balance
Practice · Calculate
Question
The fan pushes the cart with 0.3 N, and it stands still. How big is the friction force acting on it?

Correct answer
0.3 N ± 0.01 N
Setup, choices and feedback
The Scene at this step
Fan push
0.3 Nfixed
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
Off
Other settings
Friction/nodes/cart/props/friction = 0.05
Accepted within
± 0.01 N
Computed from the Scene at this step’s setting
Friction on the cart or block: static while it holds, kinetic once it moves (flat, ramp)/nodes/cart/outputs/frictionForce
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0 N ± 0.01 N
  Misconception: Thinks a body at rest has no forces on it, so there is no friction when nothing slides.
  Feedback from the catalogue
  A body standing still is not a body with no forces on it: the forces on it balance, and the net force is zero. If something pushes or pulls it and it does not move, there must be a force that balances, such as static friction. Without the friction the body would accelerate.
Feedback for a right answer
0.3 N, backwards. The cart is at rest and stays at rest, so the net force on it is zero: friction balances the push exactly.

Hint
The cart does not accelerate. What does that say about the net force along the track?

Worked solution
1. The cart stands still and does not start to move, so its acceleration is zero.
2. By the first law, the net force along the track is zero.
3. Two forces act along the track: the push forwards, 0.3 N, and friction backwards. For the net force to be zero, the friction is 0.3 N.
If you answered $0$ ⁠, you took a body that does not move to be a body with no forces. But if the friction were zero, the push would accelerate the cart.
Step 4 of 5
Open this step
Motion at a steady speed
Practice · Choose
Question
A lift rises at a steady speed. How big is the force with which the cable pulls the lift up, compared with its weight?

Correct answer
- Correct Equal to the weight
Setup, choices and feedback
The Scene at this step
Fan push
0.3 N
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0.05
Choices
Shown in a random order.
- Correct Equal to the weight
- Wrong Bigger than the weight, or the lift would not go up
  Misconception: Thinks a body moving one way must have a net force that way: the cable of a lift rising at a steady speed pulls harder than its weight.
  Feedback from the catalogue
  The net force sets the acceleration, not the direction of motion. A lift rising at a steady speed is not accelerating, so the net force on it is zero: the cable pulls up exactly as hard as the weight pulls down. The cable pulls harder than the weight only while the lift speeds up upwards, as it sets off.
- Wrong Smaller than the weight
  Feedback
  If the cable pulled less than the weight, the net force would point down, and the lift would slow down.
Feedback for a right answer
The lift is not accelerating, so the net force on it is zero: the cable pulls up exactly as hard as the weight pulls down. Like the cart standing against the push, only moving.

Hint
A steady speed: what does that say about the acceleration, and the net force?

Worked solution
1. The lift rises at a steady speed in a straight line, so its acceleration is zero.
2. By the first law, the net force on it is zero.
3. Two forces act on the lift: the cable’s pull up and the weight down. For the net force to be zero, they are equal.
If you chose a pull bigger than the weight, you took moving up to need a net force up. A net force up is needed only to speed up upwards, for instance as the lift sets off.
Step 5 of 5
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 5 · About 25 minutes

The second law: force, mass and acceleration

How much acceleration a force gives: the link between force, mass and acceleration, the newton as a unit, and what happens when several forces act on the same cart.

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The 6 steps and their answers

Step 1 of 6
Open this step
Force and acceleration
Practice · Calculate
Question
The fan pushes a 1.5 kg cart with 0.6 N, and there is no friction. What is the cart’s acceleration?

Correct answer
0.4 m/s² ± 0.005 m/s²
Setup, choices and feedback
The Scene at this step
Fan push
0.6 Nfixed
Cart mass
1.5 kgfixed
Labels
offdefault
Show the forces
Off
Other settings
Friction/nodes/cart/props/friction = 0
Accepted within
± 0.005 m/s²
Computed from the Scene at this step’s setting
Constant acceleration along the track (flat, ramp)/nodes/cart/outputs/acceleration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.9 m/s² ± 0.01 m/s²
  Misconception: Multiplies force by mass to get the acceleration.
  Feedback from the catalogue
  $F = m a$ ⁠, so the acceleration is the force divided by the mass: $a = \dfrac{F}{m}$ ⁠. Check by sense: a heavier cart should get less acceleration from the same push, and multiplying by the mass would give it more. And check the units: $\mathrm{N \cdot kg}$ ⁠ is not $\mathrm{m/s^2}$ ⁠.
- 2.5 m/s² ± 0.01 m/s²
  Misconception: Divides mass by force to get the acceleration.
  Feedback from the catalogue
  The acceleration is the force divided by the mass, $a = \dfrac{F}{m}$ ⁠, not the other way round. Check by sense: a stronger push should give a bigger acceleration, and $\dfrac{m}{F}$ ⁠ would get smaller.
Feedback for a right answer
$a = \dfrac{0.6}{1.5} = 0.4\ \mathrm{m/s^2}$ ⁠: the cart’s speed grows by 0.4 m/s every second.

Hint
$a = \dfrac{F}{m}$ ⁠.

Worked solution
1. Given: the net force $F = 0.6\ \mathrm{N}$ ⁠ (no friction), and the mass $m = 1.5\ \mathrm{kg}$ ⁠.
2. The second law: $a = \dfrac{F}{m}$ ⁠.
3. Substitute: $a = \dfrac{0.6}{1.5} = 0.4\ \mathrm{m/s^2}$ ⁠.
If you got $0.9$ ⁠, you multiplied the force by the mass; if you got $2.5$ ⁠, you divided the mass by the force. A heavier cart gets less acceleration, so the mass goes underneath.
Step 2 of 6
Open this step
Twice the mass
Practice · Predict, then look
Question
Now the acceleration is 0.8 m/s². You will load another 1 kg onto the cart, and the fan will go on pushing with 0.8 N. What will the acceleration be, in m/s²?

Correct answer
0.4 m/s² ± 0.01 m/s²
A number prediction counts as right within ± 0.01 m/s²
Setup, choices and feedback
The Scene at this step
Fan push
0.8 Nfixed
Cart mass
1 kgfixed
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0
Once the prediction is locked in, the Scene changes
Cart mass
2 kgfixed
The Scene then shows: 0.4 m/s² · Constant acceleration along the track (flat, ramp)/nodes/cart/outputs/acceleration
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Twice the mass and the same force: half the acceleration, $a = \dfrac{0.8}{2} = 0.4\ \mathrm{m/s^2}$ ⁠. The green arrow has not changed, because the net force is still 0.8 N, but its label now shows an acceleration of 0.4 m/s².

Hint
How many times bigger does the mass get?

Worked solution
1. Given: $F = 0.8\ \mathrm{N}$ ⁠, and the mass goes up from 1 kg to 2 kg.
2. $a = \dfrac{F}{m} = \dfrac{0.8}{2} = 0.4\ \mathrm{m/s^2}$ ⁠.
3. The mass grew by a factor of $2$ ⁠, and the acceleration shrank by a factor of $2$ ⁠: from 0.8 m/s² to 0.4 m/s².
If you predicted $1.6$ ⁠, you doubled the acceleration along with the mass. More mass is more inertia: the same force changes its velocity more slowly.
Step 3 of 6
Open this step
The newton
Practice · Reach the target
Question
The cart’s mass is 2 kg, and there is no friction. Change Fan push so that the acceleration is 0.25 m/s², and press Check.

Goal
- 0.25 m/s² ± 0.005 m/s²
  At the start of the step: 0 m/s²
Setup, choices and feedback
The Scene at this step
Fan push
0 N
Cart mass
2 kgfixed
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0
Read from the Scene at
Constant acceleration along the track (flat, ramp)/nodes/cart/outputs/acceleration
One setting that reaches it
Fan push
0.5 N
Labels
off
Feedback for a right answer
0.5 N on 2 kg gives $\dfrac{0.5}{2} = 0.25\ \mathrm{m/s^2}$ ⁠.

Feedback for a wrong answer
Not yet. If the acceleration is too big, lower the push; if it is too small, raise it. The label of the green arrow shows the acceleration.

Hint
$F = m a$ ⁠: what force gives two kilograms an acceleration of 0.25 m/s²?

Worked solution
1. Given: $m = 2\ \mathrm{kg}$ ⁠, and you want $a = 0.25\ \mathrm{m/s^2}$ ⁠.
2. The second law: $F = m a$ ⁠.
3. Substitute: $F = 2 \cdot 0.25 = 0.5\ \mathrm{N}$ ⁠. Set Fan push to 0.5 N.
If you set 0.25 N, you forgot the mass: a force of 0.25 N gives 0.25 m/s² only to a body of one kilogram.
Step 4 of 6
Open this step
Several forces together
Practice · Calculate
Question
The fan pushes the cart forwards with 1 N, friction holds it back with 0.2 N, and its mass is 2 kg. What is its acceleration?

Correct answer
0.4 m/s² ± 0.005 m/s²
Setup, choices and feedback
The Scene at this step
Fan push
1 Nfixed
Cart mass
2 kgfixed
Labels
offdefault
Show the forces
Off
Other settings
Friction/nodes/cart/props/friction = 0.010193679918450561
Accepted within
± 0.005 m/s²
Computed from the Scene at this step’s setting
Constant acceleration along the track (flat, ramp)/nodes/cart/outputs/acceleration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 0.5 m/s² ± 0.005 m/s²
  Misconception: Puts one force into F = ma instead of the net force, leaving out the friction against it.
  Feedback from the catalogue
  In $F = m a$ ⁠, $F$ ⁠ is the net force: all the forces along the line added with their signs. When friction acts against the motion, you subtract it from the force that pushes. Without that you get the acceleration there would be with no friction.
- 0.6 m/s² ± 0.005 m/s²
  Misconception: Adds a force that acts against the motion instead of subtracting it.
  Feedback from the catalogue
  Forces along one line add with signs. Friction acts backwards, against the push, so it goes in with a minus: $F = 1 - 0.2 = 0.8\ \mathrm{N}$ ⁠. Adding the two sizes suits only two forces in the same direction.
Feedback for a right answer
The net force is $1 - 0.2 = 0.8\ \mathrm{N}$ ⁠ forwards, and the acceleration $\dfrac{0.8}{2} = 0.4\ \mathrm{m/s^2}$ ⁠. With Show the forces set to On, the green arrow shows both.

Hint
Find the net force first: forwards positive, backwards negative.

Worked solution
1. Choose a positive direction: forwards. The push is $+1\ \mathrm{N}$ ⁠, and the friction $-0.2\ \mathrm{N}$ ⁠.
2. The net force: $F = 1 - 0.2 = 0.8\ \mathrm{N}$ ⁠.
3. The second law: $a = \dfrac{F}{m} = \dfrac{0.8}{2} = 0.4\ \mathrm{m/s^2}$ ⁠.
If you got $0.5$ ⁠, you used only the push and forgot the friction. If you got $0.6$ ⁠, you added the friction instead of subtracting it.
Step 5 of 6
Open this step
Speed after three seconds
Practice · Calculate
Question
The same cart: a push of 1 N, friction of 0.2 N and a mass of 2 kg. It sets off from rest. What is its speed after 3 s?

Correct answer
1.2 m/s ± 0.01 m/s
Setup, choices and feedback
The Scene at this step
Fan push
1 Nfixed
Cart mass
2 kgfixed
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0.010193679918450561
Other settings
Probe time/nodes/cart/props/probeTime = 3
Accepted within
± 0.01 m/s
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
- 0.4 m/s ± 0.01 m/s
  Misconception: Gives the acceleration as the speed reached after some time.
  Feedback from the catalogue
  The acceleration says how much the speed grows each second. From rest, after $t$ ⁠ seconds the speed is $v = a t$ ⁠: with an acceleration of 0.4 m/s², after 3 s the speed is 1.2 m/s, not 0.4 m/s.
- 1.5 m/s ± 0.01 m/s
  Misconception: Puts one force into F = ma instead of the net force, leaving out the friction against it.
  Feedback from the catalogue
  In $F = m a$ ⁠, $F$ ⁠ is the net force: all the forces along the line added with their signs. When friction acts against the motion, you subtract it from the force that pushes. Without that you get the acceleration there would be with no friction.
Feedback for a right answer
$v = a t = 0.4 \cdot 3 = 1.2\ \mathrm{m/s}$ ⁠. In that time the cart has covered $\tfrac12 \cdot 0.4 \cdot 3^2 = 1.8\ \mathrm{m}$ ⁠, almost the whole track.

Hint
Find the acceleration from the net force, then $v = a t$ ⁠.

Worked solution
1. The net force: $1 - 0.2 = 0.8\ \mathrm{N}$ ⁠.
2. The acceleration: $a = \dfrac{0.8}{2} = 0.4\ \mathrm{m/s^2}$ ⁠, constant as long as the forces are.
3. From rest: $v = a t = 0.4 \cdot 3 = 1.2\ \mathrm{m/s}$ ⁠.
If you answered $0.4$ ⁠, you gave the acceleration: it says how much the speed grows each second, and after three seconds it has grown three times. If you got $1.5$ ⁠, you forgot the friction.
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 3 of 5 · About 25 minutes

The third law, weight and the normal force

Forces come in pairs: the fan and the air, the Earth and the cart. Weight, the normal force, and why an action–reaction pair does not cancel itself.

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The 6 steps and their answers

Step 1 of 6
Open this step
Forces come in pairs
Practice · Choose
Question
The fan pushes the air backwards with a force of 0.5 N. With what force does the air push the fan forwards?

Correct answer
- Correct Exactly 0.5 N
Setup, choices and feedback
The Scene at this step
Fan push
0.5 N
Cart mass
1 kg
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0
Choices
Shown in a random order.
- Correct Exactly 0.5 N
- Wrong Less than 0.5 N, because the fan is the one pushing
  Misconception: Thinks the body doing the pushing exerts the larger force of an action-reaction pair.
  Feedback from the catalogue
  In an action–reaction pair neither side is stronger. The fan pushes the air, and the air pushes the fan with exactly the same size of force. “Who started it” makes no difference: the two forces arise together, at the same moment and with the same size.
- Wrong More than 0.5 N, or the cart would not move
  Misconception: Thinks something can move only if its push beats the push back, so the pair must be unequal.
  Feedback from the catalogue
  The cart does not move because its force “beats” the air’s. The two forces of a pair are always equal, but they act on different bodies: one on the air and one on the cart. To tell whether the cart accelerates, add only the forces acting on it, and there a push forwards has nothing to balance it.
- Wrong Zero: air does not push
  Feedback
  Air certainly pushes: it is what moves a sailing boat and what you feel in the wind. And the fan throws the air backwards only because it pushes it, so the air pushes it back.
Feedback for a right answer
The two arrows, the fan on the air and the air on the cart, are the same length and point opposite ways: 0.5 N each.

Hint
Compare the labels of the two arrows that Show the forces draws.

Worked solution
1. The fan exerts a force on the air: 0.5 N backwards.
2. By the third law, the air exerts on the fan a force equal in size and opposite in direction.
3. So the air pushes the fan, and the cart with it, forwards with 0.5 N.
If you chose that the cart needs a bigger force to move, you compared two forces that act on different bodies. The next question is about that.
Step 2 of 6
Open this step
Why the pair does not cancel
Practice · Choose
Question
The fan pushes the air backwards with 0.5 N, and the air pushes the cart forwards with 0.5 N. Why does the cart accelerate all the same?

Correct answer
- Correct Because the two forces act on different bodies: only one of them acts on the cart
Setup, choices and feedback
The Scene at this step
Fan push
0.5 Nfixed
Cart mass
1 kg
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0
Choices
Shown in a random order.
- Correct Because the two forces act on different bodies: only one of them acts on the cart
- Wrong They cancel each other, and the cart moves by inertia
  Misconception: Thinks the two forces of an action-reaction pair cancel each other.
  Feedback from the catalogue
  Forces cancel only when they act on the same body. The two forces of an action–reaction pair always act on two different bodies: the fan pushes the air, and the air pushes the cart. Only the air’s push acts on the cart, so it accelerates.
- Wrong Because the push on the cart is really a little bigger
  Misconception: Thinks something can move only if its push beats the push back, so the pair must be unequal.
  Feedback from the catalogue
  The cart does not move because its force “beats” the air’s. The two forces of a pair are always equal, but they act on different bodies: one on the air and one on the cart. To tell whether the cart accelerates, add only the forces acting on it, and there a push forwards has nothing to balance it.
- Wrong Because the reaction comes a moment after the action
  Feedback
  There is no delay: the two forces appear together and vanish together. And if the reaction came late and acted on the cart, it would stop the cart the moment it arrived.
Feedback for a right answer
Forces cancel only when they act on the same body. Only the air’s push acts on the cart, 0.5 N forwards, and nothing balances it, so it accelerates. The fan’s force acts on the air, and accelerates it backwards.

Hint
What does each force act on: the cart, or the air?

Worked solution
1. Write each force with the body it acts on: the fan pushes the air; the air pushes the cart.
2. To tell whether the cart accelerates, add only the forces on the cart. Along the track there is only one: 0.5 N forwards, and no friction.
3. The net force on the cart is 0.5 N, and it accelerates at $\dfrac{0.5}{1} = 0.5\ \mathrm{m/s^2}$ ⁠.
If you chose that they cancel, you added a force on the air to a force on the cart. Then nothing could ever accelerate.
Step 3 of 6
Open this step
Weight
Practice · Calculate
Question
The cart’s mass is 2.5 kg. What is its weight? Take $g = 9.81\ \mathrm{m/s^2}$ ⁠.

Correct answer
24.53 N ± 2.5%
Setup, choices and feedback
The Scene at this step
Fan push
0 Nfixed
Cart mass
2.5 kgfixed
Labels
offdefault
Show the forces
Off
Other settings
Friction/nodes/cart/props/friction = 0
Accepted within
± 2.5%
Computed from the Scene at this step’s setting
Weight m·g (every layout)/nodes/cart/outputs/weight
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 2.5 N ± 0.05 N
  Misconception: Confuses mass with weight: gives the mass in kilograms as the weight in newtons.
  Feedback from the catalogue
  Mass and weight are different quantities. Mass, in kilograms, is the amount of matter. Weight is a force, in newtons: $W = m g$ ⁠. So a 2.5 kg body weighs about 24.5 N, $9.81$ ⁠ times the number for its mass.
Feedback for a right answer
$W = 2.5 \cdot 9.81 \approx 24.5\ \mathrm{N}$ ⁠. The weight arrow that Show the forces draws shows it, rounded to 25 N.

Hint
$W = m g$ ⁠.

Worked solution
1. Given: $m = 2.5\ \mathrm{kg}$ ⁠, $g = 9.81\ \mathrm{m/s^2}$ ⁠.
2. The weight: $W = m g$ ⁠.
3. Substitute: $W = 2.5 \cdot 9.81 \approx 24.5\ \mathrm{N}$ ⁠.
If you answered $2.5$ ⁠, you gave the mass. Mass is in kilograms, and weight is a force in newtons, $g$ ⁠ times bigger.
Step 4 of 6
Open this step
Falling together
Practice · Choose
Question
A 1 kg ball and a 3 kg ball are released together from the same height. There is no air. Which reaches the floor first?

Correct answer
- Correct Both together
Setup, choices and feedback
The Scene at this step
Fan push
0 Nfixed
Cart mass
2.5 kg
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0
Choices
Shown in a random order.
- Correct Both together
- Wrong The heavy one, because its weight is three times bigger
  Misconception: Thinks a heavier body falls faster, or slips sooner on a slope, because its weight is bigger.
  Feedback from the catalogue
  A heavy body has a bigger weight, but also a bigger mass, and how hard it is to accelerate grows in the same proportion. Falling without air, $a = \dfrac{m g}{m} = g$ ⁠ for every body. On a slope both the component pulling down and the largest friction grow with the mass, so a heavy block slips at the same angle and with the same acceleration as a light block of the same material.
- Wrong The light one, because it is easier to accelerate
  Feedback
  It is easier to accelerate, but its weight is smaller too, in exactly the same proportion. The two effects cancel.
Feedback for a right answer
The heavy ball’s weight is three times bigger, but so are its mass and its inertia. Both accelerate at $g$ ⁠.

Hint
Work out the acceleration of each ball: $a = \dfrac{W}{m}$ ⁠.

Worked solution
1. The only force on each ball is its weight: 9.81 N on the light one and 29.4 N on the heavy one.
2. The second law for each: $a = \dfrac{W}{m}$ ⁠. For the light one $\dfrac{9.81}{1} = 9.81\ \mathrm{m/s^2}$ ⁠, and for the heavy one $\dfrac{29.43}{3} = 9.81\ \mathrm{m/s^2}$ ⁠.
3. The same acceleration, from the same height and from rest: they land together.
If you chose the heavy one, you looked only at the force. A bigger force on a mass bigger in the same proportion gives the same acceleration.
Step 5 of 6
Open this step
The normal force
Practice · Choose
Question
The cart’s weight is the Earth’s pull on the cart. What is its partner under the third law?

Correct answer
- Correct The cart’s pull on the Earth, upwards
Setup, choices and feedback
The Scene at this step
Fan push
0 Nfixed
Cart mass
2.5 kg
Labels
offdefault
Show the forces
On
Other settings
Friction/nodes/cart/props/friction = 0
Choices
Shown in a random order.
- Correct The cart’s pull on the Earth, upwards
- Wrong The normal force: the track’s push on the cart, upwards
  Misconception: Takes the normal force to be the third-law partner of the weight.
  Feedback from the catalogue
  The weight and the normal force both act on the same body, the cart, so they are not an action–reaction pair. The partner of the weight, the Earth’s pull on the cart, is the cart’s pull on the Earth, upwards. The partner of the normal force, the track’s push on the cart, is the cart’s push on the track, downwards. The weight and the normal force are equal here for another reason: the cart does not accelerate vertically.
- Wrong The cart’s push on the track, downwards
  Feedback
  That is the partner of the normal force: the track pushes the cart up, and the cart pushes the track down. The weight’s partner involves the Earth, because the Earth is what exerts the weight.
Feedback for a right answer
The Earth pulls the cart down, and the cart pulls the Earth up just as hard, about 24.5 N. For the Earth that is next to nothing: its mass is enormous, and the acceleration this force gives it is tiny.

Hint
The partner acts on the body that exerts the first force. Who exerts the weight?

Worked solution
1. The weight is a force of the Earth on the cart.
2. By the third law, its partner is a force of the cart on the Earth, equal in size and opposite in direction: a pull upwards.
3. The normal force is a force of the track on the cart. It acts on the cart, like the weight, so it cannot be the weight’s partner.
If you chose the normal force, that is understandable: it is equal and opposite to the weight. But that is because of the first law, since the cart does not accelerate vertically, not because of the third law.
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 4 of 5 · About 25 minutes

A free-body diagram on a slope

A wooden block resting on a slope: how to draw every force acting on it, split its weight into two components, and find the normal force and the friction.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Drawing the forces
Practice · Choose
Question
A wooden block rests without moving on a tilted board. Which forces act on it? Choose all of them.

Correct answer
- Correct The weight: the Earth pulls it down
- Correct A normal force: the board pushes it at right angles to the board
- Correct Friction: the board holds it back along the board
Setup, choices and feedback
The Scene at this step
Slope angle
20 °
Surface grip
0.4 μₛfixed
Labels
offdefault
Show the forces
Off
Choices
Shown in a random order. More than one choice is correct: every one must be selected.
- Correct The weight: the Earth pulls it down
- Correct A normal force: the board pushes it at right angles to the board
- Correct Friction: the board holds it back along the board
- Wrong A force pushing it down the slope, besides the weight
  Misconception: Draws a force down the slope as an extra force beside the weight.
  Feedback from the catalogue
  The force “down the slope” is not an extra force: it is a component of the weight. No body exerts it apart from the Earth, which is already in the diagram as the weight. In a free-body diagram every arrow has a body that exerts it; drawing both the weight and its component counts the same force twice.
- Wrong The force of inertia, holding it in place
  Feedback
  Inertia is a property of the body, not a force: no other body exerts it. What holds the block in place is friction.
- Wrong The force with which the block presses on the board
  Feedback
  That is a force the block exerts on the board, the partner of the normal force. The block’s free-body diagram shows only forces acting on it.
Feedback for a right answer
Three forces: the weight from the Earth, and two from the board, the normal force and friction. Show the forces draws these as the three arrows, and the net force is zero.

Hint
The weight, plus every force from a body that touches the block. What touches the block?

Worked solution
1. The body: the block.
2. A force without contact: the weight, from the Earth, straight down.
3. What touches the block? Only the board. It pushes at right angles to itself, the normal force, and holds back along itself, static friction, up the board.
4. No other bodies touch the block, so there are no other forces.
“A force down the slope” is a component of the weight, not an extra force: the Earth is already in the diagram once.
Step 2 of 5
Open this step
Splitting the weight
Practice · Calculate
Question
The block, 2 kg, rests on a board tilted at $30^\circ$ ⁠. What normal force does the board exert on it?

Correct answer
16.99 N ± 2.5%
Setup, choices and feedback
The Scene at this step
Slope angle
30 °fixed
Surface grip
0.8 μₛfixed
Labels
offdefault
Show the forces
Off
Other settings
Mass/nodes/ramp/props/mass = 2
Accepted within
± 2.5%
Computed from the Scene at this step’s setting
Normal force of the track or board (flat, ramp)/nodes/ramp/outputs/normal
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 19.62 N ± 2.5%
  Misconception: Thinks the normal force always equals the weight.
  Feedback from the catalogue
  The normal force is as much as the surface must push so that the body does not sink into it, and that is not always the weight. On a slope only the weight’s component at right angles, $m g \cos\theta$ ⁠, presses into the board, so $N = m g \cos\theta$ ⁠: smaller than the weight, and smaller still the steeper the slope. It equals the weight only on a level surface with no other vertical forces.
- 9.81 N ± 2.5%
  Misconception: Swaps sine and cosine for the slope: the component into the board, or the ratio tan θ.
  Feedback from the catalogue
  Check a level board, $\theta = 0$ ⁠: there all the weight presses into the board, none of it pulls along, and no grip at all is needed for the block to hold. $\cos 0 = 1$ ⁠, $\sin 0 = 0$ ⁠ and $\tan 0 = 0$ ⁠, so the component at right angles to the board is $m g \cos\theta$ ⁠, the component along it is $m g \sin\theta$ ⁠, and the block holds as long as $\tan\theta = \dfrac{\sin\theta}{\cos\theta} \le \mu_s$ ⁠.
Feedback for a right answer
$N = 2 \cdot 9.81 \cdot \cos 30^\circ \approx 17.0\ \mathrm{N}$ ⁠: less than the weight, 19.6 N.

Hint
$N = m g \cos\theta$ ⁠.

Worked solution
1. Given: $m = 2\ \mathrm{kg}$ ⁠, $\theta = 30^\circ$ ⁠, $g = 9.81\ \mathrm{m/s^2}$ ⁠. The weight: $m g = 19.62\ \mathrm{N}$ ⁠.
2. At right angles to the board the block does not accelerate, so the normal force balances the weight’s component at right angles: $N = m g \cos\theta$ ⁠.
3. Substitute: $N = 19.62 \cdot \cos 30^\circ \approx 19.62 \cdot 0.866 \approx 17.0\ \mathrm{N}$ ⁠.
If you answered 19.6 N, you gave the whole weight: on a slope only part of it presses into the board. If you got 9.8 N, you used the sine instead of the cosine.
Step 3 of 5
Open this step
The friction that holds
Practice · Calculate
Question
The same block, 2 kg, rests on the board at $30^\circ$ ⁠. Surface grip is $\mu_s = 0.8$ ⁠. How big is the friction force acting on it?

Correct answer
9.81 N ± 2.5%
Setup, choices and feedback
The Scene at this step
Slope angle
30 °fixed
Surface grip
0.8 μₛfixed
Labels
offdefault
Show the forces
Off
Other settings
Mass/nodes/ramp/props/mass = 2
Accepted within
± 2.5%
Computed from the Scene at this step’s setting
Friction on the cart or block: static while it holds, kinetic once it moves (flat, ramp)/nodes/ramp/outputs/frictionForce
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 13.59 N ± 2.5%
  Misconception: Thinks static friction always equals μs N, its largest value.
  Feedback from the catalogue
  $\mu_s N$ ⁠ is the limit of static friction, not its value. Static friction grows only as much as needed to keep the block from moving: on a slope it equals the component pulling down, $m g \sin\theta$ ⁠. Only at the moment the block is about to slip does it reach $\mu_s N$ ⁠.
- 0 N ± 0.05 N
  Misconception: Thinks a body at rest has no forces on it, so there is no friction when nothing slides.
  Feedback from the catalogue
  A body standing still is not a body with no forces on it: the forces on it balance, and the net force is zero. If something pushes or pulls it and it does not move, there must be a force that balances, such as static friction. Without the friction the body would accelerate.
Feedback for a right answer
Friction balances the component pulling down the slope: $m g \sin 30^\circ = 19.62 \cdot 0.5 \approx 9.8\ \mathrm{N}$ ⁠, up the board.

Hint
Along the board the block does not accelerate. What force must friction balance?

Worked solution
1. The block stays put, so the net force along the board is zero.
2. Two forces act along the board: the weight’s component down it, $m g \sin\theta$ ⁠, and friction up it.
3. So $f = m g \sin\theta = 2 \cdot 9.81 \cdot \sin 30^\circ \approx 9.81\ \mathrm{N}$ ⁠.
If you answered about 13.6 N, you multiplied $0.8$ ⁠ by the normal force. $\mu_s N$ ⁠ is the limit of static friction, the most it can hold, not its value: the next lesson shows what happens when it is reached.
Step 4 of 5
Open this step
A steeper slope
Practice · Predict, then look
Question
The board will go up from $30^\circ$ ⁠ to $38^\circ$ ⁠, and the block will still hold. What will happen to the normal force?

Correct answer
- Correct It will get smaller
Setup, choices and feedback
The Scene at this step
Slope angle
30 °
Surface grip
0.8 μₛfixed
Labels
offdefault
Show the forces
On
Other settings
Mass/nodes/ramp/props/mass = 2
Choices
- Correct It will get smaller
- Wrong It will stay the same, because the weight does not change
  Misconception: Thinks the normal force always equals the weight.
  Feedback from the catalogue
  The normal force is as much as the surface must push so that the body does not sink into it, and that is not always the weight. On a slope only the weight’s component at right angles, $m g \cos\theta$ ⁠, presses into the board, so $N = m g \cos\theta$ ⁠: smaller than the weight, and smaller still the steeper the slope. It equals the weight only on a level surface with no other vertical forces.
- Wrong It will get bigger, because the block presses harder into the board
  Feedback
  On a steeper slope the block presses into the board less: more of the weight pulls along the board, and less of it presses into it. On a vertical wall it would not press at all.
Once the prediction is locked in, the Scene changes
Slope angle
38 °
The Scene then shows: 15.46 N · Normal force of the track or board (flat, ramp)/nodes/ramp/outputs/normal
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$N = m g \cos\theta$ ⁠, and the cosine gets smaller as the angle grows: from 17.0 N at $30^\circ$ ⁠ to 15.5 N at $38^\circ$ ⁠. At the same time the friction grows, because more of the weight pulls down the slope.

Worked solution
1. $N = m g \cos\theta$ ⁠, and $m g = 19.62\ \mathrm{N}$ ⁠ does not change.
2. At $30^\circ$ ⁠: $19.62 \cdot \cos 30^\circ \approx 17.0\ \mathrm{N}$ ⁠.
3. At $38^\circ$ ⁠: $19.62 \cdot \cos 38^\circ \approx 19.62 \cdot 0.788 \approx 15.5\ \mathrm{N}$ ⁠.
The weight is the same, but the part of it that presses into the board shrinks the steeper the board. If you chose that the normal force stays the same, you took it to always equal the weight.
Step 5 of 5
Open this step
In summary
No practice question in this step.
The Scene at this step
Slope angle
20 °default
Surface grip
0.4 μₛdefault
Labels
offdefault
Show the forces
Offdefault

Lesson 5 of 5 · About 25 minutes

Static and kinetic friction

How much friction can hold, at what angle a block starts to slide, and how fast it then speeds up.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The limit of static friction
Practice · Calculate
Question
The board is tilted at $30^\circ$ ⁠. What is the smallest $\mu_s$ ⁠ that holds the block in place? Answer to two decimal places.

Correct answer
0.577 ± 0.01
Setup, choices and feedback
The Scene at this step
Slope angle
30 °fixed
Surface grip
0.1 μₛ
Labels
offdefault
Show the forces
On
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 1.732 ± 0.02
  Misconception: Swaps sine and cosine for the slope: the component into the board, or the ratio tan θ.
  Feedback from the catalogue
  Check a level board, $\theta = 0$ ⁠: there all the weight presses into the board, none of it pulls along, and no grip at all is needed for the block to hold. $\cos 0 = 1$ ⁠, $\sin 0 = 0$ ⁠ and $\tan 0 = 0$ ⁠, so the component at right angles to the board is $m g \cos\theta$ ⁠, the component along it is $m g \sin\theta$ ⁠, and the block holds as long as $\tan\theta = \dfrac{\sin\theta}{\cos\theta} \le \mu_s$ ⁠.
Feedback for a right answer
$\mu_s = \tan 30^\circ \approx 0.577$ ⁠. Surface grip moves in steps of $0.05$ ⁠, so its smallest value that holds is $0.6$ ⁠: try it, and $0.55$ ⁠.

Hint
The block holds as long as $\tan\theta \le \mu_s$ ⁠.

Worked solution
1. The block holds as long as the component down the board does not pass the limit of static friction: $m g \sin\theta \le \mu_s\, m g \cos\theta$ ⁠.
2. Divide by $m g \cos\theta$ ⁠: $\tan\theta \le \mu_s$ ⁠.
3. At $30^\circ$ ⁠: $\mu_s \ge \tan 30^\circ \approx 0.577$ ⁠.
If you got about $1.73$ ⁠, you turned the ratio upside down: you divided the component at right angles by the component along the board. Check a level board: there no grip is needed at all, and $\tan 0 = 0$ ⁠.
Step 2 of 6
Open this step
A heavier block
Practice · Predict, then look
Question
At $21^\circ$ ⁠ on wood, $\mu_s = 0.4$ ⁠, the 1 kg block holds. You will swap it for a 2 kg block of the same wood. What will happen?

Correct answer
- Correct The heavy block will hold too
Setup, choices and feedback
The Scene at this step
Slope angle
21 °fixed
Surface grip
0.4 μₛfixed
Labels
offdefault
Show the forces
On
Choices
- Correct The heavy block will hold too
- Wrong The heavy block will slide, because its weight pulls down the slope harder
  Misconception: Thinks a heavier body falls faster, or slips sooner on a slope, because its weight is bigger.
  Feedback from the catalogue
  A heavy body has a bigger weight, but also a bigger mass, and how hard it is to accelerate grows in the same proportion. Falling without air, $a = \dfrac{m g}{m} = g$ ⁠ for every body. On a slope both the component pulling down and the largest friction grow with the mass, so a heavy block slips at the same angle and with the same acceleration as a light block of the same material.
- Wrong The heavy block will hold, and now on a much steeper slope too
  Feedback
  The largest friction does double, but so does the pull down the slope. The ratio between them, and so the angle at which the block slips, does not change.
Once the prediction is locked in, the Scene changes
Other settings
Mass/nodes/ramp/props/mass = 2
The Scene then shows: 7.031 N · Friction on the cart or block: static while it holds, kinetic once it moves (flat, ramp)/nodes/ramp/outputs/frictionForce
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The block holds. Every arrow has doubled: the pull down the slope, the normal force and the static friction, from 3.5 N to 7.0 N. The limit, $\mu_s N$ ⁠, has doubled too, so the angle at which the block slips has not moved.

Worked solution
1. The block slides when $m g \sin\theta > \mu_s\, m g \cos\theta$ ⁠, that is when $\tan\theta > \mu_s$ ⁠. The mass is on both sides and cancels.
2. $\tan 21^\circ \approx 0.384$ ⁠, less than $0.4$ ⁠, so the block holds, whatever its mass.
3. At 2 kg: the pull down the slope is $19.62 \sin 21^\circ \approx 7.0\ \mathrm{N}$ ⁠, and the limit $0.4 \cdot 19.62 \cos 21^\circ \approx 7.3\ \mathrm{N}$ ⁠. The static friction, 7.0 N, is still under the limit.
If you chose that it slides, you looked only at the pull down the slope. A heavy block also presses harder into the board, and that raises the friction the board can exert in the same proportion.
Step 3 of 6
Open this step
Kinetic friction
Practice · Calculate
Question
The block, 1 kg, slides down a wooden board tilted at $35^\circ$ ⁠. The coefficient of kinetic friction is $\mu_k = 0.3$ ⁠. How big is the friction force on it?

Correct answer
2.411 N ± 2.5%
Setup, choices and feedback
The Scene at this step
Slope angle
35 °fixed
Surface grip
0.4 μₛfixed
Labels
offdefault
Show the forces
Off
Accepted within
± 2.5%
Computed from the Scene at this step’s setting
Friction on the cart or block: static while it holds, kinetic once it moves (flat, ramp)/nodes/ramp/outputs/frictionForce
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 3.214 N ± 2.5%
  Misconception: Uses the coefficient of static friction for a block that is already sliding.
  Feedback from the catalogue
  A block that is already sliding feels kinetic friction, $f_k = \mu_k N$ ⁠, with the kinetic coefficient. $\mu_s$ ⁠ only decides how long the block holds before it starts to slide. In the Scene $\mu_k$ ⁠ is three quarters of $\mu_s$ ⁠: for wood, $\mu_k = 0.3$ ⁠, not $0.4$ ⁠.
- 5.627 N ± 2.5%
  Misconception: Thinks friction on a sliding block still equals the pull down the slope.
  Feedback from the catalogue
  Friction equals the pull down the slope only while the block stands still, because then the net force is zero. A block that slides and speeds up feels kinetic friction, $f_k = \mu_k N$ ⁠, which is smaller than the pull: the difference between them is the net force that accelerates it.
Feedback for a right answer
$f_k = 0.3 \cdot 9.81 \cdot \cos 35^\circ \approx 2.41\ \mathrm{N}$ ⁠, up the board, against the sliding.

Hint
$f_k = \mu_k N$ ⁠, and $N = m g \cos\theta$ ⁠.

Worked solution
1. Given: $m = 1\ \mathrm{kg}$ ⁠, $\theta = 35^\circ$ ⁠, $\mu_k = 0.3$ ⁠.
2. The normal force: $N = m g \cos\theta = 9.81 \cdot 0.819 \approx 8.04\ \mathrm{N}$ ⁠.
3. Kinetic friction: $f_k = \mu_k N = 0.3 \cdot 8.04 \approx 2.41\ \mathrm{N}$ ⁠.
If you got about 3.2 N, you used $\mu_s = 0.4$ ⁠, but the block is already sliding. If you got about 5.6 N, you set the friction equal to the component pulling down the slope, as for a block at rest; a block that slides and speeds up gets less friction than that.
Step 4 of 6
Open this step
Down the slope
Practice · Calculate
Question
The same block slides down the board at $35^\circ$ ⁠, with $\mu_k = 0.3$ ⁠. What is its acceleration?

Correct answer
3.216 m/s² ± 2.5%
Setup, choices and feedback
The Scene at this step
Slope angle
35 °fixed
Surface grip
0.4 μₛfixed
Labels
offdefault
Show the forces
Off
Accepted within
± 2.5%
Computed from the Scene at this step’s setting
Constant acceleration along the track (flat, ramp)/nodes/ramp/outputs/acceleration
Any unit of the same quantity is accepted; it is converted before checking.
Wrong answers it recognises
- 5.627 m/s² ± 2.5%
  Misconception: Puts one force into F = ma instead of the net force, leaving out the friction against it.
  Feedback from the catalogue
  In $F = m a$ ⁠, $F$ ⁠ is the net force: all the forces along the line added with their signs. When friction acts against the motion, you subtract it from the force that pushes. Without that you get the acceleration there would be with no friction.
- 2.413 m/s² ± 2.5%
  Misconception: Uses the coefficient of static friction for a block that is already sliding.
  Feedback from the catalogue
  A block that is already sliding feels kinetic friction, $f_k = \mu_k N$ ⁠, with the kinetic coefficient. $\mu_s$ ⁠ only decides how long the block holds before it starts to slide. In the Scene $\mu_k$ ⁠ is three quarters of $\mu_s$ ⁠: for wood, $\mu_k = 0.3$ ⁠, not $0.4$ ⁠.
Feedback for a right answer
$a = 9.81\,(\sin 35^\circ - 0.3 \cos 35^\circ) \approx 3.22\ \mathrm{m/s^2}$ ⁠. Without friction it would be about 5.6 m/s².

Hint
$a = g\,(\sin\theta - \mu_k \cos\theta)$ ⁠.

Worked solution
1. Along the board: the component down it, $m g \sin 35^\circ \approx 5.63\ \mathrm{N}$ ⁠, and kinetic friction up it, about 2.41 N.
2. The net force: $5.63 - 2.41 = 3.22\ \mathrm{N}$ ⁠ down the slope.
3. The second law: $a = \dfrac{3.22}{1} \approx 3.22\ \mathrm{m/s^2}$ ⁠. As a formula: $a = g\,(\sin\theta - \mu_k \cos\theta)$ ⁠.
If you got about 5.6 m/s², you forgot the friction: that is the acceleration on a smooth slope. If you got about 2.4 m/s², you used the static coefficient.
Step 5 of 6
Open this step
Choosing an acceleration
Practice · Reach the target
Question
On a wooden board, $\mu_k = 0.3$ ⁠. Change Slope angle so that the block slides down with an acceleration of about 2 m/s², and press Check.

Goal
- 2 m/s² ± 0.05 m/s²
  At the start of the step: 0 m/s²
Setup, choices and feedback
The Scene at this step
Slope angle
20 °
Surface grip
0.4 μₛfixed
Labels
offdefault
Show the forces
On
Read from the Scene at
Constant acceleration along the track (flat, ramp)/nodes/ramp/outputs/acceleration
One setting that reaches it
Slope angle
28 °
Labels
off
Feedback for a right answer
At $28^\circ$ ⁠: $a = 9.81\,(\sin 28^\circ - 0.3 \cos 28^\circ) \approx 2.0\ \mathrm{m/s^2}$ ⁠.

Feedback for a wrong answer
Not yet. If the block does not move, raise the angle; if it speeds up too fast, lower it.

Hint
At $35^\circ$ ⁠ the acceleration was about 3.2 m/s². You need a gentler slope, but not too gentle: below $22^\circ$ ⁠ the block does not slide at all.

Worked solution
1. The acceleration down the slope: $a = g\,(\sin\theta - \mu_k \cos\theta)$ ⁠, with $\mu_k = 0.3$ ⁠.
2. Try angles: at $25^\circ$ ⁠, $9.81\,(0.423 - 0.3 \cdot 0.906) \approx 1.48\ \mathrm{m/s^2}$ ⁠; at $30^\circ$ ⁠, $9.81\,(0.5 - 0.3 \cdot 0.866) \approx 2.36\ \mathrm{m/s^2}$ ⁠. The answer lies between them.
3. At $28^\circ$ ⁠: $9.81\,(0.469 - 0.3 \cdot 0.883) \approx 2.01\ \mathrm{m/s^2}$ ⁠.
The block also has to start sliding: below $22^\circ$ ⁠, where $\tan\theta \le 0.4$ ⁠, static friction holds it and the acceleration is zero.
Step 6 of 6
Open this step
In summary
No practice question in this step.
The Scene at this step
Slope angle
20 °default
Surface grip
0.4 μₛdefault
Labels
offdefault
Show the forces
Offdefault

Across the unit · 22 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

The first law: force and motion

A steady push

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

When the fan switches off

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Forces that balance

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Motion at a steady speed

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

In summary

The second law: force, mass and acceleration

Force and acceleration

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Twice the mass

Question

Correct answer

The Scene at this step

Once the prediction is locked in, the Scene changes

Hint

Worked solution

The newton

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Several forces together

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Speed after three seconds

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

In summary

The third law, weight and the normal force

Forces come in pairs

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Why the pair does not cancel

Question

Correct answer

The Scene at this step

Choices