Newton's first law — inertia, operational definition of force: Worked examples

Before anything, recall the one tool we use everywhere:

The scenario matrix

Every problem this topic can throw is one of these cells. The table lists the class, what changes, and which example nails it.

Cell	What is happening to $v$	Net force?	Example
A. True rest	$v = 0$ , stays $0$	$0$	Ex 1
B. Constant velocity, forces cancel	$v$ constant, $\neq = 0$	$0$ (but many forces)	Ex 2
C. Speeding up (same direction)	magnitude of $v$ grows	$\neq = 0$ , along motion	Ex 3
D. Slowing down (same direction)	magnitude of $v$ shrinks	$\neq = 0$ , against motion	Ex 3
E. Direction change, constant speed	$v$ turns	$\neq = 0$ , sideways (centripetal)	Ex 4
F. Non-inertial frame illusion	looks like accel, no real force	$0$ really	Ex 5
G. Degenerate: zero time / zero interval	no time to change	$0$ net effect	Ex 6
H. Real-world word problem	mixed	depends	Ex 7
I. Exam twist: partial cancellation	one axis balanced, other not	$\neq = 0$ on one axis	Ex 8

Cells C and D are two signs of the same "along-the-line" case, so Ex 3 does both.

Worked examples

Worked example 1 — Cell A: the block that just sits there

A steel block rests on a level floor. You do nothing. Describe the net force and predict its motion in the next second.

Forecast: Guess before reading — is the net force zero because nothing touches it, or is there something going on?

List the real forces. Gravity pulls down with weight $W$ ; the floor pushes up with a normal force $N$ .
- Why this step? "Nothing is happening" is a trap — two forces are present. The first law cares only about their sum, so we must list them before summing.
Sum vertically. The block does not sink into the floor and does not fly up, so $N = W$ and the vertical net is $N - W = 0$ .
- Why this step? Observed motion (staying put) tells us the net must be zero; the first law is being read backwards — from " $v$ constant" to " $F_{net} = 0$ ".
Conclude. $F_{net} = 0 \Rightarrow v$ stays $0$ . It sits.

Verify: Rest is " $v = constant$ " with the constant being $0$ — fully consistent. Units: $N$ and $W$ are both forces (newtons), so their difference is a force; setting it to $0 N$ is dimensionally fine. See Mass vs weight: weight is a force, not the block's inertia.

Worked example 2 — Cell B: the plane that cruises (many forces, zero net)

A jet flies straight and level at a steady $250 m/s$ . Thrust $= 200000 N$ forward, and it is in equilibrium. What is drag? What is the net force?

Forecast: Engines are roaring — surely the net force is huge and forward?

Read the velocity. Straight and level, steady speed ⇒ $v$ is constant.
- Why this step? The first law reads velocity first. Constant $v$ forces the conclusion below no matter how loud the engines are.
Apply the law. $v$ constant $\Rightarrow F_{net} = 0$ .
- Why this step? This is the operational content: constant velocity is the symptom of zero net force.
Find drag. Horizontally, thrust and drag are the only forces, so $drag = thrust = 200000 N$ (backward).
- Why this step? Net $= 0$ on each axis. Horizontal net $= 200000 - drag = 0$ .
Vertically lift balances weight the same way — not asked, but it's why it stays level.

Verify: Net force $= (200000 - 200000) N = 0 N$ . Speed of $250 m/s$ never entered the force balance — correct, because the first law needs only that $v$ doesn't change, not its value.

Worked example 3 — Cells C & D: pushing and dragging a puck (both signs)

On frictionless ice a puck first moves at $+ 3 m/s$ , then $2 s$ later at $+ 9 m/s$ (Case C). In a second run it goes $+ 9 m/s$ then $2 s$ later $+ 3 m/s$ (Case D). In which cases did a net force act, and which way did it point?

Forecast: Two runs — guess the sign (direction) of the net force in each.

Case C, compute the velocity change. $Δ v = 9 - 3 = + 6 m/s$ (positive).
- Why this step? The law's trigger is any change in $v$ . The sign of $Δ v$ tells us the direction of the net force.
Case C verdict. $Δ v \neq = 0 \Rightarrow$ net force acted, pointing in $+$ direction (same as motion — that's why it sped up). Look at the blue arrow in the figure pointing with the puck.
Case D, compute. $Δ v = 3 - 9 = - 6 m/s$ (negative).
- Why this step? Same trigger, opposite sign. Negative $Δ v$ ⇒ net force points $-$ , against the motion (pink arrow).
Case D verdict. Net force acted, pointing backward (a braking force). Speed dropped.

Verify: $∣Δ v ∣ = 6 m/s \neq = 0$ in both runs — the first law says a force acted in both, which is correct (frictionless ice won't slow or speed a puck by itself). The sign flips ( $+ 6$ vs $- 6$ ): the tool distinguishes "speeding up" from "slowing down" purely by sign, exactly as promised. Zero friction here means the change can only come from an external push.

Worked example 4 — Cell E: constant speed, turning (direction counts)

A ball on a string swings in a horizontal circle of radius $R = 0.5 m$ at a constant speed $4 m/s$ . Its velocity at the "east" point is $(0, + 4) m/s$ (pointing north); a quarter-circle later at "north" it is $(- 4, 0) m/s$ (pointing west). Did a net force act?

Forecast: Speed never changed — so surely net force is zero?

Compare the two velocity vectors, not the speeds. Start $(0, + 4)$ , later $(- 4, 0)$ .
- Why this step? The first law demands the whole vector $v$ be constant, not just its length. This is the entire point of the "straight line" clause in the law.
Subtract them. $Δ v = (- 4, 0) - (0, + 4) = (- 4, - 4) m/s$ .
- Why this step? Same trigger as Ex 3, but now in 2-D. A nonzero $Δ v$ means velocity changed even though $∣ v ∣$ did not.
Verdict. $Δ v \neq = 0 \Rightarrow$ a net force acted — the inward string tension (centripetal). See the yellow arrow pointing toward the centre in the figure.

Verify: $∣ v ∣$ at start $= 0^{2} + 4^{2} = 4$ ; at end $= (- 4)^{2} + 0^{2} = 4$ — equal speeds, yet $∣Δ v ∣ = (- 4)^{2} + (- 4)^{2} = 32 \approx 5.66 m/s \neq = 0$ . Constant speed, changed velocity — the first law correctly demands a force. More in Uniform circular motion.

Worked example 5 — Cell F: the braking-bus illusion (non-inertial frame)

Standing in a bus at $10 m/s$ , the driver brakes and you lurch forward. A passenger claims "a force pushed me forward." From the ground (inertial) frame, is there a real forward force on you?

Forecast: You definitely felt thrown forward — real force or fake?

Switch to the honest frame. Watch from the ground, an inertial frame where the first law holds.
- Why this step? The first law is only guaranteed in an inertial frame. The bus, while braking, is accelerating — a non-inertial frame — so we can't trust "apparent" motion there.
Apply the law to your torso. No horizontal force pushes your upper body forward; by the first law it keeps moving at $10 m/s$ .
- Why this step? Inertia: constant $v$ needs no cause. Your torso simply continues.
Explain the "lurch." The bus (and your feet) slow to, say, $6 m/s$ ; your torso still at $10$ ⇒ relative to the bus it moves forward. No real forward force needed.

Verify: Relative velocity of torso w.r.t. bus $= 10 - 6 = + 4 m/s$ forward — matches the lurch, produced with zero real forward force. In the bus frame you'd invent a pseudo-force of size $ma$ forward; the first law is precisely the flag that this frame is non-inertial.

Worked example 6 — Cell G: the degenerate "zero interval" (tablecloth)

A cloth is yanked from under a plate. Friction from the cloth acts on the plate with force $F = 2 N$ for only $Δ t = 0.05 s$ ; the plate's mass is $0.4 kg$ . How much does the plate's velocity change? Effectively, does the plate move?

Forecast: A force does act — so shouldn't the plate slide off?

Impulse changes velocity. Change in velocity $Δ v = \frac{F Δ t}{m}$ .
- Why this step? When a force acts for a tiny time, the velocity change is the force $\times$ time divided by mass. As $Δ t \to 0$ , $Δ v \to 0$ : the "no time to change" degenerate case the matrix demands.
Plug in. $Δ v = \frac{2 \times 0.05}{0.4} = 0.25 m/s$ .
- Why this step? We turn "tiny" into a number to prove it's negligible.
Interpret. A quarter of a metre per second, gained over $0.05 s$ , moves the plate about $\frac{1}{2} \cdot 0.25 \cdot 0.05 \approx 0.006 m$ — six millimetres. It stays.

Verify: $Δ v = 0.25 m/s$ ; distance during the pull $\approx \frac{1}{2} Δ v Δ t = 0.00625 m$ . As $Δ t \to 0$ the plate's motion $\to 0$ : the limiting behaviour the first law promises — with essentially no interval, inertia wins and nothing changes. See Friction.

Worked example 7 — Cell H: real-world word problem (crate in a truck)

A truck carries a crate ( $m = 50 kg$ ) on a rough bed. The truck accelerates from $0$ to $12 m/s$ in $6 s$ on a straight road. For the crate to move with the truck (not slide back), what forward force must friction supply, and is the crate's frame inertial while accelerating?

Forecast: Guess — does the crate need a force, and where does it come from?

Find the crate's required velocity change. It must go $0 \to 12 m/s$ , so $Δ v = 12 m/s$ over $6 s$ , i.e. acceleration $a = 12/6 = 2 m/s^{2}$ .
- Why this step? The crate's $v$ changes ⇒ the first law says a net force must act on it. We quantify how much change is required.
Name the force. The only forward horizontal agent is static friction from the truck bed, so friction $= ma = 50 \times 2 = 100 N$ forward.
- Why this step? The first law names the culprit (something must change $v$ ); the value comes from the change needed. Without enough friction the crate would keep its old velocity and slide backward relative to the truck — pure inertia.
Frame question. The crate rides an accelerating truck — that co-moving frame is non-inertial; in the road frame (inertial) the first law and this force analysis hold.

Verify: $a = 2 m/s^{2}$ , friction $= 100 N$ . Sanity: if the bed were frictionless (Cell A of Ex 1 applied horizontally), friction $= 0$ , the crate keeps $v = 0$ , and the truck slides out from under it — exactly the tablecloth logic scaled up. Consistent.

Worked example 8 — Cell I: exam twist (one axis balanced, one not)

A box slides down a frictionless ramp. Along the surface (call it $x$ ) nothing balances the downhill pull; perpendicular to the surface ( $y$ ) the normal force balances the perpendicular part of gravity. Is $F_{net} = 0$ ? Does the box keep constant velocity?

Forecast: One direction is balanced — is that "enough" for the first law?

Check each axis separately. Perpendicular ( $y$ ): normal force $=$ perpendicular gravity ⇒ net $y = 0$ . Along-ramp ( $x$ ): the downhill component of gravity has nothing to cancel it ⇒ net $x \neq = 0$ .
- Why this step? The first law needs the whole vector $F_{net}$ to vanish. A vector is zero only if every component is zero. Balancing one axis is not enough.
Combine. Net $y = 0$ but net $x \neq = 0 \Rightarrow F_{net} \neq = 0$ .
- Why this step? This is the classic exam trap: "the normal force balances gravity, so net force is zero." It balances only the perpendicular part.
Verdict. $F_{net} \neq = 0 \Rightarrow v$ is not constant — the box speeds up down the slope.

Verify: If $F_{net}$ were truly $0$ , the box would glide at constant velocity; experience says it accelerates downhill, so the net must be nonzero. Only the perpendicular axis is balanced. The first law's demand — all components zero — is exactly what catches this.

Recall Quick self-check on the whole matrix

Every cell reduces to one question ::: "Did $v$ (magnitude or direction) change?" — if yes, a net force acted. Why can "many forces" still mean zero net force? ::: Because the first law reads only the sum; forces that cancel (Ex 2) leave $v$ constant. Why is balancing one axis not enough? ::: $F_{net} = 0$ requires every component zero; Ex 8's ramp balances $y$ but not $x$ . Why does the braking-bus "force" feel real but isn't? ::: The bus is a non-inertial frame; from the ground your torso simply keeps its velocity (inertia), no real forward force.

Connections

Newton's second law — F=ma — turns the "a force acted" verdict into a number (used in Ex 3, 6, 7).
Newton's third law — action-reaction — the friction on the crate (Ex 7) has a reaction on the truck.
Inertial vs non-inertial frames — Ex 5 and Ex 7 hinge on which frame is honest.
Friction — the hidden force in Ex 3 (its absence), Ex 6, Ex 7.
Uniform circular motion — the force revealed in Ex 4.
Pseudo-forces — what you'd invent inside the braking bus (Ex 5).
Mass vs weight — weight as a force (Ex 1, Ex 8) vs mass as inertia.

Worked examples — Newton's first law — inertia, operational definition of force

The scenario matrix

Worked examples

Connections

Case Map