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Test 2 — Forces & Circular Motion

Chapters 4–5 · opens Jun 19, due Jun 20.

Opens Jun 19

Plan: master the Test 2 tab (essentials → worked examples → traps), then drill Practice. Your deadlines, grades, and tasks live in Profile.

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TestChaptersOpensDue
Test 1 doneCh 2–3 · KinematicsJun 11
Test 2Ch 4–5 · Forces & CircularJun 19Jun 20
Test 3Ch 6–7 · Energy & MomentumJun 27Jun 28
Test 4Ch 8–9Jul 5Jul 6
Test 5Ch 10–11Jul 13Jul 14
Test 6Ch 12–13
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Test 2 — Chapter 4 (Forces) + Chapter 5 (Circular Motion)

This is your test this week. Work top to bottom: read the essentials, then solve each example yourself before revealing.

Game plan
  • Now → Jun 18: read both essentials boxes; solve Ch 4 examples 1–7, then 13–14; solve Ch 5 examples 8–12, then 15–17. Reveal only after trying.
  • Day before: redo the Practice tab cold (P1–P12). Re-check the "traps" list below.
  • Test day: for every problem write the equation first, plug in with units, box the answer. Draw a quick free-body diagram for force problems.
  • If stuck: name the type (forces? circular?), list what's given, pick the formula that has your knowns + one unknown.

Chapter 4 essentials — Newton's Laws

The one idea: a net force changes motion. No net force → no change in velocity (Newton's 1st law). A net force → acceleration in the same direction (Newton's 2nd law).

On your formula sheet (Ch 4):F_net = m·a F_net,x = m·a_x

In plain words: the total push/pull (F_net) equals mass times acceleration. The second line is the same rule for just the sideways direction. New to the letters? The Decoder tab defines every one.

Good to know: your sheet's Chapter 4 list is just the two above. f = μ·N is not printed on it (for circles it gives μ = v²/(rg) instead), and W = m·g isn't a standalone line though mg appears in Chapter 6. So know f = μN and W = mg from memory, or ask if they're allowed.

  • Net force = the single force you'd get by adding all forces (as vectors). Solve x and y separately.
  • Weight is a force: W = m·g, pointing down, with g = 9.8 m/s². Mass (kg) is not weight (N).
  • Normal force N is the surface pushing back. On flat ground with nothing vertical happening, N = m·g.
  • Friction opposes sliding: f = μ·N. Bigger normal force → more friction.

Chapter 4 worked examples

1A 1200 kg car speeds up at 2.5 m/s². What net force acts on it?
1 · What kind of problemNewton's 2nd law — given mass and acceleration, find net force.
2 · Why this equationF_net = m·a. It directly links the two things you're given (m, a) to the one you want (force).
3 · Step by step
F_net = m · a F_net = (1200 kg)(2.5 m/s²) F_net = 3000 kg·m/s²
4 · AnswerF_net = 3000 N  (1 N = 1 kg·m/s²)
5 · Common mistakeForgetting that newtons are kg·m/s². If your units don't collapse to N, you set it up wrong.
6 · TakeawayNet force and acceleration always point the same way; F = ma is the bridge between them.
2A 50 N net force pushes a 10 kg cart. What is its acceleration?
1 · What kind of problemNewton's 2nd law, solving for acceleration this time.
2 · Why this equationF_net = m·a rearranged to a = F_net / m — you have force and mass, want acceleration.
3 · Step by step
a = F_net / m a = (50 N) / (10 kg) a = 5 N/kg = 5 m/s²
4 · Answera = 5 m/s²
5 · Common mistakeMultiplying instead of dividing. If you're solving for a, mass goes under the force.
6 · TakeawayN/kg and m/s² are the same unit — a quick way to check you divided correctly.
3What is the weight of an 8.0 kg backpack?
1 · What kind of problemWeight from mass — a force calculation, not just "the mass."
2 · Why this equationW = m·g. Weight is the pull of gravity, which is mass times g. (Your sheet uses mg in Chapter 6, e.g. PE = mgh.)
3 · Step by step
W = m · g W = (8.0 kg)(9.8 m/s²) W = 78.4 kg·m/s²
4 · AnswerW = 78.4 N (downward)
5 · Common mistakeReporting 8.0 as the weight. 8.0 kg is the mass; weight is a force in newtons.
6 · Takeaway"How heavy" (newtons) ≠ "how much stuff" (kilograms). Multiply by g to switch from mass to weight.
4A 20 kg box on a flat floor (μ = 0.30) is pulled horizontally with 80 N. Find its acceleration.
1 · What kind of problemNewton's 2nd law with friction. Two horizontal forces: your pull and friction fighting back.
2 · Why this equationFirst friction f = μ·N (not printed on your sheet — see Formulas), with N = m·g on flat ground. Then net force, then a = F_net/m.
3 · Step by step
N = m·g = (20)(9.8) = 196 N f = μ·N = (0.30)(196) = 58.8 N F_net = pull − friction = 80 − 58.8 = 21.2 N a = F_net / m = 21.2 / 20 = 1.06 m/s²
4 · Answera = 1.06 m/s² (in the pull direction)
5 · Common mistakeUsing the 80 N pull as the net force. Friction (58.8 N) subtracts first — only what's left over accelerates the box.
6 · TakeawayOn flat ground: N = mgf = μmg. Net force is "push minus friction," then divide by mass.
5A 6.0 kg crate feels 30 N to the right and 12 N to the left. Find its acceleration.
1 · What kind of problemNet force from forces in opposite directions, then Newton's 2nd law.
2 · Why this equationAdd forces as signed numbers (right +, left −) to get F_net, then a = F_net/m.
3 · Step by step
F_net = (+30) + (−12) = 18 N to the right a = F_net / m = 18 / 6.0 = 3.0 m/s²
4 · Answera = 3.0 m/s² to the right
5 · Common mistakeAdding 30 + 12 = 42. Opposite directions subtract; pick a positive direction and keep signs.
6 · Takeaway"Net" means after the tug-of-war. Direction of the net force is the direction of acceleration.
6A 5.0 kg lamp hangs at rest from a cord. What is the tension in the cord?
1 · What kind of problemEquilibrium (at rest → a = 0). The cord's tension balances gravity.
2 · Why this equationAt rest, F_net = 0, so up = down: T = W = m·g.
3 · Step by step
F_net = 0 → T − W = 0 → T = m·g T = (5.0 kg)(9.8 m/s²) = 49 N
4 · AnswerT = 49 N
5 · Common mistakeAdding an extra force. "At rest" means the forces already cancel — tension just equals the weight.
6 · TakeawayIf something isn't accelerating, every direction balances. Up-forces = down-forces.
7A 60 kg person stands in an elevator accelerating upward at 2.0 m/s². What does the floor push up with (their apparent weight)?
1 · What kind of problemNewton's 2nd law in the vertical direction — the floor's normal force vs. gravity, with a real acceleration.
2 · Why this equationUp is positive. Net force = N − W = m·a, so N = m·g + m·a = m(g + a).
3 · Step by step
N − m·g = m·a N = m·g + m·a = m(g + a) N = (60)(9.8 + 2.0) = (60)(11.8) = 708 N
4 · AnswerN = 708 N (more than their 588 N weight — they feel heavier)
5 · Common mistakeThinking gravity "increased." It didn't — accelerating up just means the floor must push harder than gravity.
6 · TakeawayAccelerating up → feel heavier (N > mg). Accelerating down → feel lighter (N < mg).
8You push on a wall with 50 N. How hard does the wall push back on you?
1 · What kind of problemNewton's 3rd law — action and reaction.
2 · Why this equation3rd law: every force comes in an equal-and-opposite pair, acting on the two different objects.
3 · Reasoning
You push wall: 50 N on the wall. Wall pushes you: 50 N back on you (opposite direction). Equal sizes, opposite directions, different objects.
4 · Answer50 N, back toward you
5 · Common mistakeThinking the pair cancels. They don't — they act on different objects (you and the wall), so neither cancels the other.
6 · TakeawayForces always come in pairs: A on B equals B on A, opposite directions, different objects.
9Two blocks (2.0 kg and 3.0 kg) sit touching on a frictionless table. A 10 N force pushes on the 2.0 kg block, which then pushes the 3.0 kg block. Find the acceleration and the force the 2.0 kg block exerts on the 3.0 kg block.
1 · What kind of problemConnected objects — treat the system together, then look inside.
2 · Why this equationWhole system: a = F/(m_total). Then for the 3.0 kg block alone, only the contact push acts: F_contact = m·a.
3 · Step by step
System: a = F / (m1+m2) = 10 / (2.0+3.0) = 10/5.0 = 2.0 m/s² On the 3.0 kg block: F_contact = m·a = (3.0)(2.0) = 6.0 N
4 · Answera = 2.0 m/s²; the 2.0 kg block pushes the 3.0 kg block with 6.0 N
5 · Common mistakeUsing the full 10 N on the back block. Only part of the push gets transmitted — exactly enough to give that block the same acceleration.
6 · TakeawayFor connected objects: find the shared acceleration from the whole system, then isolate one piece.

Chapter 5 essentials — Circular Motion

The one idea: moving in a circle at steady speed still counts as accelerating, because the direction of velocity keeps changing. That acceleration points toward the center (centripetal = "center-seeking").

Circular-motion relations (all verified on your sheet):a_c = v² / r (centripetal acceleration, toward center) F_c = m·a_c = m·v² / r (net inward force) v = 2·π·r / T (speed for one full circle) T = period (s), f = 1/T (frequency, Hz)

In plain words: the inward acceleration (a_c) is speed² ÷ radius; the inward force (F_c) is that times mass; speed is once-around-the-circle ÷ time-per-loop. Every symbol is defined in the Decoder tab.

Confirm these on your Ch 5 sheet. They are the standard circular-motion formulas, but verify the exact forms your sheet allows before the test.

  • Centripetal force is not a new force. It's the name for the net inward force, supplied by something real — a string's tension, friction, gravity, a wall.
  • There is no outward "centrifugal" force on the object. You feel flung out, but the real force on you points inward.
  • r is the radius (center to edge), not the diameter.

Chapter 5 worked examples

1A car rounds a curve of radius 45 m at 15 m/s. What is its centripetal acceleration?
1 · What kind of problemPure circular motion — find centripetal acceleration from speed and radius.
2 · Why this equationa_c = v²/r. It uses exactly what you're given (v and r) and gives the inward acceleration.
3 · Step by step
a_c = v² / r a_c = (15 m/s)² / (45 m) a_c = 225 / 45 a_c = 5.0 m/s² (toward the center)
4 · Answera_c = 5.0 m/s² (toward center)
5 · Common mistakeForgetting to square v. The speed is squared — that's why going faster around a curve feels so much stronger.
6 · TakeawaySteady speed in a circle still accelerates — always pointed at the center.
2A 1000 kg car goes 20 m/s around a curve of radius 80 m. What centripetal force is needed?
1 · What kind of problemCircular motion force — how much inward force keeps the car on the curve.
2 · Why this equationF_c = m·v²/r — it's just F = m·a with a = a_c = v²/r.
3 · Step by step
F_c = m · v² / r F_c = (1000)(20)² / (80) F_c = (1000)(400) / 80 F_c = 400000 / 80 F_c = 5000 N (toward the center)
4 · AnswerF_c = 5000 N (inward)
5 · Common mistakeCalling F_c an extra force on the free-body diagram. It's the net inward force — here it's supplied by friction between tires and road.
6 · Takeaway"Centripetal force needed" = m v²/r. Then ask: what real force provides it?
3A 0.50 kg ball on a 1.2 m string is whirled in a circle. The string can pull with 24 N before it needs to. How fast is the ball moving when the inward force is 24 N?
1 · What kind of problemCircular motion, solving for speed. The tension is the centripetal force.
2 · Why this equationStart from F_c = m·v²/r and rearrange for v: v = √(F_c·r / m).
3 · Step by step
F_c = m·v²/r → v² = F_c · r / m v² = (24 N)(1.2 m) / (0.50 kg) v² = 28.8 / 0.50 = 57.6 v = √57.6 = 7.59 m/s
4 · Answerv ≈ 7.6 m/s
5 · Common mistakeForgetting the square root at the end. You solved for v², so take the root.
6 · TakeawayWhen a string whirls a ball, tension = the centripetal force. Set them equal and solve for what you need.
4A stone goes around a circle of radius 2.0 m, completing one loop every 4.0 s. Find its speed and centripetal acceleration.
1 · What kind of problemCircular motion using the period (time per loop) instead of a given speed.
2 · Why this equationSpeed = distance/time, and one loop is the circumference 2πr: v = 2πr/T. Then a_c = v²/r.
3 · Step by step
v = 2·π·r / T = 2π(2.0) / 4.0 v = 12.566 / 4.0 = 3.14 m/s a_c = v² / r = (3.14)² / 2.0 a_c = 9.87 / 2.0 = 4.93 m/s²
4 · Answerv ≈ 3.14 m/s, a_c ≈ 4.93 m/s²
5 · Common mistakeUsing r as the distance per loop. One loop is the whole circumference, 2πr, not r.
6 · TakeawayIf you're given a period, find speed first with 2πr/T, then everything else follows.
5A car turns on a flat road, radius 50 m, where friction has μ = 0.80. What is the fastest it can go without sliding? Ch 4 + Ch 5 combo
1 · What kind of problemA combo: friction (Ch 4) provides the centripetal force (Ch 5). Classic Test-2 question.
2 · Why this equationThe most friction available is f = μ·N = μ·m·g. At max speed that equals the needed centripetal force m·v²/r; mass cancels → v = √(μ·g·r). (This is your sheet's μ = v²/(rg) solved for v.)
3 · Step by step
μ·m·g = m·v² / r (mass cancels) v² = μ · g · r v² = (0.80)(9.8)(50) = 392 v = √392 = 19.8 m/s
4 · Answerv_max ≈ 19.8 m/s
5 · Common mistakeThinking a heavier car can go faster. Mass cancels — the max safe speed doesn't depend on the car's mass, only on μ, g, and r.
6 · TakeawayOn a flat curve, friction is the centripetal force. v_max = √(μgr). Bigger μ or radius → faster you can corner.
6A wheel of radius 0.50 m spins at 60 rpm (revolutions per minute). Find the speed of a point on the rim and its centripetal acceleration.
1 · What kind of problemCircular motion where the rate is given in rpm — convert to a period first.
2 · Why this equation60 rpm = 60 loops per 60 s = 1 loop per second, so T = 1.0 s. Then v = 2πr/T and a_c = v²/r.
3 · Step by step
T = 60 s / 60 rev = 1.0 s per rev v = 2·π·r / T = 2π(0.50)/1.0 = 3.14 m/s a_c = v²/r = (3.14)²/0.50 = 9.87/0.50 = 19.7 m/s²
4 · Answerv ≈ 3.14 m/s, a_c ≈ 19.7 m/s²
5 · Common mistakePlugging "60" straight into a formula. rpm must become a period (seconds per loop) first.
6 · Takeawayrpm → period: T = 60 / (rpm), in seconds. Then it's a normal circular-motion problem.
7A 2.0 kg mass moves at 10 m/s and needs 200 N of centripetal force. What is the radius of its circle?
1 · What kind of problemCircular motion, solving for radius.
2 · Why this equationRearrange F_c = m·v²/r for r: r = m·v²/F_c.
3 · Step by step
r = m·v² / F_c r = (2.0)(10)² / 200 r = (2.0)(100) / 200 = 200/200 = 1.0 m
4 · Answerr = 1.0 m
5 · Common mistakePutting F_c on top. Solving for r, the force goes underneath: r = m v²/F_c.
6 · TakeawayMore force at the same speed → tighter circle (smaller r). Less force → wider circle.
8Conceptual: For each, name the real force that provides the centripetal force — (a) a car turning a corner, (b) a ball on a string, (c) the Moon orbiting Earth.
1 · What kind of problemThe key Ch 5 idea: centripetal force is a job done by a real force.
2 · Why this mattersThere's no separate "centripetal force" to draw — you must identify what physically pulls/pushes toward the center.
3 · Reasoning
(a) car turning → FRICTION between tires and road (b) ball on string → TENSION in the string (c) Moon orbiting → GRAVITY from Earth
4 · Answer(a) friction, (b) tension, (c) gravity
5 · Common mistakeWriting "centripetal force" as the source. That's the role, not the source — always name the real force.
6 · TakeawayOn any circular problem, first ask: "what real force points toward the center?" Set that equal to m v²/r.

Test 2 traps to memorize

  • Net force first. Subtract friction (and other opposing forces) before dividing by mass.
  • Signs are directions. Choose what counts as positive; anything opposite that direction is negative. A negative answer is not automatically wrong — it often just means left, down, backward, or slowing.
  • Mass ≠ weight. kg is mass; multiply by g for weight in N.
  • Circular motion is accelerated motion even at constant speed — the direction changes.
  • No centrifugal force. The real force points inward; "being flung out" is just inertia.
  • Square the speed in v²/r. Doubling speed quadruples the needed force.
  • Use radius, not diameter, in every circular formula.
✓ Completed Jun 11

Test 1 — Chapter 2–3 (Kinematics)

This one's done — kept here for review, since Ch 4–5 build on it.

Essentials

The five motion equations (constant acceleration):v_avg = Δx / Δt a = (v_f − v_0) / t v_f = v_0 + a·t d = v_0·t + ½·a·t² v_f² = v_0² + 2·a·d d = ½·(v_0 + v_f)·t Projectiles (x and y are independent; a = −9.8 m/s² up-positive):x = v_0x·t y = v_0y·t + ½·a·t² v_0x = v_0·cos(θ) v_0y = v_0·sin(θ) speed = √(v_x² + v_y²) direction = tan⁻¹(v_y / v_x)

In plain words: these connect starting speed, final speed, acceleration, time, and distance — pick the one that has your three knowns plus the unknown. For projectiles, do sideways and up/down separately and share the time. The Formulas and Decoder tabs translate each one fully.

Pick a positive direction and keep signs. "Dropped/from rest" → v_0 = 0. "Comes to rest / highest point" → v_f = 0 (vertically).

Worked examples

1A sled starts from rest and accelerates at 3.0 m/s² for 5.0 s. What is its final speed?
1 · What kind of problemStraight-line constant acceleration; find final velocity from a and t.
2 · Why this equationv_f = v_0 + a·t — it has exactly v_0, a, t (knowns) and v_f (unknown), no distance needed.
3 · Step by step
v_f = v_0 + a·t v_f = 0 + (3.0)(5.0) v_f = 15 m/s
4 · Answerv_f = 15 m/s
5 · Common mistakeReaching for a distance equation. You weren't asked about distance — pick the equation missing the variable you don't have.
6 · Takeaway"From rest" means v_0 = 0. Choose the equation that contains your knowns and your one unknown.
2A ball is dropped and falls for 3.0 s. How far does it fall? (g = 9.8 m/s²)
1 · What kind of problemFree fall (a special constant-acceleration case), find distance from time.
2 · Why this equationd = v_0·t + ½·a·t² with v_0 = 0 and a = g.
3 · Step by step
d = v_0·t + ½·a·t² d = 0 + ½(9.8)(3.0)² d = ½(9.8)(9.0) = 44.1 m
4 · Answerd = 44.1 m
5 · Common mistakeSquaring only the number, not the time-with-units, or forgetting the ½. Both halves matter: ½ · a · t².
6 · TakeawayDropped object: v_0 = 0, a = 9.8 m/s². Distance grows with time squared.
3A ball is thrown straight up at 20 m/s. How high does it go?
1 · What kind of problemFree fall going up; find the height where it momentarily stops.
2 · Why this equationNo time given or wanted → use the no-time equation v_f² = v_0² + 2·a·d. At the top, v_f = 0.
3 · Step by step
v_f² = v_0² + 2·a·d (up +, a = −9.8) 0 = (20)² + 2(−9.8)d 0 = 400 − 19.6 d d = 400 / 19.6 = 20.4 m
4 · Answerd = 20.4 m
5 · Common mistakeThinking acceleration is zero at the top. Velocity is zero there, but gravity (a = −9.8) never stops.
6 · Takeaway"Highest point" → v_f = 0 vertically. No time in the problem → the no-time equation.
4A ball rolls off a 45 m cliff horizontally at 20 m/s. How far from the base does it land?
1 · What kind of problemProjectile, horizontal launch. Vertical fall sets the time; horizontal speed sets the distance.
2 · Why this equationx and y are independent. Find fall time from y = ½·g·t², then x = v_0x·t.
3 · Step by step
Vertical: 45 = ½(9.8)t² → t² = 90/9.8 = 9.18 t = √9.18 = 3.03 s Horizontal: x = v_0x · t = (20)(3.03) = 60.6 m
4 · Answerx ≈ 60.6 m
5 · Common mistakePutting the 20 m/s into the vertical equation. Horizontal speed never affects fall time — they're separate.
6 · TakeawayHorizontal launch: time comes from the drop, range = horizontal speed × that time.
5A ball is kicked at 25 m/s at 30° above the ground. Find its maximum height and range.
1 · What kind of problemProjectile launched at an angle — split into x and y components first.
2 · Why this equationComponents: v_0x = v_0 cos θ, v_0y = v_0 sin θ. Height from the no-time equation (v_y = 0 at top); time of flight from 2·v_0y/g; range = v_0x · t.
3 · Step by step
v_0x = 25·cos30° = 25(0.866) = 21.65 m/s v_0y = 25·sin30° = 25(0.500) = 12.5 m/s Max height: v_y² = v_0y² − 2g·H, v_y = 0 H = v_0y² / (2g) = (12.5)² / 19.6 = 156.25/19.6 = 7.97 m Time of flight: t = 2·v_0y / g = 25/9.8 = 2.55 s Range: x = v_0x · t = (21.65)(2.55) = 55.2 m
4 · AnswerMax height ≈ 7.97 m, Range ≈ 55.2 m
5 · Common mistakeUsing the full 25 m/s in the vertical equations. Only the vertical component (12.5 m/s) drives height and flight time.
6 · TakeawayAngle launch = two problems sharing one clock. Cos for across, sin for up.

Practice — fresh problems (Ch 2–3)

Same idea as the Test 2 drill: try it, tap Show solution, then New problem for a fresh version of the same type.

Final velocity

A runner starts at 3 m/s and accelerates at 2 m/s² for 6 s. Find final speed.

v_f = v_0 + at v_f = 3 + 2·6 = 15 m/s

Free fall

A dropped object falls for 4 s. How far does it fall?

d = 1/2 gt² d = 0.5·9.8·4² d = 78.4 m

Thrown up

A ball is thrown upward at 18 m/s. What max height?

H = v_0²/(2g) H = 18²/19.6 H = 16.5 m

Projectile range

A ball rolls off a 20 m cliff at 12 m/s horizontally. Find range.

t = √(2h/g) = √(40/9.8) = 2.02 s x = vt = 12·2.02 = 24.2 m
Test 2 practice

Practice — fresh problems anytime

Each card below is one problem type. Try it on paper, tap Show solution to check your work, then tap New problem for a brand-new version of the same type — different numbers, same idea. Drill a type until it feels automatic.

Chapter 4 — Forces

F = m·a

A 9 kg cart accelerates at 4 m/s². What net force is needed?

F_net = m·a F_net = (9)(4) F_net = 36 N

Friction first, then acceleration

A 12 kg box on a flat floor has μ = 0.25 and is pulled with 50 N. Find acceleration.

N = mg = 12·9.8 = 117.6 N f = μN = 0.25·117.6 = 29.4 N F_net = 50 - 29.4 = 20.6 N a = F_net/m = 20.6/12 = 1.72 m/s²

Opposite forces

A 6 kg crate has 38 N right and 14 N left. Find acceleration.

F_net = 38 - 14 = 24 N right a = F_net/m = 24/6 a = 4.0 m/s² right

Weight is a force

What is the weight of a 7 kg backpack?

W = mg = 7·9.8 W = 68.6 N downward

Chapter 5 — Circular Motion

a_c = v²/r

A car goes 16 m/s around a curve of radius 40 m. Find centripetal acceleration.

a_c = v²/r a_c = 16²/40 = 256/40 a_c = 6.4 m/s² inward

F_c = mv²/r

A 1000 kg car goes 18 m/s around a radius 60 m curve. What inward force is needed?

F_c = mv²/r F_c = 1000·18²/60 F_c = 5400 N inward

Speed from period

A ball makes one loop every 5 s on a radius 2 m circle. Find speed.

v = 2πr/T v = 2π·2/5 v = 2.51 m/s

Flat curve max speed

On a flat curve, r = 50 m and μ = 0.70. Find max speed before sliding.

v = √(μgr) v = √(0.70·9.8·50) v = 18.5 m/s
Write your own answer first, then check — that's where the learning happens. The New problem button never runs out, so keep drilling the types that feel shaky.

Formula sheet

Grouped by chapter, each with a plain-English translation. These match your actual formula sheet: Chapter 5 is fully verified against it, and Chapter 4 calls out the two relations (f = μN, W = mg) that aren't printed on the sheet so you're never surprised.

New to the symbols? Read the Decoder tab once — it explains every letter and unit. Each formula below also has a plain-English translation right under it.

Chapter 2–3 · Kinematics (Test 1)

v_avg = Δx / ΔtAverage speed = how far you moved ÷ how long it took.
a = (v_f − v_0) / tAcceleration = (ending speed − starting speed) ÷ time. In words: how much the speed changed each second.
v_f = v_0 + a·tFinal speed = starting speed + the speed you gained (acceleration × time).
d = v_0·t + ½·a·t²Distance = travel from your starting speed + the extra distance from speeding up.
v_f² = v_0² + 2·a·dThe "no-time" equation — use it when the problem never mentions time. Links the two speeds, acceleration, and distance.
d = ½·(v_0 + v_f)·tDistance = the average of the start and end speed, times the time.
g = 9.8 m/s²Gravity speeds up any falling object by 9.8 m/s every second.

Chapter 2–3 · Projectiles (2-D motion)

x = v_0x·tSideways distance = sideways speed × time. (Sideways speed never changes.)
y = v_0y·t + ½·a·t²Up/down distance, with gravity pulling — this is the falling part of the motion.
v_0x = v_0·cos(θ) v_0y = v_0·sin(θ)Split a launch into its sideways part (use cos) and its up part (use sin).
speed = √(v_x² + v_y²) direction = tan⁻¹(v_y / v_x)Recombine the sideways and up pieces into one overall speed and angle.

Chapter 4 · Forces (Test 2)

Your sheet's Chapter 4 list is exactly these two. Everything else in a forces problem is built from them.

F_net = m·aNet force = mass × acceleration. Push harder → it speeds up faster; more mass → it speeds up slower.
F_net,x = m·a_xThe same rule, applied to just the sideways (x) direction on its own.
Not printed on your sheet — but you'll likely still use them: f = μ·N — friction = grip-number × normal force. (Your sheet gives μ for circles as μ = v²/(rg), but not f = μN for flat surfaces — so for a flat-floor friction problem you'd need to know this one. Ask your instructor if it's allowed.) W = m·g — weight. (The sheet uses mg in Chapter 6, e.g. PE = mgh, but doesn't list "W = mg" by itself.)

Chapter 5 · Circular Motion (Test 2)

All of these are on your sheet (verified). The μ and tanθ ones are the friction/banked-curve relations.

a_c = v² / rCentripetal (inward) acceleration = speed² ÷ radius. Faster, or a tighter turn, makes it much stronger.
F_c = m·a_c = m·v² / rInward force needed = mass × speed² ÷ radius. (A real force — friction, tension, gravity — provides it.)
v = 2·π·r / TSpeed = the distance around the circle (2πr) ÷ the time for one full loop (T).
μ = v² / (r·g)Friction needed for a flat curve. Solve it for speed and you get v = √(μ·g·r) — the fastest you can corner before sliding.
tan θ = v² / (r·g)Banked curve: the angle θ a road should be tilted so a car at speed v can round a curve of radius r without relying on friction.
Start here if it looks like another language

Symbol Decoder — every letter, explained

Physics formulas are just shorthand. Each letter stands for a real, everyday thing. This page translates every symbol, unit, and word you will see on Test 1 and Test 2 into plain English — with a quick memory hook for each. Skim it once, then come back whenever a formula looks scary.

How to read any formula

  • Letters are stand-ins for numbers. "v" just means "whatever the speed is." You swap in the number from the problem.
  • Units tell you what kind of thing it is. Meters = a distance, seconds = a time, newtons = a force. If your units do not match, something is off.
  • Little letters below (subscripts) are labels, not multiplication. v_0 is "the starting speed," v_f is "the final speed." Same v, just tagged.
  • Read it as a sentence. F = m·a literally says "force equals mass times acceleration." That is all a formula is — a sentence in shorthand.

Notation key (the squiggles)

squared — a number times itself. means v × v. (4² = 16.)
square root — undoes squaring. √9 = 3 because 3² = 9.
v_0
subscript — the small "0" is a label meaning "at the start." Say "v-naught." Not a multiply.
Δ
delta — "the change in." Δv = final speed − starting speed.
approximately equals — "about." Used when we round.
θ
theta — a stand-in for an angle (like a launch or ramp angle), in degrees.
π
pi — a fixed number, about 3.14. Shows up with circles.
μ
mu — the friction number (how grippy two surfaces are). Just a number, no unit.

Motion symbols (Test 1, and used everywhere)

d, x, y
distance / position ("dee, ex, why") — how far, or where something is. x is usually sideways, y up-and-down. m (meters). Hook: a ruler measurement.
t
time ("tee") — how long something takes. s (seconds). Hook: a stopwatch.
v
velocity / speed ("vee") — how fast something moves (and which way). m/s (meters per second). Hook: a speedometer reading.
v_0
starting velocity ("v-naught") — the speed at the very beginning. m/s. Hook: the "0" = the start.
v_f
final velocity ("v-eff") — the speed at the end. m/s. Hook: "f" = finish.
a
acceleration ("ay") — how fast the speed is changing (speeding up, slowing, or turning). m/s². Hook: the gas pedal or the brake.
g
gravity's acceleration ("gee") — how fast falling things speed up on Earth, 9.8 m/s². m/s². Hook: Earth's built-in downward gas pedal.

Force symbols (Test 2 — Chapter 4)

F
force ("eff") — a push or a pull. N (newtons). Hook: how hard you shove something.
F_net
net force — all the pushes and pulls added together into one. N. Hook: the winner of the tug-of-war. This is what actually moves things.
m
mass ("em") — how much stuff something is made of. kg (kilograms). Hook: stays the same on the Moon.
W
weight — the downward force of gravity on a mass, W = m·g. N. Hook: what a scale pushes back with.
N
normal force ("en") — the surface pushing back, straight out of the surface. N. Hook: the floor holding you up. Not always equal to weight.
f
friction (lowercase "eff") — the force that resists sliding, f = μ·N. N. Hook: the grippy drag between two surfaces.
μ
coefficient of friction ("mu") — a number for how grippy the surfaces are (bigger = grippier). none (just a number). Hook: ice ≈ small, rubber ≈ big.
T
tension ("tee") — the pull carried through a rope or string. N. Hook: how hard the rope is pulling. (Same letter as period — see below.)

Circular-motion symbols (Test 2 — Chapter 5)

r
radius ("are") — distance from the center to the edge of the circle. m. Hook: half the width — not all the way across.
a_c
centripetal acceleration ("ay-sub-c") — the inward acceleration that keeps something turning, a_c = v²/r. m/s². Hook: centripetal = center-seeking.
F_c
centripetal force ("eff-sub-c") — the net inward force that bends the path into a circle, F_c = m·v²/r. N. Hook: the inward tug — supplied by a real force like tension or friction.
T
period ("tee") — the time for one full lap around the circle. s. Hook: one lap on the clock. (Same letter as tension — units tell you which.)
π
pi — about 3.14; appears in v = 2πr/T because circles. none.

Units, in plain words

m
meter — how far. About one big step.
s
second — how long. One "one-Mississippi."
m/s
meters per second — speed. 10 m/s is a fast run; 30 m/s is highway driving.
m/s²
meters per second, per second — how much the speed changes each second. 9.8 m/s² means a falling object gets 9.8 m/s faster every second.
kg
kilogram — how much stuff (mass). A liter of water ≈ 1 kg.
N
newton — a push or pull. About the weight of a small apple in your hand. (1 N = 1 kg·m/s².)

Words that trip everyone up

  • Mass vs weight: mass = how much stuff (kg, same everywhere). Weight = gravity pulling on it (N, less on the Moon).
  • Speed vs velocity: speed = how fast. Velocity = how fast and which direction.
  • Acceleration: any change in velocity — speeding up, slowing down, or turning.
  • Net force: what is left after all the forces fight it out. It is the only thing that changes motion.
  • Normal force: the surface pushing back, straight out of the surface. On flat ground it equals weight; on a ramp or in an elevator it does not.
  • Friction: the force that fights sliding. More normal force → more friction (f = μ·N).
  • Tension: the pull carried through a rope, string, or cable.
  • Centripetal force: not a new force — it is the name for the net inward force that keeps something circling (a real force like friction or tension does the job).
  • Period vs frequency: period = seconds per loop. Frequency = loops per second. They are flips of each other.
  • Equilibrium: forces balanced, net force = 0 → no acceleration (sitting still, or moving at steady speed).
  • Inertia: object's resistance to changing its motion. More mass = more inertia = harder to start or stop.

Phrase decoder — turn words into numbers

Test questions hide the numbers in phrases. Here is the translation:

  • "starts from rest" / "dropped" → v_0 = 0
  • "comes to rest" / "stops" → v_f = 0
  • "at the highest point" → vertical velocity = 0 (but gravity still pulls — acceleration is NOT zero)
  • "constant speed" / "constant velocity" → a = 0, and net force = 0
  • "smooth" / "frictionless" → ignore friction (f = 0)
  • "how far" → find distance d  ·  "how long" → find time t  ·  "how fast" → find velocity v

Same letter, different meaning

A few letters get reused. The units (and the topic) always tell you which one:

  • T = tension (a force, in N) and period (a time, in s). Force problem or circle problem?
  • a = plain acceleration; a_c = the special centripetal (inward) acceleration.
  • F = a force; F_net = all forces combined; F_c = the inward force in a circle.
  • v = speed now; v_0 = speed at the start; v_f = speed at the end.

Explain it like I'm new

Plain-language versions of the ideas that trip people up. No jargon, just intuition.

Why moving in a circle is "accelerating" even at steady speed

Acceleration doesn't only mean "speeding up." It means velocity is changing — and velocity includes direction. Going around a circle, your direction is constantly turning, so your velocity is constantly changing, so you're accelerating, even if the speedometer never moves.

Analogy: swinging a ball on a string. Your hand keeps tugging it inward the whole time. That inward tug is what bends its path into a circle. Stop tugging (string snaps) and it shoots off in a straight line — it was never being pushed outward.

That inward acceleration is a_c = v²/r, and the inward force causing it is F_c = m v²/r.

What "centripetal force" really is

It's not a new kind of force you add to a diagram. It's a job title. Whatever real force happens to point toward the center — the string's tension, friction on tires, gravity on a satellite — that force is the centripetal force for that situation.

So on a test: first ask "what real thing pushes/pulls toward the center here?" Then set that force equal to m v²/r.

And there's no outward "centrifugal" force on the object. The feeling of being flung outward in a turning car is just your body's inertia wanting to go straight while the car turns under you.

Mass vs. weight (kilograms vs. newtons)

Mass is how much stuff you're made of — measured in kilograms. It's the same on Earth, the Moon, or in space.

Weight is how hard gravity pulls on that stuff — a force, measured in newtons. It changes with gravity: W = m·g.

Analogy: a brick has the same mass everywhere, but it "weighs" less on the Moon because the Moon pulls more gently. On a test, if you're given kg and need a force, multiply by g; if given newtons and need mass, divide by g.

"Net force" — the tug-of-war idea

Objects don't respond to each force separately; they respond to the one combined force left after all the pushes and pulls fight it out. Push right with 30 N, friction pulls back 12 N → the object only "feels" 18 N to the right.

Recipe: pick a positive direction, add forces with signs, get the net force, then use a = F_net/m. Only the leftover force accelerates anything.

➕➖ Signs — when is something negative (and why)?

A sign isn't "less than zero" — it's a direction. First pick which way is positive and stick with it the whole problem. Standard: up = +, down = − and right = +, left = −. The way something moves is usually +, which makes friction and "slowing down" negative.

Quantity+ means− means / why
Velocitymoving the + waymoving the − way — the sign just says which direction.
Accelerationspeeding up the + waypoints the − way. Same sign as v → speeding up; opposite sign → slowing down.
Displacementended on the + side of startended on the − side (e.g. below the launch point).
Force / net forcepushes the + waypushes the − way. Weight is − (down); friction is − (opposes motion).
gacceleration from gravity is −9.8 m/s² when up is +. 9.8 is the size, − is "down."
Mass · speed · distance · timealways +never negative — these are sizes only.

The one trick that explains most of it: acceleration is negative whenever it points opposite your chosen + direction — braking, gravity, any backward push. Entering it: "how fast / how far / how much" wants the size → positive; a net force or component with a direction keeps the − when it points the negative way.

📐 sin vs cos — which one do I use?

The #1 mix-up, fixed by one rule: the side next to the angle uses cos; the side across from it uses sin (SOH-CAH-TOA: sin = opp/hyp, cos = adj/hyp).

For a force F at angle θ from the horizontal: F_x = F·cos θ (horizontal — next to the angle) and F_y = F·sin θ (vertical — across from it).

⚠️ If the angle is measured from the vertical instead, it flips (F_x = F·sin θ, F_y = F·cos θ). Always check which axis the angle opens from — the side touching the angle gets cos.

On a ramp tilted θ: gravity down the slope = mg·sin θ (steeper → slides more), into the ramp = mg·cos θ. Banked curve: tan θ = v²/(r·g). Worked: a 50 N pull at 30° → F_x = 50·cos30° = 43.3 N, F_y = 50·sin30° = 25 N.

➕ Adding vectors — components → resultant (your Wiley Ch 1)

To add A and B into R = A + B you never add the arrows directly — break each into x/y parts, add those, then rebuild:

1. components (watch signs): Vx = V·cos θ, Vy = V·sin θ — a part pointing left is −x, down is −y.  2. add: Rx = Ax + Bx, Ry = Ay + By (a left-pointing part subtracts → that's how you get a −4.97).  3. rebuild: R = √(Rx² + Ry²), θ = tan⁻¹(Ry/Rx).

The angle trap: tan⁻¹ only returns −90°…90°, so check the signs of Rx, Ry to place the arrow — if Rx is negative (points left), add 180° to the angle. Worked: Rx = 8, Ry = 6 → R = √(8²+6²) = 10 m, θ = tan⁻¹(6/8) = 36.9° above +x.

Positive and negative signs — what they actually mean

A minus sign is usually a direction label, not a mistake. First choose your positive direction, then every vector gets a sign compared with that choice.

Standard choice unless told otherwise: up = + and down = −; right = + and left = −; if the problem is one-dimensional, call the direction of motion + so friction/braking becomes −.

  • Position / displacement: positive is one side of your origin; negative is the other side.
  • Velocity: the sign tells which way the object is moving. If right is positive, moving right means v > 0; moving left means v < 0.
  • Acceleration: the sign tells which way velocity is changing, or which way the net force points. Moving right but slowing down means v > 0 and a < 0.
  • Forces: add pushes and pulls with signs. If right is positive, 30 N right + 12 N left becomes +30 + (-12) = +18 N.
  • Circular motion: centripetal acceleration and force point inward. If inward is left or down on your diagram, the inward value can be negative in that axis.
  • Mass, speed, distance, and time: these are sizes, so they are never negative. If the problem asks "how fast," "how far," or "how much force," it usually wants a positive size plus a direction word.
One trick:same sign for v and a = speeding up; opposite signs for v and a = slowing down
Enter a negative when the value itself points left, down, or backward compared with your chosen + direction: gravity with up positive is -9.8; braking acceleration with forward positive is negative.
Do not enter a negative when the question asks only for magnitude: speed, distance, mass, time, or "how much force." Give the positive number and write the direction in words.

Best habit: write the sign key before the math: right/up = + or toward center = +. Then the answer's sign tells the direction.

sin, cos, x, and y signs

Cos and sin give sizes. The diagram gives the sign. The side next to the angle uses cos; the side across from the angle uses sin.

Angle measured from horizontal:F_x = F·cos(θ) F_y = F·sin(θ) Angle measured from vertical:F_x = F·sin(θ) F_y = F·cos(θ) Right/up launch or force:x = +v·cos(θ) y = +v·sin(θ) Right/down:x = +v·cos(θ) y = −v·sin(θ) Left/up:x = −v·cos(θ) y = +v·sin(θ) Left/down:x = −v·cos(θ) y = −v·sin(θ) Ramp tilted θ at the base:down-slope gravity = m·g·sin(θ) into-ramp gravity = m·g·cos(θ) Banked curve relation:tan(θ) = v² / (r·g)
Quick check: a 50 N pull at 30° above the ground has F_x = 50cos30° = 43.3 N right and F_y = 50sin30° = 25 N up, so both signs are positive.

If the calculator gives a positive component size, attach the sign from the words or picture: left = negative x, down = negative y, backward = negative if forward was chosen positive.

Free-body diagrams — draw the forces, nothing else

Shrink the object to a dot. Draw an arrow for every real force touching it: gravity down (weight), the surface pushing back (normal), any pull/push, friction. Do not draw an arrow for "motion" or "centrifugal force" — those aren't forces.

Then split arrows into x and y, and write F_net,x = m·a_x and F_net,y = m·a_y. The diagram turns a word problem into equations.

Why velocity can be zero while acceleration isn't (review from Test 1)

Throw a ball straight up. At the very top it stops for an instant — velocity = 0. But gravity is still pulling it down the entire time (a = −9.8 m/s²); that's exactly why it doesn't hang there and instead falls back. Zero velocity is a single instant; the acceleration never paused.

Calculator + test speed

For the TI-30 (TI-30X IIS or TI-30XS MultiView). Set it up right, learn the few button patterns Test 2 needs, and use a steady checking habit so you move faster and trust your answers.

Do this first — 20 seconds

Put it in DEGREE mode

Any problem with an angle (cos, sin, tan) is wrong if the calculator is in radians. Don't guess the mode — test it:

Type this:cos ( 60 ) =
  • If you get 0.5 → you're in degrees.
  • If you get −0.952… → you're in radians. Fix it: press MODE, choose DEG (or use the DRG key), then re-test until cos(60) = 0.5.

Sanity values to remember: cos(60) = 0.5, sin(30) = 0.5, cos(0) = 1. If those come out right, your mode is right.

The 6 keys you actually need

  • Square: the key. To square v, press v then .
  • Square root: the key (usually 2nd then ). It opens √( — type what's inside, then close ).
  • Negative sign: the (-) key (bottom row), NOT the big minus. Use (-) for things like a = -9.8.
  • Parentheses ( ): use them around any bottom of a fraction and around everything under a √.
  • π: 2nd then the π key. (If unsure, just type 3.14159 — same answer.)
  • Last answer: the Ans key reuses your previous result so you don't re-type it.

When to type a negative number

Use the calculator's (-) key when the value itself is negative. Use the big minus key only to subtract one value from another.

Gravity when up is positive:(-) 9.8 Net force with right positive:30 + (-) 12 = → 18 N Braking acceleration, then force:1200 × (-) 3 = → -3600 N Right/down component:25 × sin(30) = 12.5, then write y = -12.5 m/s

A negative final answer means "opposite my positive direction." If the problem asks for a magnitude, report the positive size plus the direction in words.

Button sequences for Test 2 formulas

Type these left to right, then press =. (Example numbers in parentheses.)

Net force / 2nd law — F = m·a (1200 kg, 2.5 m/s²)1200 × 2.5 = → 3000 N Acceleration — a = F / m (50 N, 10 kg)50 ÷ 10 = → 5 m/s² Weight — W = m·g (8 kg)8 × 9.8 = → 78.4 N Friction then net — f = μ·N, with N = m·g (μ=0.30, m=20, pull=80)20 × 9.8 = → 196 (that's N) 0.30 × 196 = → 58.8 (that's f) 80 − 58.8 = → 21.2 (net force) 21.2 ÷ 20 = → 1.06 m/s² Centripetal accel — a_c = v² / r (v=15, r=45)15 x² ÷ 45 = → 5 m/s² Centripetal force — F_c = m·v² / r (m=1000, v=20, r=80)1000 × 20 x² ÷ 80 = → 5000 N Solve for speed — v = √(F·r / m) (F=24, r=1.2, m=0.5)√ ( 24 × 1.2 ÷ 0.5 ) = → 7.59 m/s Speed from period — v = 2·π·r / T (r=2, T=4)2 × π × 2 ÷ 4 = → 3.14 m/s Flat-curve max speed — v = √(μ·g·r) (μ=0.8, r=50)√ ( 0.8 × 9.8 × 50 ) = → 19.8 m/s A component — v0·cos(θ) (25 m/s at 30°, DEGREE mode!)25 × cos ( 30 ) = → 21.65 m/s
Two killers to avoid: (1) for a √, put the WHOLE expression in parentheses — √(0.8×9.8×50), not √0.8 ×9.8×50. (2) Keep extra decimals as you go; only round the final answer.

Go faster (without rushing)

  • Write the equation first, numbers second. Don't re-derive on the test — pick the formula, then plug in. The Formulas tab is your map.
  • List givens with units before touching the calculator: "m = 20 kg, μ = 0.30, F = 80 N, find a." Half the work is just sorting that out.
  • Estimate first. Round to easy numbers in your head ("≈ 80/20 = 4"). If the calculator says 0.4 or 40, you fat-fingered something.
  • One clean keystroke chain per number. Type it once, press =. If it looks off, re-key once slowly — don't stare and second-guess.
  • Multiple choice = work backward. Eliminate answers with wrong units or impossible size; if two are close, the math decides. You don't always need the full solve.
  • Stuck for 2 minutes? Flag it and move on. Come back after the easy points are banked. (Test 2 is 20 MC pts + 80 problem pts — don't lose easy ones to a hard one.)
  • Show your work on the problem part. The equation + plugged-in numbers earns partial credit even if the final arithmetic slips.

Check & verify every answer (the 30-second pass)

  • Units: did it come out in the right unit? Force → N, acceleration → m/s², speed → m/s. Wrong unit = wrong setup.
  • Size: is it believable? A car's acceleration is a few m/s², not 5000. A centripetal force on a ball is a few N to tens of N.
  • Sign: does the direction make sense? A braking force points backward; "up" forces are positive if you chose up as positive.
  • Plug it back: put your answer into the original equation and see if both sides match.
  • Did you square / root correctly? Most circular-motion errors are a missed or a missed . Re-check those two keys.
  • Angle mode: if a problem used cos/sin, re-confirm cos(60)=0.5 in your head — that's the radians trap.

Homework links

Direct links to your Canvas assignments. Test 2 covers Chapters 4 & 5 — those are the ones to prioritize this week.

Test 2 chapters (this week)

Chapter 4 — Concept HomeworkOpen
Chapter 4 — Computational HomeworkOpen
Chapter 5 — Concept HomeworkOpen
Chapter 5 — Computational HomeworkOpen

The Test 2 assignments themselves

Test 2 — Multiple Choice (Honorlock)Open
Test 2N — Problem PartOpen
Test 2 — Problem Work UploadOpen

Earlier chapters (review)

Chapter 2 — Concept HomeworkOpen
Chapter 2 — Computational HomeworkOpen
Chapter 3 — Concept HomeworkOpen
Chapter 3 — Computational HomeworkOpen
All homework above is listed on Canvas as due Jul 22, but do the Ch 4–5 sets now — they're your Test 2 practice.

Sample Quiz

A fresh mini-exam every time — 3 multiple choice (2 conceptual, 1 numerical) plus 3 fill-in numerical problems. Answer all, then submit to see your score and full worked solutions.

Pick a week and tap New quiz to start.

1 · Multiple choice

If a car doubles its speed around the same curve, the centripetal force needed:

Answer: quadruples. Why: F_c = mv²/r, so doubling v makes v² four times bigger.

2 · Multiple choice

A truck moves in a straight line at constant velocity. The net force is:

Answer: zero. Why: constant velocity means a = 0, so F_net = ma = 0.

3 · Numerical multiple choice

A 10 kg box accelerates at 3 m/s². The net force is:

F_net = ma = 10·3 = 30 N

4 · Fill-in number

A car goes 12 m/s around a 36 m curve. Find centripetal acceleration.

a_c = v²/r = 12²/36 = 4.0 m/s² inward

5 · Fill-in number

A 7 kg object has what weight? (g = 9.8 m/s²)

W = mg = 7·9.8 = 68.6 N down

6 · Fill-in number

On a flat curve, r = 50 m and μ = 0.70. Find max speed before sliding.

v = √(μgr) = √(0.70·9.8·50) = 18.5 m/s
Preview backup bank

If the text-message preview blocks buttons, use these as the backup question set. The hosted link gives the same types with fresh numbers.

Concept · Net force

An object moves at constant velocity. What is F_net?

Answer: 0 N. Why: constant velocity means a = 0, so F_net = ma = 0.

Concept · 3rd law

You push a wall with 45 N. How hard does it push back?

Answer: 45 N opposite you. The forces are equal in size and act on different objects.

Concept · Circular direction

Which way does centripetal acceleration point?

Answer: toward the center. It changes direction, not necessarily speed.

Numbers · F = ma

A 14 kg crate accelerates at 2.5 m/s². Find net force.

F = ma = 14·2.5 = 35 N

Numbers · weight

A 5.5 kg bag has what weight?

W = mg = 5.5·9.8 = 53.9 N down

Numbers · friction

A 20 kg box, μ = 0.30, pull = 90 N. Find acceleration.

N = 20·9.8 = 196 N f = 0.30·196 = 58.8 N F_net = 90 - 58.8 = 31.2 N a = 31.2/20 = 1.56 m/s²

Numbers · a_c

A runner goes 8 m/s around a 16 m curve. Find a_c.

a_c = v²/r = 8²/16 = 4.0 m/s² inward

Numbers · F_c

A 2 kg object moves 6 m/s in a 3 m circle. Find F_c.

F_c = mv²/r = 2·6²/3 = 24 N inward

Numbers · period

An object moves at 5 m/s around radius 2 m. Find one-loop time.

T = 2πr/v = 2π·2/5 = 2.51 s

Mastery checklist

Check things off as you nail them. In Safari/Chrome, this saves on this device automatically.

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Chapter 4 — Forces

Chapter 5 — Circular Motion

Chapter 2–3 — Kinematics (review)

Chapter 4

F_net = m·a, opposite forces, mass vs weight, friction, tension, elevator normal force.

Chapter 5

a_c = v²/r, F_c = mv²/r, v = 2πr/T, flat curve speed, no real outward force.

Review

Pick the right kinematic equation, solve free fall, split projectiles into x/y, recombine speed and direction.

Test habit

Write the equation first, plug numbers with units, box the final answer, then check size and units.