2.4 Newton's First Law

Picture stories

Press Next (or Play) to walk through each story one small step at a time. The numbers come last.

1. A hockey puck on smooth ice

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A car stalls on a flat road. You and a friend get out and push. Getting it rolling is hard. Once it rolls at a slow walking speed, you still have to keep pushing, or it slowly stops. So it seems that moving things need a push.

But a hockey puck, hit once, slides across the whole rink with nobody pushing it. It barely slows down at all.

The question this topic answers: does motion need a force? What happens to an object when the forces on it balance?

2. Pushing a stalled car

3. A ball in a wagon

Quick check. Lena and Hugo push their stalled car along at a steady walking speed. How does their push compare with the backward drag on the car?

4. Kai on a braking bus

5. Stopping a bike and a truck

6. Mia in a steady elevator

Quick check. Mia's elevator now moves down at a steady speed. How does the cable's pull compare with the weight of the elevator and Mia?

7. Check yourself

Think of your answer first, then tap to see it.

a) Far from any star, a spaceship fires its engines, then shuts them off. What does it do next?

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It keeps moving in a straight line at a constant speed. No net force is needed to keep moving.

b) An elevator moves up at a steady 3 m/s. Compare the cable's pull with the elevator's weight.

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They are equal. Steady speed in a straight line means zero net force, so up balances down.

c) Ana pushes a crate with 150 N and it slides at a steady speed. How big is the friction on it?

Show answer

Steady speed means balanced forces, so friction = 150 N, pointing backward.

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Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.

1. A puck slides across frictionless ice at 5 m/s. What net force is needed to keep it moving at 5 m/s?

2. Omar pushes a 50 kg crate across the floor with a 120 N horizontal force. It slides at a constant velocity. How big is the friction force on the crate?

3. A bus brakes suddenly and a standing passenger tips forward. Why?

The idea in plain words

Newton's first law: if the net force on an object is zero, its velocity does not change. An object at rest stays at rest. A moving object keeps moving at the same speed in the same direction (a straight line).

ΣF = 0  ⇔  a = 0  ⇔  v is constant

So a force is not needed to keep something moving. A force is needed to change its motion: to speed it up, slow it down, or turn it. The car slows down because friction and air push backward on it. The puck keeps going because the ice pushes back almost not at all.

Inertia is this "keep doing what you are doing" behavior. Mass measures inertia: a loaded truck is harder to start and harder to stop than a bicycle. Inertia is not a force; never draw it on a free body diagram.

Equilibrium (translational equilibrium) means the net force is zero. An object in equilibrium is either at rest or moving at constant velocity. "Constant velocity" does not mean "no forces": it means the forces balance, like the pushed car at steady speed (push forward = friction backward).

Trap: saying a ball at the top of its flight is in equilibrium because it is at rest for an instant. Instead: equilibrium means zero net force; at the top gravity still acts, so a = 9.8 m/s² down. Zero velocity is not zero acceleration.

Inertial reference frames. The first law works when you watch from a frame that is not accelerating (the ground, or a train moving at constant velocity). In a car that brakes, a loose water bottle slides forward although nothing pushes it forward: the car's frame is accelerating, so it is not an inertial frame. From the ground, the bottle simply keeps going while the car slows.

Worked numbers: pushing the stalled car

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The car has mass 1200 kg. On this road, rolling friction and air together push back on it with 300 N whenever it moves. Forward is positive.

  1. You want the car to roll at a steady 1.0 m/s. That means a = 0. Why: constant velocity means no acceleration.
  2. First law: a = 0 means ΣF = 0, so Fpush − 300 N = 0 and Fpush = 300 N. Why: the push only has to cancel the backward forces, not "carry" the car.
  3. Vertically: weight 1200 × 9.8 = 11 760 N down, and the road pushes up 11 760 N. They balance, so no vertical motion.
  4. If you push with 400 N instead: ΣF = 400 − 300 = 100 N forward, so the car speeds up (a = 100 / 1200 = 0.083 m/s², topic 2.5).
  5. If you stop pushing: ΣF = −300 N, so the car slows down. It does not stop because "the force ran out"; it stops because friction acts backward. Why: on frictionless ice (the puck) the net force would be zero and it would roll forever.

Lab: puck on ice

Two hockey sticks push the puck: one to the right, one to the left. At the time you choose, both sticks let go. The ice is frictionless, so after that the net force is zero. Positive is to the right. Try equal forces too: what happens then?

Examples

Example 1: the gliding puck basic

A puck slides on frictionless ice at 10 m/s to the right. How far does it go in 3.0 s, and what horizontal force keeps it moving?

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  1. No horizontal force acts, so the velocity stays 10 m/s (first law).
  2. Distance = v t = 10 × 3.0 = 30 m.
  3. No force keeps it moving. Zero force is exactly what lets the speed stay the same.
Trap: thinking a forward force is needed to keep the puck moving. Instead: with zero net force the velocity stays constant; a force is needed only to change the motion.

Example 2: highway cruising medium

A car cruises at a steady 25 m/s on a straight, level highway. The road pushes it forward (through the tires) with 500 N. How big is the air drag plus rolling friction?

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  1. Steady speed in a straight line means a = 0, so the net force is zero.
  2. So the backward forces total 500 N. The car is in equilibrium while moving fast.

Example 3: elevator at constant speed medium

A 60 kg person rides an elevator going up at a constant 2.0 m/s. What force does the floor exert on the person?

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  1. Constant velocity, so ΣF = 0, even though the elevator is moving.
  2. FN = mg = 60 × 9.8 = 588 N up. Same as standing on the ground. Why: the speed and the direction of motion do not matter; only acceleration would change the normal force (topic 2.5).

Example 4: a sign on two ropes AP

A 20 kg sign hangs at rest from two ropes. Each rope makes 30° with the horizontal, one to the left and one to the right. (a) Find the tension in each rope. (b) The ropes are changed to make only 10° with the horizontal. Find the new tension, and explain why it changed.

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  1. At rest: equilibrium. By symmetry both tensions are equal, T.
  2. Horizontal: T cos 30° to the left and T cos 30° to the right cancel. Good.
  3. Vertical: 2 T sin 30° = mg = 20 × 9.8 = 196 N → T = 196 / (2 × 0.50) = 196 N.
  4. (b) 2 T sin 10° = 196 N → T = 196 / (2 × 0.174) = 564 N.
  5. Flatter ropes have a smaller upward part (sin 10° is small), so they must pull much harder to hold up the same weight. That is why a perfectly horizontal rope can never hold up a weight.

Practice (AP style)

1. A puck slides at a constant 5 m/s across frictionless ice. Which statement is correct?

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First law: with zero net force, velocity does not change. The stick's force ended when contact ended, and inertia is not a force.

2. A car moves at a constant 20 m/s along a straight, level road. What is the net force on the car?

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Constant velocity means a = 0, so ΣF = 0. The forward push from the road and the backward drag balance.

3. A ball on a string is swung in a horizontal circle on a frictionless table. The string breaks. What path does the ball follow?

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With no horizontal force, the ball keeps its velocity at that instant: a straight line along the tangent to the circle.

4. An elevator moves downward at constant speed. Compared with the elevator's weight, the tension in its cable is

Show answer

Constant velocity, so equilibrium: tension up = weight down. The direction of motion does not matter.

5. A bus brakes suddenly and the standing passengers lurch forward. Which explanation is correct, as seen from the ground?

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From the ground (an inertial frame) no forward force acts on the passengers; they keep their velocity while the floor slows down. The braking bus is not an inertial frame, which is why it looks like a mysterious force from inside.

6. Two forces act on an object: 30 N to the east and 40 N to the north. What single extra force would put it in equilibrium?

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The two forces add to √(30² + 40²) = 50 N pointing north of east. The extra force must cancel it: 50 N in the exact opposite direction, which is south of west (about 53° south of west).

7. Which object is in translational equilibrium?

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Only the skydiver has constant velocity (air drag up equals weight down). The ball at the top is momentarily at rest but gravity still accelerates it; the car's direction changes; the stone speeds up.

8. (Short free response: argument) While pushing the stalled car at a steady speed, a student says: "I must be pushing harder than friction, otherwise the car would not be moving forward." Is the student right? Explain using Newton's first law, and say what would happen if the push really were larger than friction.

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The student is wrong. Steady speed in a straight line means zero acceleration, so by the first law the net force is zero: the push equals friction. Motion does not need an unbalanced force. If the push were larger than friction there would be a forward net force, and the car would speed up instead of moving at a steady speed.

9. (Short free response: experimental design) You have a cart, a long track whose tilt you can adjust, and a motion detector that records velocity against time. Friction in the wheels is small but not zero. Describe how to show that with zero net force the cart keeps a constant velocity. What graph shows success?

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1. Give the cart a gentle push on the level track and record v-t; it slowly slows down because of friction. 2. Raise one end very slightly, push again and repeat until the v-t graph is a flat horizontal line: now the small part of gravity along the track cancels friction, so the net force is zero. 3. Repeat with different starting pushes (different speeds), and in both directions only where the tilt helps (downhill).

Success: for every push, the velocity-time graph is a horizontal line (constant velocity) at the speed the push gave it. The size of the starting speed does not matter, only that the net force is zero.

10. (Short free response: mathematical routines) A 15 kg traffic light hangs at rest from two cables. Cable 1 is horizontal. Cable 2 pulls up and away at 50° above the horizontal. (a) Calculate the tension in cable 2. (b) Calculate the tension in cable 1. (c) Explain why the tension in cable 2 is larger than the light's weight.

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  1. Weight: mg = 15 × 9.8 = 147 N. At rest, so ΣF = 0 in both directions.
  2. (a) Vertical: only cable 2 has an upward part. T2 sin 50° = 147 N, so T2 = 147 / 0.766 = 192 N.
  3. (b) Horizontal: T1 = T2 cos 50° = 192 × 0.643 = 123 N.
  4. (c) Only part of cable 2's pull is upward, and that part alone must equal 147 N. The rest of its pull is sideways, balanced by cable 1. So the whole pull must be bigger than 147 N.

Point guide (4 points):

  • 1 point: equilibrium (net force zero) in both directions.
  • 1 point: T2 ≈ 192 N.
  • 1 point: T1 ≈ 123 N.
  • 1 point: explanation using the vertical component.

11. (Short free response: translation between representations) A 60 kg person rides an elevator. Its velocity-time graph (up positive) is: a straight line from 0 to 3.0 m/s between t = 0 and 2 s; flat at 3.0 m/s from 2 s to 8 s; a straight line back to 0 between 8 s and 10 s. (a) In which time intervals is the person in equilibrium? (b) For each interval, pick the matching free body diagram: (A) FN arrow longer than the weight arrow; (B) equal arrows; (C) FN arrow shorter. (c) Calculate FN during 2 s to 8 s.

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  1. (a) Only 2 s to 8 s (constant velocity, a = 0). Before 0 s and after 10 s at rest also counts.
  2. (b) 0 to 2 s: speeding up going up, a points up: (A). 2 to 8 s: a = 0: (B). 8 to 10 s: slowing down going up, a points down: (C).
  3. (c) a = 0, so FN = mg = 60 × 9.8 = 588 N, even though the elevator moves at 3.0 m/s.

Point guide (4 points):

  • 1 point: equilibrium only while the slope of the v-t graph is zero.
  • 1 point: (A) for 0 to 2 s and (C) for 8 to 10 s.
  • 1 point: (B) for 2 to 8 s.
  • 1 point: FN = 588 N.

12. (Short free response: qualitative/quantitative translation) An 80 kg skydiver falls at a steady (terminal) speed of 55 m/s. Then she opens her parachute. A while later she falls at a new steady speed of 5.0 m/s. (a) Without numbers, compare the air drag force with her weight at 55 m/s, just after the parachute opens, and at 5.0 m/s. (b) Calculate the drag force at each steady speed. (c) A student says "drag must be smaller at 5.0 m/s because she is slower." Use your answers to explain what is wrong.

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  1. (a) At 55 m/s: drag = weight (constant velocity, net force zero). Just after opening: drag is much bigger than weight, so the net force is up and she slows down. At 5.0 m/s: drag = weight again.
  2. (b) Both steady speeds: drag = mg = 80 × 9.8 = 784 N up.
  3. (c) At any steady speed the first law requires drag = weight, so drag is 784 N at both speeds. The parachute makes the same 784 N at a much lower speed; slower motion does not mean less force.

Point guide (4 points):

  • 1 point: drag = weight at both steady speeds.
  • 1 point: drag greater than weight while slowing.
  • 1 point: 784 N at both speeds.
  • 1 point: uses the first law (constant velocity means zero net force) to refute the claim.

Common mistakes

The mistake: "A moving object must have a force pushing it forward."

Why it is wrong: forces change velocity; they are not needed to keep it. Zero net force means constant velocity.

How to spot it: if you drew a forward arrow on a gliding object, name the object exerting it. Usually you cannot.

The mistake: "Constant velocity means no forces act."

Why it is wrong: it means the forces cancel. A car at steady speed has big forward and backward forces.

How to spot it: write ΣF = 0, then list all the forces that add to zero.

The mistake: "At the top of its flight the ball is at rest, so it is in equilibrium."

Why it is wrong: its velocity is zero for an instant, but gravity still acts, so its velocity keeps changing.

How to spot it: equilibrium is about acceleration (zero), not velocity.

The mistake: drawing "inertia" or "force of motion" as an arrow.

Why it is wrong: inertia is a property of the object (measured by mass), not a push from another object.

How to spot it: every arrow needs an "X on object" label with a real X.

The mistake: using the first law from inside an accelerating car, bus or elevator.

Why it is wrong: the first law holds only in inertial (non accelerating) frames. Objects seem to move "by themselves" in an accelerating frame.

How to spot it: if something seems to accelerate with no force, switch to the ground's point of view.

More explanations: free resources.