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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Course calendar
| Test | Chapters | Opens | Due |
|---|---|---|---|
| Test 1 done | Ch 2–3 · Kinematics | — | Jun 11 |
| Test 2 | Ch 4–5 · Forces & Circular | Jun 19 | Jun 20 |
| Test 3 | Ch 6–7 · Energy & Momentum | Jun 27 | Jun 28 |
| Test 4 | Ch 8–9 | Jul 5 | Jul 6 |
| Test 5 | Ch 10–11 | Jul 13 | Jul 14 |
| Test 6 | Ch 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.
- 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_xIn 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.
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, withg = 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?
F_net = m·a. It directly links the two things you're given (m, a) to the one you want (force).2A 50 N net force pushes a 10 kg cart. What is its acceleration?
F_net = m·a rearranged to a = F_net / m — you have force and mass, want acceleration.3What is the weight of an 8.0 kg backpack?
W = m·g. Weight is the pull of gravity, which is mass times g. (Your sheet uses mg in Chapter 6, e.g. PE = mgh.)4A 20 kg box on a flat floor (μ = 0.30) is pulled horizontally with 80 N. Find its acceleration.
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.N = mg → f = 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.
F_net, then a = F_net/m.6A 5.0 kg lamp hangs at rest from a cord. What is the tension in the cord?
F_net = 0, so up = down: T = W = m·g.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)?
N = m·g + m·a = m(g + a).8You push on a wall with 50 N. How hard does the wall push back on you?
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.
a = F/(m_total). Then for the 3.0 kg block alone, only the contact push acts: F_contact = m·a.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.
- 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.
ris 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?
a_c = v²/r. It uses exactly what you're given (v and r) and gives the inward acceleration.2A 1000 kg car goes 20 m/s around a curve of radius 80 m. What centripetal force is needed?
F_c = m·v²/r — it's just F = m·a with a = a_c = v²/r.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?
F_c = m·v²/r and rearrange for v: v = √(F_c·r / m).4A stone goes around a circle of radius 2.0 m, completing one loop every 4.0 s. Find its speed and centripetal acceleration.
2πr: v = 2πr/T. Then a_c = v²/r.2πr, not r.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
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.)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.
T = 1.0 s. Then v = 2πr/T and a_c = v²/r.7A 2.0 kg mass moves at 10 m/s and needs 200 N of centripetal force. What is the radius of its circle?
F_c = m·v²/r for r: r = m·v²/F_c.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.
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.
Test 1 — Chapter 2–3 (Kinematics)
This one's done — kept here for review, since Ch 4–5 build on it.
Essentials
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.
Worked examples
1A sled starts from rest and accelerates at 3.0 m/s² for 5.0 s. What is its final speed?
v_f = v_0 + a·t — it has exactly v_0, a, t (knowns) and v_f (unknown), no distance needed.2A ball is dropped and falls for 3.0 s. How far does it fall? (g = 9.8 m/s²)
d = v_0·t + ½·a·t² with v_0 = 0 and a = g.3A ball is thrown straight up at 20 m/s. How high does it go?
v_f² = v_0² + 2·a·d. At the top, v_f = 0.4A ball rolls off a 45 m cliff horizontally at 20 m/s. How far from the base does it land?
y = ½·g·t², then x = v_0x·t.5A ball is kicked at 25 m/s at 30° above the ground. Find its maximum height and range.
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.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/sFree 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 mThrown up
A ball is thrown upward at 18 m/s. What max height?
H = v_0²/(2g) H = 18²/19.6 H = 16.5 mProjectile 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 mPractice — 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 NFriction 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/s²Opposite 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 rightWeight is a force
What is the weight of a 7 kg backpack?
W = mg = 7·9.8 W = 68.6 N downwardChapter 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 inwardF_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 inwardSpeed 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/sFlat 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/sFormula 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.
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.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.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_0is "the starting speed,"v_fis "the final speed." Same v, just tagged. - Read it as a sentence.
F = m·aliterally says "force equals mass times acceleration." That is all a formula is — a sentence in shorthand.
Notation key (the squiggles)
v² means v × v. (4² = 16.)√9 = 3 because 3² = 9.Δv = final speed − starting speed.Motion symbols (Test 1, and used everywhere)
x is usually sideways, y up-and-down. m (meters). Hook: a ruler measurement.9.8 m/s². m/s². Hook: Earth's built-in downward gas pedal.Force symbols (Test 2 — Chapter 4)
W = m·g. N. Hook: what a scale pushes back with.f = μ·N. N. Hook: the grippy drag between two surfaces.Circular-motion symbols (Test 2 — Chapter 5)
a_c = v²/r. m/s². Hook: centripetal = center-seeking.F_c = m·v²/r. N. Hook: the inward tug — supplied by a real force like tension or friction.v = 2πr/T because circles. none.Units, in plain words
9.8 m/s² means a falling object gets 9.8 m/s faster every second.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 timet· "how fast" → find velocityv
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 |
|---|---|---|
| Velocity | moving the + way | moving the − way — the sign just says which direction. |
| Acceleration | speeding up the + way | points the − way. Same sign as v → speeding up; opposite sign → slowing down. |
| Displacement | ended on the + side of start | ended on the − side (e.g. below the launch point). |
| Force / net force | pushes the + way | pushes the − way. Weight is − (down); friction is − (opposes motion). |
| g | — | acceleration from gravity is −9.8 m/s² when up is +. 9.8 is the size, − is "down." |
| Mass · speed · distance · time | always + | 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 meansv < 0. - Acceleration: the sign tells which way velocity is changing, or which way the net force points. Moving right but slowing down means
v > 0anda < 0. - Forces: add pushes and pulls with signs. If right is positive,
30 N right + 12 N leftbecomes+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.
-9.8; braking acceleration with forward positive is negative.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)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.
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 theDRGkey), then re-test untilcos(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
x²key. To square v, pressvthenx². - Square root: the
√key (usually2ndthenx²). It opens√(— type what's inside, then close). - Negative sign: the
(-)key (bottom row), NOT the big−minus. Use(-)for things likea = -9.8. - Parentheses
( ): use them around any bottom of a fraction and around everything under a √. - π:
2ndthen theπkey. (If unsure, just type3.14159— same answer.) - Last answer: the
Anskey 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.
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.)
√(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
x²or a missed√. Re-check those two keys. - Angle mode: if a problem used cos/sin, re-confirm
cos(60)=0.5in 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)
The Test 2 assignments themselves
Earlier chapters (review)
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 N4 · 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 inward5 · Fill-in number
A 7 kg object has what weight? (g = 9.8 m/s²)
W = mg = 7·9.8 = 68.6 N down6 · 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/sIf 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 NNumbers · weight
A 5.5 kg bag has what weight?
W = mg = 5.5·9.8 = 53.9 N downNumbers · 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/s²Numbers · 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 inwardNumbers · 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 inwardNumbers · period
An object moves at 5 m/s around radius 2 m. Find one-loop time.
T = 2πr/v = 2π·2/5 = 2.51 sMastery checklist
Check things off as you nail them. In Safari/Chrome, this saves on this device automatically.
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.