Newton's First and Second Laws
Inertia, net force, and F = ma
Newton's First and Second Laws
Newton's First Law (Law of Inertia)
Statement: An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net external force.
Key Concepts
Inertia is the tendency of an object to resist changes in its motion.
- More massive objects have more inertia
- Inertia is measured by mass (kg)
Equilibrium occurs when the net force is zero.
- Static equilibrium: object at rest ()
- Dynamic equilibrium: object moving at constant velocity (, )
Common Misconceptions
❌ Wrong: "A force is needed to keep an object moving" ✓ Correct: A force is only needed to change motion (accelerate)
❌ Wrong: "If velocity is zero, force must be zero" ✓ Correct: Multiple forces can balance to give zero net force
Newton's Second Law
Statement: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
Or in component form:
Understanding the Equation
Net force (): Vector sum of all forces
- Measured in Newtons (N)
Mass (): Measure of inertia
- Measured in kilograms (kg)
- Scalar quantity (no direction)
- Not the same as weight!
Acceleration (): Rate of change of velocity
- Measured in m/s²
- Vector quantity (same direction as net force)
- If , then
Key Relationships
-
Direct relationship with force: If you double the force, you double the acceleration (same mass)
-
Inverse relationship with mass: If you double the mass, you halve the acceleration (same force)
-
Direction: Acceleration is always in the same direction as the net force
Mass vs. Weight
Mass
- Mass is the amount of matter in an object
- Scalar quantity
- Measured in kg
- Same everywhere in the universe
- Measure of inertia
Weight
- Weight is the gravitational force on an object
- Vector quantity (points toward Earth's center)
- Measured in Newtons (N)
- Depends on location (gravity varies)
- Formula:
On Earth's surface:
On the Moon: m/s² (about 1/6 of Earth)
- Same mass, but weight is 1/6 as much!
Free Body Diagrams (FBDs)
A free body diagram shows all forces acting on a single object.
Steps to Draw an FBD
- Isolate the object (draw a dot or box)
- Draw all forces as arrows starting from the object
- Length represents magnitude
- Direction shows force direction
- Label each force (e.g., , , )
- Do NOT include:
- Motion (velocity, acceleration)
- Forces the object exerts on other things
Common Forces
- Weight ( or ): Always points downward, magnitude
- Normal force ( or ): Perpendicular to surface, pushes away from surface
- Tension ( or ): Along rope/string, pulls toward rope
- Friction (): Parallel to surface, opposes motion/attempted motion
- Applied force (): Push or pull from external agent
Problem-Solving Strategy
- Draw a free body diagram
- Choose coordinate system (align with motion when possible)
- Write and
- List all forces in each direction (use + and - signs)
- Solve for unknowns
- Check units and reasonableness
Newton's Laws Summary
| Law | Statement | Equation | |-----|-----------|----------| | First | Object maintains velocity unless net force acts | | | Second | Net force causes acceleration | | | Third | Forces come in equal/opposite pairs | |
Special Cases
Zero Acceleration
If , then (equilibrium)
- Object at rest OR moving at constant velocity
- All forces balance
Constant Acceleration
If , then
- Can use kinematic equations
- Common in free fall, inclined planes
Variable Acceleration
If changes, then changes
- More complex analysis required
- Need calculus or numerical methods
📚 Practice Problems
1Problem 1medium
❓ Question:
A 5.0 kg box is pushed across a frictionless floor with a force of 20 N. (a) What is the acceleration of the box? (b) If the box starts from rest, what is its velocity after 3.0 seconds? (c) How far has it traveled in this time?
💡 Show Solution
Solution:
Given: m = 5.0 kg, F = 20 N, friction = 0
(a) Acceleration: Using Newton's 2nd Law: F_net = ma 20 = 5.0a a = 4.0 m/s²
(b) Velocity after 3.0 s: Using v = v₀ + at v = 0 + 4.0(3.0) v = 12 m/s
(c) Distance traveled: Using Δx = v₀t + ½at² Δx = 0 + ½(4.0)(3.0)² Δx = 2.0(9.0) Δx = 18 m
2Problem 2easy
❓ Question:
A kg box experiences a net force of N to the right. What is its acceleration?
💡 Show Solution
Given:
- Mass: kg
- Net force: N (to the right)
Find: Acceleration
Use Newton's Second Law:
Solve for acceleration:
Direction: To the right (same as net force)
Answer: The acceleration is 4 m/s² to the right.
Check:
- Units: N/kg = (kg·m/s²)/kg = m/s² ✓
- Larger force → larger acceleration ✓
- Direction matches force ✓
3Problem 3medium
❓ Question:
A 5.0 kg box is pushed across a frictionless floor with a force of 20 N. (a) What is the acceleration of the box? (b) If the box starts from rest, what is its velocity after 3.0 seconds? (c) How far has it traveled in this time?
💡 Show Solution
Solution:
Given: m = 5.0 kg, F = 20 N, friction = 0
(a) Acceleration: Using Newton's 2nd Law: F_net = ma 20 = 5.0a a = 4.0 m/s²
(b) Velocity after 3.0 s: Using v = v₀ + at v = 0 + 4.0(3.0) v = 12 m/s
(c) Distance traveled: Using Δx = v₀t + ½at² Δx = 0 + ½(4.0)(3.0)² Δx = 2.0(9.0) Δx = 18 m
4Problem 4hard
❓ Question:
Three forces act on a 2.0 kg object: F₁ = 10 N east, F₂ = 8.0 N north, and F₃ = 6.0 N west. (a) Find the net force (magnitude and direction). (b) Find the acceleration of the object. (c) If the object starts from rest, what is its velocity after 5.0 seconds?
💡 Show Solution
Solution:
(a) Net force: Choose coordinate system: +x = east, +y = north
F_net,x = F₁ - F₃ = 10 - 6.0 = 4.0 N (east) F_net,y = F₂ = 8.0 N (north)
Magnitude: F_net = √(4.0² + 8.0²) = √(16 + 64) = √80 = 8.9 N
Direction: θ = tan⁻¹(F_y/F_x) = tan⁻¹(8.0/4.0) = tan⁻¹(2) = 63° north of east
(b) Acceleration: F_net = ma 8.9 = 2.0a a = 4.5 m/s² (in direction of F_net)
(c) Velocity after 5.0 s: v = v₀ + at = 0 + 4.5(5.0) = 22.5 m/s at 63° north of east
Components: vₓ = 4.0/2.0 × 5.0 = 10 m/s, vᵧ = 8.0/2.0 × 5.0 = 20 m/s
5Problem 5hard
❓ Question:
Three forces act on a 2.0 kg object: F₁ = 10 N east, F₂ = 8.0 N north, and F₃ = 6.0 N west. (a) Find the net force (magnitude and direction). (b) Find the acceleration of the object. (c) If the object starts from rest, what is its velocity after 5.0 seconds?
💡 Show Solution
Solution:
(a) Net force: Choose coordinate system: +x = east, +y = north
F_net,x = F₁ - F₃ = 10 - 6.0 = 4.0 N (east) F_net,y = F₂ = 8.0 N (north)
Magnitude: F_net = √(4.0² + 8.0²) = √(16 + 64) = √80 = 8.9 N
Direction: θ = tan⁻¹(F_y/F_x) = tan⁻¹(8.0/4.0) = tan⁻¹(2) = 63° north of east
(b) Acceleration: F_net = ma 8.9 = 2.0a a = 4.5 m/s² (in direction of F_net)
(c) Velocity after 5.0 s: v = v₀ + at = 0 + 4.5(5.0) = 22.5 m/s at 63° north of east
Components: vₓ = 4.0/2.0 × 5.0 = 10 m/s, vᵧ = 8.0/2.0 × 5.0 = 20 m/s
6Problem 6medium
❓ Question:
A kg car accelerates from rest to m/s in seconds on a straight road. What is the net force on the car?
💡 Show Solution
Given:
- Mass: kg
- Initial velocity: m/s
- Final velocity: m/s
- Time: s
Find: Net force
Step 1: Find acceleration
Step 2: Apply Newton's Second Law
Answer: The net force on the car is 4800 N (in the direction of motion).
Alternative: N kN (kilonewtons)
Physical interpretation: This is a significant force—equivalent to the weight of about 490 kg! The engine must provide this force (minus friction) to accelerate the car.
7Problem 7hard
❓ Question:
A kg book rests on a table. Draw a free body diagram and find the normal force. Then, if someone pushes down on the book with a force of N, what is the new normal force?
💡 Show Solution
Part 1: Book at rest
Free Body Diagram:
- Weight pointing down
- Normal force pointing up
Given:
- Mass: kg
- m/s² (approximate)
- Acceleration: (at rest)
Apply Newton's Second Law (vertical direction):
Choose up as positive.
Part 2: Someone pushes down with 10 N
Updated Free Body Diagram:
- Weight N pointing down
- Normal force pointing up
- Applied force N pointing down
Apply Newton's Second Law:
Answers:
- Part 1: Normal force = 20 N (equals weight)
- Part 2: Normal force = 30 N (equals weight + push)
Key insight: Normal force is NOT always equal to weight! It adjusts to prevent the object from accelerating through the surface. Here, the table "pushes back harder" to support both the weight and the downward push.
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