Newton's laws and forces AP Physics 1 · Lesson 1 of 7
The start of Unit 2. You will learn to draw free-body diagrams the way the AP exam wants, use Newton's three laws, find the net force, work with friction, explain why your weight seems to change in an elevator, and use Hooke's law for springs.
On this page
- In this lesson
- Forces and free-body diagrams
- Newton's three laws of motion
- Net force with horizontal forces
- Introduction to friction
- Vertical forces and elevators (apparent weight)
- Spring forces and Hooke's law
- Lab: motion detectors and force probes
- Key equations
- Practice problems
- Free-response practice
- Common mistakes
- FAQs
- Keep going
In this lesson
This lesson covers net force with horizontal forces, an introduction to friction, vertical forces and elevators, spring forces and Hooke's law, and the motion detector and force probe lab. Work through the topics in order:
- Forces and free-body diagrams
- Newton's three laws
- Net force (horizontal)
- Introduction to friction
- Vertical forces and elevators
- Spring forces and Hooke's law
- Lab: motion detectors and force probes
All numbers use g = 9.8 m/s², as on the AP Physics 1 equation sheet. Every formula used here is on that sheet: AP Physics 1 equation sheet.
Forces and free-body diagrams
A force is a push or a pull that one object exerts on another. Every force has a size (in newtons, N), a direction, and two objects: the one exerting it and the one it acts on. Contact forces need touching (normal force, friction, tension, spring force). Gravity acts at a distance.
| Force | Symbol | Direction | What causes it |
|---|---|---|---|
| Gravitational force (weight) | Fg = mg | Straight down, toward Earth's centre | Earth's gravity |
| Normal force | FN | Perpendicular to the surface, pushing away from it | A surface resisting being pressed into |
| Tension | FT | Along the string or rope, pulling away from the object | A taut string, rope or cable |
| Friction | Ff | Parallel to the surface, opposing sliding (or the tendency to slide) | Rough surfaces in contact |
| Spring force | Fs | Back toward the spring's natural length | A stretched or compressed spring |
| Applied force | Fapp | Whichever way the push or pull acts | A hand, a motor, a push |
A free-body diagram (FBD) shows every force acting on one object, and nothing else. On the AP exam, draw the object as a dot, start every arrow on the dot, point it in the direction of the force, and make the lengths show relative sizes. Label each arrow with a clear symbol. Do not draw components or the net force on the FBD itself; those go in a separate sketch if you need them.
How to draw a free-body diagram
- Choose the object (the system). Draw it as a dot.
- Gravity first: one arrow straight down, Fg = mg.
- Find every contact: each surface gives a normal force (and maybe friction); each string gives a tension; each spring gives a spring force.
- Check lengths: if the object has no acceleration in a direction, the arrows in that direction must balance.
- Never add a force for motion itself. There is no “force of motion” or “force of velocity”.
Newton's three laws of motion
Newton's first law: an object's velocity stays constant (including staying at rest) unless a net external force acts on it. This is inertia. A puck sliding on frictionless ice keeps going at the same speed in the same direction, with no force pushing it forward.
Newton's second law: the acceleration of a system is the net external force divided by its mass:
a = ΣF / m or ΣF = ma
Acceleration is in the same direction as the net force. It is a vector law, so you apply it separately in each direction: ΣFx = max and ΣFy = may.
Newton's third law: if object A exerts a force on object B, then B exerts a force on A that is equal in size and opposite in direction. The two forces in a third-law pair act on different objects and are always the same type of force.
Common trap
A book resting on a table has FN = mg, but these two forces are not a third-law pair. They act on the same object (the book) and they are different types of force. The pair to the table's normal force on the book is the book's normal force on the table. The pair to Earth's gravity on the book is the book's gravity on Earth.
Worked example: a third-law pair
A 1,500 kg car and a 15,000 kg truck collide head-on. Compare the force the truck exerts on the car with the force the car exerts on the truck, and compare their accelerations.
- By Newton's third law, the forces are equal in size and opposite in direction, at every instant of the collision.
- Acceleration is F/m. The car's mass is one tenth of the truck's, so the car's acceleration is 10 times bigger.
- Equal forces, very different effects: this is why the people in the car are at more risk.
Net force with horizontal forces
When the forces are all horizontal (or all vertical), the net force is found by adding them with signs. Choose a positive direction, usually the direction of motion or acceleration, and write forces that way as positive and the other way as negative.
ΣFx = F1 + F2 + … = max
Method for any Newton's second law problem
- Draw the free-body diagram.
- Choose a positive direction (the way it accelerates, if you know it).
- Write ΣF = ma for that direction, with each force signed.
- Solve with symbols first, then put the numbers in.
Worked example 1: two forces
Two students pull a 25 kg crate across frictionless ice. One pulls east with 80 N, the other pulls west with 30 N. Find the crate's acceleration.
- Take east as positive. ΣF = 80 − 30 = 50 N.
- a = ΣF / m = 50 / 25 = 2.0 m/s² east.
Worked example 2: finding a force from the motion
A 1,200 kg car speeds up from 0 to 18 m/s in 6.0 s on a level road. Air resistance and rolling friction add up to 600 N backward. What forward force does the road exert on the car's tyres?
- Acceleration: a = Δv/Δt = 18 / 6.0 = 3.0 m/s².
- Forward positive: F − 600 = 1200 × 3.0 = 3600.
- F = 4,200 N forward. (This forward force is static friction from the road on the tyres.)
Worked example 3: constant velocity
A worker pushes a 40 kg box at a steady 1.5 m/s across a floor with a horizontal 120 N push. What is the friction force?
- Constant velocity means a = 0, so the net force is zero (Newton's first law).
- Friction balances the push: Ff = 120 N backward. The speed (1.5 m/s) is not needed.
Introduction to friction
Friction is the force a surface exerts parallel to itself. It opposes the sliding of the two surfaces over each other. When an object slides, kinetic friction acts opposite to its velocity relative to the surface, and its size is:
Ff = μkFN
The coefficient of kinetic friction μk has no units and depends on the two surfaces (rubber on dry concrete is about 0.7; steel on ice is about 0.02). On a level floor with only vertical forces besides friction, FN = mg, so Ff = μkmg.
When an object is not sliding, static friction can act. It is whatever size it needs to be, up to a maximum, to stop sliding starting. You will meet static friction in detail in Lesson 2.
Worked example: sliding to a stop
A 2.0 kg book slides across a table at 3.0 m/s. The coefficient of kinetic friction is 0.30. How far does it slide before stopping?
- Vertical: FN = mg = 2.0 × 9.8 = 19.6 N. Friction: Ff = 0.30 × 19.6 = 5.88 N.
- Taking the direction of motion as positive, a = −5.88 / 2.0 = −2.94 m/s². (Notice a = −μkg: the mass cancels.)
- v² = v0² + 2aΔx gives 0 = 9.0 − 2(2.94)Δx, so Δx = 1.5 m.
Key idea
Friction does not depend on the area of contact or (in AP Physics 1) on speed. It depends on the surfaces (μ) and on how hard they are pressed together (FN).
Vertical forces and elevators (apparent weight)
For vertical motion, take up as positive and use ΣFy = may. A bathroom scale measures the normal force it exerts on you, so the scale reading is often called your apparent weight.
FN − mg = ma ⇒ FN = m(g + a)
The sign of a decides everything. The direction of the velocity does not.
| Elevator is… | Acceleration | Scale reads |
|---|---|---|
| At rest, or moving at constant speed (up or down) | 0 | mg (true weight) |
| Speeding up going up | Up (+) | More than mg |
| Slowing down going down (about to stop at a lower floor) | Up (+) | More than mg |
| Speeding up going down | Down (−) | Less than mg |
| Slowing down going up (about to stop at a higher floor) | Down (−) | Less than mg |
| Cable snaps (free fall, a = −g) | −g | Zero: “weightless” |
Worked example 1: scale reading
A 65 kg student stands on a scale in an elevator that accelerates downward at 1.8 m/s². What does the scale read?
- Up positive, so a = −1.8 m/s².
- FN = m(g + a) = 65(9.8 − 1.8) = 65 × 8.0 = 520 N, less than the true weight of 637 N.
Worked example 2: cable tension
An 800 kg elevator car rises and slows down at 1.2 m/s² as it nears the top floor. What is the tension in the cable?
- The car is moving up but slowing, so its acceleration is down: a = −1.2 m/s².
- ΣF = T − mg = ma gives T = m(g + a) = 800(9.8 − 1.2) = 6,880 N.
Worked example 3: working backward
A 50 kg student's scale reads 600 N in an elevator. Find the elevator's acceleration.
- a = FN/m − g = 600/50 − 9.8 = 12 − 9.8 = 2.2 m/s² upward.
- You cannot tell whether the elevator is going up and speeding up, or going down and slowing down. Both have an upward acceleration.
Spring forces and Hooke's law
An ideal spring pulls back when stretched and pushes back when compressed. The size of the spring force is proportional to how far the spring is stretched or compressed from its natural length. This is Hooke's law:
Fs = kΔx (as a vector: Fs = −kΔx)
k is the spring constant in N/m: the force needed per metre of stretch. A stiff spring has a large k. The minus sign in the vector version means the force always points back toward the natural length: it is a restoring force.
On a graph of spring force against stretch, the gradient is k. On a graph of stretch against force, the gradient is 1/k. Watch which way the axes go.
Worked example 1: hanging mass
A 0.40 kg mass hangs at rest from a spring with k = 50 N/m. How far is the spring stretched?
- At rest, the net force is zero: kΔx = mg.
- Δx = mg/k = (0.40)(9.8)/50 = 0.078 m (7.8 cm).
Worked example 2: spring pulling a cart
A spring with k = 80 N/m, stretched by 0.15 m, pulls horizontally on a 3.0 kg cart on a level, frictionless track. What is the cart's acceleration at that moment?
- Fs = kΔx = 80 × 0.15 = 12 N.
- a = F/m = 12 / 3.0 = 4.0 m/s², toward the spring's natural length.
Key idea
Hooke's law only works while the spring is elastic. Stretch it too far and it stays permanently deformed, and the graph curves away from a straight line. AP questions say “ideal spring” when Hooke's law applies.
Lab: motion detectors and force probes
The classic lab for Newton's second law uses a cart on a level track, a force probe (which measures the pull on the cart) and a motion detector (which measures the cart's velocity many times a second, so its acceleration is the gradient of the velocity-time graph).
How the experiment works
- Keep the cart's mass fixed. Pull it with the force probe, or attach a string over a pulley to a small hanging mass.
- Record the force and find the acceleration from the gradient of the v-t graph from the motion detector.
- Repeat for at least five different forces.
- Plot acceleration (vertical) against net force (horizontal). A straight line through the origin shows a ∝ Fnet.
- The gradient is 1/m. Repeat with different masses at a fixed force to show a ∝ 1/m.
On the AP exam, the Experimental Design and Analysis (LAB) question often asks you to say which quantities to graph so that the result is a straight line, and how to get a physical quantity from the gradient. For ΣF = ma: plot a against Fnet and the gradient is 1/m, or plot Fnet against a and the gradient is m.
Worked example: reading the gradient
In the graph above, the best-fit line passes through (2.5 N, 2.08 m/s²). Find the cart's mass.
- Gradient = 2.08 / 2.5 = 0.832 kg−1.
- m = 1 / gradient = 1 / 0.832 = 1.2 kg.
Lab tip
If friction is not negligible, the line does not pass through the origin: the cart needs some force before it starts to accelerate. The horizontal intercept is about the size of the friction force.
Key equations
| Newton's second law | a = ΣF/m |
|---|---|
| Weight (gravitational force near Earth) | Fg = mg, g = 9.8 m/s² |
| Kinetic friction | Ff = μkFN |
| Scale reading in an elevator (up positive) | FN = m(g + a) |
| Hooke's law | Fs = kΔx |
| Kinematics (for stopping distances) | v² = v0² + 2aΔx, v = v0 + at |
Practice problems
17 problems, from warm-up to exam level. Answer each one before you look: use the hint if you are stuck, then reveal the worked solution one step at a time. Number answers are marked within 2.5%, so g = 10 m/s² still counts.
Hint
A third-law pair is the same type of force, acting on the other object.
Worked solution
- The force is: Earth pulls the book, by gravity.
- Swap the objects and keep the type: the book pulls Earth, by gravity.
- The normal force is a different type of force, and the table's normal force acts on the same object (the book). The answer is B.
Hint
Net force first, then divide by the mass.
Worked solution
- Right positive: ΣF = 30 − 18 = 12 N.
- a = ΣF/m = 12 / 4.0 = 3.00 m/s² to the right.
Hint
Newton's first law: what does constant velocity tell you about the net force?
Worked solution
- Constant velocity means zero acceleration.
- Zero acceleration means zero net force. No force is needed to keep it moving. The answer is B.
- Inertia is a property of the puck, not a force.
Hint
Friction is μkFN, and FN = mg on a level floor with a horizontal push.
Worked solution
- FN = mg = 8.0 × 9.8 = 78.4 N.
- Ff = 0.25 × 78.4 = 19.6 N.
- a = (40 − 19.6) / 8.0 = 2.55 m/s².
Hint
Up positive: FN − mg = ma, with a negative.
Worked solution
- a = −2.0 m/s² (down).
- FN = m(g + a) = 50(9.8 − 2.0) = 390 N.
Hint
Which way does the acceleration point when something moving up slows down?
Worked solution
- Moving up and slowing down means the acceleration is downward.
- FN = m(g + a) with a negative, so FN < mg. The answer is B.
Hint
Rearrange FN = m(g + a) for a.
Worked solution
- a = FN/m − g.
- a = 800/70 − 9.8 = 11.4 − 9.8 = 1.63 m/s², directed upward.
Hint
At rest, the spring force equals the weight.
Worked solution
- kΔx = mg.
- k = mg/Δx = (0.50 × 9.8)/0.14 = 35.0 N/m.
Hint
Rearrange Fs = kΔx.
Worked solution
- Δx = F/k = 30/200 = 0.15 m.
Hint
What does the gradient of F against Δx represent?
Worked solution
- The gradient of Fs against Δx is k.
- Steeper line, larger k: X is stiffer and stretches less for the same force. The answer is A.
Hint
Up positive: T − mg = ma.
Worked solution
- T = m(g + a).
- T = 1200(9.8 + 0.80) = 1200 × 10.6 = 12720 N.
Hint
These two forces are a Newton's third law pair.
Worked solution
- The car pushes the truck and the truck pushes the car: a third-law pair.
- Third-law forces are always equal in size, whatever the masses or speeds. The answer is C.
- The accelerations are different: the car's is larger because its mass is smaller.
Hint
Find the spring force, then use Newton's second law.
Worked solution
- Fs = kΔx = 120 × 0.050 = 6.0 N.
- a = 6.0/1.5 = 4.00 m/s².
Hint
a = F/m, so the gradient of a against F is 1/m.
Worked solution
- Gradient = 1/m = 2.5 kg−1.
- m = 1/2.5 = 0.40 kg. The answer is A.
Hint
Find the acceleration from the speed and time first.
Worked solution
- a = Δv/Δt = 3.0/2.0 = 1.5 m/s² upward.
- FN = m(g + a) = 55(9.8 + 1.5) = 622 N.
Hint
Constant speed in a straight line means zero acceleration.
Worked solution
- Zero acceleration means zero net force.
- So T − mg = 0: tension equals the weight, even though the bucket is moving down. The answer is C.
Hint
The deceleration is μkg, then use a kinematics equation.
Worked solution
- Ff = μkmg, so a = −μkg = −0.20 × 9.8 = −1.96 m/s². The mass cancels.
- 0 = 4.0² + 2(−1.96)Δx, so Δx = 16/3.92 = 4.08 m.
Free-response practice
Original questions in the style of the AP exam's free-response section. Write a full answer on paper, with words as well as equations, then open the scoring guide and tick each point you earned.
Experimental Design and Analysis (LAB) style · 8 points · about 15 minutes
A student wants to find the spring constant k of a spring. The spring hangs vertically from a clamp. The student hangs different masses from it, waits for each to come to rest, and measures how far the spring has stretched from its natural length. The results are below.
| Hanging mass (kg) | 0.050 | 0.100 | 0.150 | 0.200 | 0.250 |
|---|---|---|---|---|---|
| Stretch Δx (m) | 0.021 | 0.039 | 0.061 | 0.081 | 0.099 |
- (a) Explain why the spring force equals the weight of the hanging mass when the mass is at rest.
- (b) Say which quantities the student should plot on each axis to get a straight line whose gradient lets k be found, and how k is found from the gradient.
- (c) Use the data to find a value for k.
- (d) The student says a 0.40 kg mass would stretch the spring by about 0.16 m. State one assumption this relies on, and how the student could check it.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
The mass is at rest, so its acceleration is zero and the net force on it is zero. The only forces on it are the upward spring force and the downward weight, so they are equal in size: kΔx = mg.
Part (b)
Plot the weight mg (spring force) on the vertical axis and the stretch Δx on the horizontal axis. Since mg = kΔx, this is a straight line through the origin with gradient k. (Plotting Δx against mg also works; then k = 1/gradient.)
Part (c)
Weights: 0.49, 0.98, 1.47, 1.96, 2.45 N. A best-fit line through the origin passes close to (0.100 m, 2.45 N), so k ≈ 2.45/0.100 ≈ 25 N/m. (Answers from 23 to 26 N/m earn the point.)
Part (d)
It assumes the spring still obeys Hooke's law (stays elastic, with a straight-line graph) beyond the largest mass tested. To check, hang 0.30, 0.35 and 0.40 kg masses and see whether the points stay on the same straight line, then remove the masses and check the spring returns to its original length.
Your mark: 0 of 8
Mathematical Routines (MR) style · 7 points · about 12 minutes
A 52 kg student stands on a bathroom scale in an elevator. The graph shows the elevator's vertical velocity, with up as positive, from the moment it starts moving.
- (a) Draw a free-body diagram for the student between t = 0 and t = 2 s. Draw arrows to show relative sizes.
- (b) Calculate the scale reading during each of the three stages of the trip.
- (c) The student says: “The scale reads less than my weight at the end because the elevator is moving slower.” Explain what is wrong with this reasoning.
Write your full answer on paper first, then open the scoring guide and tick each point you earned.
Scoring guide and model answer
Part (a)
A dot with two arrows: the normal force from the scale, upward, and the gravitational force, downward. The upward arrow must be longer, because the student accelerates upward.
Part (b)
0 to 2 s: a = 3.0/2 = 1.5 m/s², FN = 52(9.8 + 1.5) = 588 N. 2 to 8 s: a = 0, FN = 52 × 9.8 = 510 N. 8 to 11 s: a = −3.0/3 = −1.0 m/s², FN = 52(9.8 − 1.0) = 458 N.
Part (c)
The scale reading depends on the acceleration, not the speed. From 8 to 11 s the elevator is still moving up but its velocity is decreasing, so its acceleration is downward. A downward acceleration needs a net downward force, so FN must be less than mg. A slow elevator moving at constant speed would read exactly mg.
Your mark: 0 of 7
Common mistakes
- Drawing the normal force and weight as a Newton's third law pair. They act on the same object, so they cannot be.
- Adding a “force of motion” in the direction an object moves. Moving objects need no force to keep moving; only changes in velocity need a net force.
- Using the direction of the velocity, not the acceleration, to decide whether a scale reads more or less than mg.
- Writing FN = mg automatically. It is only true when nothing else pushes or pulls vertically and there is no vertical acceleration.
- Mixing up the gradient of a force-stretch graph (k) and a stretch-force graph (1/k).
- Forgetting to convert grams to kilograms or centimetres to metres before using Hooke's law.
- Drawing components or the net force on the free-body diagram. On the AP exam, the FBD shows only the forces themselves.
Frequently asked questions
Is weight the same as mass?
No. Mass (kg) measures how much matter an object has and how hard it is to accelerate. Weight is the gravitational force on it, Fg = mg, measured in newtons. A 60 kg person has a weight of about 590 N on Earth and about 97 N on the Moon, but the same mass in both places.
If forces in a Newton's third law pair are equal and opposite, why don't they cancel?
They act on different objects. Forces can only cancel when they act on the same object. When a horse pulls a cart, the cart pulls back on the horse equally, but the cart accelerates because of the forces on the cart alone.
Does a moving object need a force to keep moving?
No. With zero net force, an object keeps the same velocity (Newton's first law). In everyday life things slow down because friction and air resistance act on them, which is what makes it seem as if a force is needed.
What exactly does a bathroom scale measure?
The normal force between you and the scale. It equals your weight only when your vertical acceleration is zero. That is why it reads more as an elevator starts up and less as it starts down.
Why do I feel weightless in free fall?
In free fall, a = −g, so FN = m(g + a) = 0. Nothing pushes up on you. Gravity is still acting (that is why you fall), but there is no normal force for you to feel.
Do I have to draw free-body diagrams as dots on the AP exam?
The AP exam asks you to represent the object as a dot, with each force drawn as an arrow starting on the dot. A box is usually accepted, but the dot is the safest choice. Label every force and make the lengths show relative sizes when the question asks.
Is g = 9.8 or 10 m/s² on the AP exam?
The equation sheet gives g = 9.8 m/s². Using 10 m/s² is generally accepted on the AP exam to keep the arithmetic simple. Our answer checker accepts either.
Does a spring's force depend on the mass hanging from it?
Not directly. The spring force depends only on k and Δx. A heavier mass stretches the spring further, and that larger stretch is what makes the spring force bigger.
Is Hooke's law on the AP Physics 1 equation sheet?
Yes, as Fs = −kΔx. The minus sign shows the spring force is opposite to the displacement from the natural length. In calculations, use its size, Fs = kΔx, and decide the direction from the diagram.
Keep going
Related guides and practice:
- Newton's laws of motion explained
- How to draw free-body diagrams
- Hooke's law guide
- AP Physics 1 forces summary
- AP Physics 1 forces practice problems
All lessons in the AP Physics 1 forces and energy course · AP Physics 1 past papers