Get Ready: Module 2 — Forces & Newton's Laws
Everyone starts somewhere. Before we draw free-body diagrams and slide blocks down ramps, let's build the floor before the building — one small skill at a time. Go at your own pace; there's no rush and nothing to prove here.
Skills to Build First
Five small building blocks. Master these and Module 2 will feel like the next natural step, not a wall.
Multiplying & Dividing Quantities with Units
What it is & why you need it: A physics quantity is a number and a unit, and the units travel through the math with the numbers. When you compute \(W = mg\) you multiply kilograms by \(\text{m/s}^2\) and the answer comes out in newtons \((\text{kg}\cdot\text{m/s}^2 = \text{N})\). Keeping units attached catches mistakes before they happen.
Worked example: compute \(W = mg\) for \(m = 6\,\text{kg}\)
- Write the quantities with their units: \(m = 6\,\text{kg}\) and \(g = 9.8\,\text{m/s}^2\).
- Multiply the numbers: \(6 \times 9.8 = 58.8\).
- Multiply the units: \(\text{kg} \cdot \text{m/s}^2\), which is the definition of a newton \((\text{N})\).
- Combine: \(W = 58.8\,\text{N}\). The unit confirms it's a force.
Tip: carry the unit through every line. If you ever divide a distance (m) by a time (s) you should see m/s pop out — a speed — without thinking about it.
1. A \(3\,\text{kg}\) mass has weight \(W = mg\). Compute it (use \(g = 9.8\)).
2. An object travels \(120\,\text{m}\) in \(8\,\text{s}\). Divide to find its speed, and name the unit.
Show answer
2. \(\dfrac{120\,\text{m}}{8\,\text{s}} = 15\). Dividing metres by seconds gives m/s, so the speed is \(15\,\text{m/s}\).
Recall \(g = 9.8\ \text{m/s}^2\)
What it is & why you need it: Near Earth's surface, gravity gives every falling object the same acceleration: \(g = 9.8\,\text{m/s}^2\) (often rounded to \(10\) for quick estimates). It's the constant that turns a mass into a weight \((W = mg)\) and that sets how fast things speed up as they fall. You'll use it on nearly every page of this module, so commit it to memory.
Worked example: how fast is a dropped ball after \(2\,\text{s}\)?
- From rest, a falling object gains \(g\) of speed every second: \(v = gt\).
- Substitute \(g = 9.8\,\text{m/s}^2\) and \(t = 2\,\text{s}\): \(v = 9.8 \times 2\).
- Multiply: \(v = 19.6\).
- Attach the unit: \(v = 19.6\,\text{m/s}\) downward.
Remember: \(g\) is an acceleration (m/s²), not a force. It's the same for a feather and an anvil in a vacuum — mass does not change \(g\).
1. State the value of \(g\) near Earth's surface, with its unit.
2. A dropped object falls from rest. Find its speed after \(3\,\text{s}\) using \(v = gt\).
Show answer
2. \(v = gt = 9.8 \times 3 = 29.4\,\text{m/s}\).
Decomposing a Vector into Components
What it is & why you need it: A vector that points at an angle can be replaced by two perpendicular pieces — usually a horizontal part and a vertical part — that together do the same job. With a magnitude \(F\) at angle \(\theta\), the components are \(F\cos\theta\) (adjacent) and \(F\sin\theta\) (opposite). On an incline this is exactly how the weight splits into "along the slope" and "into the surface."
Worked example: split a \(50\,\text{N}\) force at \(30^\circ\)
- Draw the force as the hypotenuse of a right triangle, angle \(30^\circ\) at the corner.
- The adjacent (horizontal) component uses cosine: \(F_x = F\cos\theta = 50\cos 30^\circ\).
- Compute: \(50 \times 0.866 \approx 43.3\,\text{N}\).
- The opposite (vertical) component uses sine: \(F_y = F\sin\theta = 50\sin 30^\circ = 25\,\text{N}\).
Memory hook: the component next to the angle uses cos; the component across from it uses sin. "SOH-CAH-TOA" still rules.
1. A \(20\,\text{N}\) force points \(60^\circ\) above the horizontal. Find its horizontal component \(F\cos\theta\).
2. For the same force, find its vertical component \(F\sin\theta\).
Show answer
2. \(F_y = 20\sin 60^\circ = 20 \times 0.866 \approx 17.32\,\text{N}\).
Telling Mass Apart from Weight
What it is & why you need it: Mass \((m)\) is how much matter an object contains, measured in kilograms; it never changes. Weight \((W)\) is the gravitational force on that mass, measured in newtons, and it depends on \(g\). Same mass, different planet, different weight. Free-body diagrams show the weight arrow \((W = mg)\), not the mass — so you must convert between them fluently.
Worked example: a \(10\,\text{kg}\) box on Earth and the Moon
- Mass is fixed: the box is \(10\,\text{kg}\) everywhere — Earth, Moon, deep space.
- Weight on Earth: \(W = mg = 10 \times 9.8 = 98\,\text{N}\).
- Weight on the Moon (where \(g \approx 1.6\,\text{m/s}^2\)): \(W = 10 \times 1.6 = 16\,\text{N}\).
- So the mass stayed \(10\,\text{kg}\) but the weight dropped from \(98\,\text{N}\) to \(16\,\text{N}\).
To go backward — weight to mass — divide: \(m = \dfrac{W}{g}\). A \(49\,\text{N}\) object on Earth has mass \(\dfrac{49}{9.8} = 5\,\text{kg}\).
1. A bag has a mass of \(4\,\text{kg}\). What is its weight on Earth?
2. An object weighs \(78.4\,\text{N}\) on Earth. Find its mass using \(m = W/g\).
Show answer
2. \(m = \dfrac{W}{g} = \dfrac{78.4}{9.8} = 8\,\text{kg}\). Notice the unit flips from newtons to kilograms.
Reading a Labeled Force Diagram
What it is & why you need it: A free-body diagram shows one object as a dot or box with an arrow for each force. Arrow length shows magnitude, arrow direction shows which way the force pushes or pulls, and the labels name them \((W, N, f, T, \dots)\). Before you can solve a problem you have to read the picture: which forces point which way, and which ones cancel.
Worked example: a book resting on a table
- Identify the object: the book. Draw it as a box.
- Gravity pulls it down: a weight arrow \(W\) pointing down from the center.
- The table pushes back up: a normal force arrow \(N\) pointing up, the same length as \(W\).
- Since \(W\) and \(N\) are equal and opposite, they cancel — the net force is zero, and the book stays at rest.
Reading rule: equal-length arrows pointing opposite ways cancel. If all the arrows cancel, the object is in equilibrium (at rest or moving steadily).
1. In the book-on-table diagram, which two forces are equal and opposite, and what is the net force?
2. Newton's third law pairs a force with its reaction. A book pushes down on a table — what is the reaction force?
Show answer
2. By Newton's third law, the table pushes up on the book with an equal and opposite force.
Quick Readiness Check
Six short questions across all five skills. Try each one first, then peek at the answer.
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Compute the weight of a \(7\,\text{kg}\) object on Earth, with its unit.
Show answer
\(W = mg = 7 \times 9.8 = 68.6\,\text{N}\). -
State the value of \(g\) near Earth's surface.
Show answer
\(g = 9.8\,\text{m/s}^2\) (about \(10\,\text{m/s}^2\) for quick estimates). -
A \(40\,\text{N}\) force points \(45^\circ\) above the horizontal. Find its horizontal component.
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\(F_x = 40\cos 45^\circ = 40 \times 0.707 \approx 28.28\,\text{N}\). -
An object weighs \(58.8\,\text{N}\) on Earth. What is its mass?
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\(m = \dfrac{W}{g} = \dfrac{58.8}{9.8} = 6\,\text{kg}\). -
True or false: a \(5\,\text{kg}\) mass has the same weight on the Moon as on Earth.
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False. The mass is \(5\,\text{kg}\) both places, but the weight is smaller on the Moon because \(g\) is smaller there. -
A box sits at rest on a table. Name the two forces on it and state the net force.
Show answer
Weight \(W\) down and normal force \(N\) up; they're equal and opposite, so the net force is \(0\,\text{N}\).
If these feel comfortable, you're ready for the module. If a couple felt tricky, that's fine — revisit that skill card above and drill a little with the free links below. No shame, just steps.
Practice More — Free
Want extra reps? These free sites let you drill each skill until it clicks. (Links open in a new tab.)
Units & dimensional analysis
Weight, mass & \(g\)
Vector components
Free-body diagrams
When You're Ready →
No rush. When the skills above feel solid, step into Module 2.
Instructor: Dr. Goodluck Ijezie-Desbois, PharmD · Beta Academy · Room: TBA
Foundations support for the Physics enrichment track —
not a TEKS-graded course. Offered alongside the graded mathematics pathway.