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Mechanics · Newton's Laws of Motion

Forces and Motion

Push on an object, add friction, and change its mass to see how forces produce acceleration. Identify the net force from a free-body diagram, verify Newton's second law F = ma, and confirm that acceleration is proportional to the net force and inversely proportional to the mass.

Theory — Newton's Laws and the Net Force

Force and the Free-Body Diagram

A force is a push or a pull, measured in newtons (N). Usually several forces act on an object at once — an applied push, friction, gravity, the support of the ground. A free-body diagram draws each of these as an arrow. The single most important quantity is the net force: the vector sum of all the forces acting on the object.

Net Force (one dimension) F_net = ΣF = F_applied − F_friction − …
Forces in the direction of motion are positive; opposing forces are negative.

Newton's First Law — Inertia

An object at rest stays at rest, and an object in motion continues at constant velocity, unless a net force acts on it. So if the net force is zero, the object's velocity does not change — it is either stationary or coasting at constant speed. A moving object slowing down is not "running out of force"; it is being acted on by friction.

Newton's Second Law — F = ma

When the net force is not zero, the object accelerates. The acceleration is proportional to the net force and inversely proportional to the mass. This is the central equation of mechanics.

Newton's Second Law F_net = m · a  →  a = F_net / m

Units: 1 newton (N) = 1 kg·m/s²
a ∝ F_net (fixed m) · a ∝ 1/m (fixed F_net)

Newton's Third Law — Action–Reaction

For every force there is an equal and opposite reaction force: if you push on a wall with 50 N, the wall pushes back on you with 50 N. The two forces act on different objects, which is why they don't simply cancel.

Friction

Friction is a force that opposes motion (or attempted motion) between surfaces in contact. Kinetic friction — the friction on a sliding object — is proportional to the normal force pressing the surfaces together. On a level surface the normal force equals the weight, mg.

Kinetic Friction F_friction = μ · N = μ · m · g  (level surface)
μ = coefficient of friction (no units), g = 9.8 m/s²
Net force = F_applied − F_friction, then a = F_net / m

Net force zero

Forces are balanced. Acceleration is zero. The object stays at rest or moves at constant velocity (Newton's first law).

Net force non-zero

Forces are unbalanced. The object accelerates in the direction of the net force at a = F_net/m (Newton's second law).

ChangeEffect on accelerationReason
Increase net force (fixed m)increases proportionallya = F/m, a ∝ F
Increase mass (fixed F)decreases (a ∝ 1/m)a = F/m
Increase frictionreduces net force → less aF_net = F_applied − F_friction

Instructions — Running the Virtual Experiment

The Force Lab tab lets you feel how forces, friction, and mass combine; the Measurements tab provides the quantitative test of F = ma. Record every reading in your lab notebook.

Experiment 1 — Net Force and Balanced Forces (Force Lab tab)
1
Open Simulation → Force Lab. Set the mass and applied force with the sliders (or use the scenario buttons — e.g. 50 kg · 150 N), keep friction coefficient μ = 0.20, and watch the free-body arrows and the net-force readout. Note the friction force shown in the readout — you will need it for the theoretical calculations.
2
Find the applied force at which the object just begins to move — this is where the applied force first exceeds the friction force. Below it, the net force is zero and the object stays still (Newton's first law).
3
Raise the applied force further and watch the object accelerate. Confirm that the acceleration readout equals (F_applied − F_friction)/m.
Experiment 2 — Acceleration vs Net Force (Measurements tab)
1
Open Measurements. The surface is frictionless and the mass is fixed at 2 kg, so the net force equals the applied force.
2
Click each force button (2–10 N) and then Record reading to log the acceleration produced.
3
Click Plot & fit. The acceleration-versus-force points should lie on a straight line through the origin (a ∝ F), and the slope should equal 1/m = 0.5 kg⁻¹. Confirm slope × mass = 1.
Experiment 3 — The Effect of Mass (Force Lab tab)
1
Set a fixed applied force (say 150 N) with friction off (μ = 0). Record the acceleration.
2
Double the mass and record the new acceleration. Confirm that doubling the mass halves the acceleration (a ∝ 1/m). Tabulate several masses in your report.

Simulation — Net Force, Friction & Acceleration

Forces and Motion Virtual LabApply a force and watch the object respond
applied force
friction force
net force

Controls

Readout
Applied force— N
Friction force— N
Net force— N
Acceleration— m/s²
Velocity— m/s
Each point is one (force, acceleration) reading.
Frictionless · mass fixed at 2 kg · slope of a-vs-F = 1/m.

Frictionless · m = 2 kg

Set applied force

Current reading (2 N)
Acceleration— m/s²
Applied force (N)Acceleration (m/s²)a / F (kg⁻¹)
No readings yet — pick a force and click "Record reading".

Team Questions

Question 1. A net force of 20 N acts on a 4 kg object on a frictionless surface. What is its acceleration? Use a = F_net/m. (Type just the number in m/s²)
Question 2. A box is pushed with an applied force of 30 N while friction resists with 12 N. What is the net force on the box? (Type just the number in N)
Question 3. That 18 N net force acts on a 6 kg box. What is the box's acceleration? (Type just the number in m/s²)
Question 4. If the net force on an object is doubled while its mass stays the same, what happens to its acceleration — does it double, halve, or stay the same? (One word: double / halve / same)
Question 5. If the mass of an object is doubled while the same net force is applied, what happens to its acceleration? (One word: double / halve / same)
Question 6. A hockey puck slides across frictionless ice at a constant velocity. What is the net force acting on it? (Type just the number in N)
Question 7 — Challenge. A 5 kg crate is pushed across the floor with a force of 15 N and moves at constant velocity. What must the friction force be? (Type just the number in N)

Example Lab Report

Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.

Forces and Motion: Verifying Newton's Second Law

Physics | Section: [Your Section] | Date: [Date]

Lab Members: [Names of all members present]

Purpose

To investigate the relationship between net force, mass, and acceleration, and to verify Newton's second law (F = ma) by measuring the acceleration produced by a range of net forces on a fixed mass. The lab also examines how friction reduces the net force and how mass affects acceleration.

Theory

The net force on an object is the vector sum of all forces acting on it. By Newton's second law, this net force produces an acceleration proportional to the force and inversely proportional to the mass. On a level surface, kinetic friction opposes motion with a force μmg.

F_net = F_applied − F_friction · F_friction = μmg
F_net = m·a → a = F_net / m
a ∝ F (fixed m) · a ∝ 1/m (fixed F)

A graph of acceleration versus net force (at fixed mass) is a straight line through the origin whose slope equals 1/m, so multiplying the slope by the mass should give 1.

Calculations — Sample: net force 8 N on a 2 kg mass (frictionless)

Acceleration: a = F_net/m = 8/2 = 4.0 m/s²

Slope of a-vs-F: from (2 N, 1 m/s²) and (10 N, 5 m/s²): slope = (5 − 1)/(10 − 2) = 4/8 = 0.50 kg⁻¹

Check: slope × mass = 0.50 × 2 = 1.00 ✓ (confirms F = ma)

With friction (μ = 0.20, m = 4 kg, F_applied = 20 N): F_friction = μmg = 0.20 × 4 × 9.8 = 7.84 N; F_net = 20 − 7.84 = 12.16 N; a = 12.16/4 = 3.04 m/s²

Results Table — Acceleration vs Applied Force (frictionless, m = 2 kg)

Applied force (N)Acceleration (m/s²)a / F (kg⁻¹)
21.00.50
42.00.50
63.00.50
84.00.50
105.00.50

Mass trial (F = 20 N, frictionless): doubling the mass from 2 kg to 4 kg halved the acceleration from 10 m/s² to 5 m/s², confirming a ∝ 1/m.

Discussion

The acceleration was directly proportional to the net force: each reading of a/F gave the same value, 0.50 kg⁻¹, and the a-versus-F graph was a straight line through the origin. Its slope (0.50 kg⁻¹) multiplied by the mass (2 kg) gave 1.00, confirming F = ma. When the mass was doubled at fixed force, the acceleration was halved, verifying the inverse relationship a ∝ 1/m.

Adding friction reduced the net force and therefore the acceleration. Below the point where the applied force exceeded the friction force, the net force was zero and the object did not move, illustrating Newton's first law. A crate moving at constant velocity likewise has zero net force, meaning the applied push exactly balances friction.

Conclusion

The experiment verified Newton's second law. Acceleration was proportional to net force (slope × mass = 1) and inversely proportional to mass, and friction was shown to subtract from the applied force to set the net force. Zero net force produced zero acceleration, consistent with Newton's first law.

Practice Questions

Show all work and include units in your answers.

Question 1
A 1200 kg car accelerates from rest to 24 m/s in 8.0 s on a level road. What is its acceleration, and what net force does the engine and road provide?
Hint: a = Δv/Δt, then F_net = ma.
Question 2
A 10 kg box is pushed with a horizontal force of 40 N across a floor with a coefficient of kinetic friction of 0.25. Find the friction force, the net force, and the acceleration. (Use g = 9.8 m/s².)
Hint: F_friction = μmg; F_net = F_applied − F_friction; a = F_net/m.
Question 3
Two forces act on a 5 kg object along a line: 30 N to the right and 12 N to the left. Find the net force and the acceleration, including direction.
Hint: net force = 30 − 12 to the right; a = F_net/m in that direction.
Question 4
An object accelerates at 3.0 m/s² when a 6.0 N net force is applied. What is its mass? What acceleration would a 9.0 N net force give the same object?
Hint: m = F/a, then a = F/m with the new force.
Question 5
An elevator cable holds a 500 kg elevator. State the cable tension when the elevator is (a) at rest, and (b) accelerating upward at 1.5 m/s². (Use g = 9.8 m/s².)
Hint: (a) T = mg. (b) T − mg = ma → T = m(g + a).
Question 6 — Challenge
A 3.0 kg block on a frictionless surface is pulled by two ropes: one pulls with 12 N at 0° (along +x) and the other with 5.0 N at 90° (along +y). Find the magnitude and direction of the acceleration.
Hint: net force components are 12 N (x) and 5 N (y). Magnitude = √(12² + 5²) = 13 N; a = 13/3. Direction = arctan(5/12).