Mechanics

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Mechanics is the branch of physics that studies motion and the forces that cause motion. It forms the foundation for understanding everything from how objects fall to how rockets launch into space.

Kinematics: Describing Motion

Kinematics describes motion without considering the forces that cause it. The key quantities are:

Quantity	Symbol	Unit	Definition
Displacement	Δx or d	meters (m)	Change in position (vector—has direction)
Velocity	v	m/s	Rate of change of displacement
Acceleration	a	m/s²	Rate of change of velocity
Time	t	seconds (s)	Duration of motion

Kinematic Equations (Constant Acceleration)

These equations relate displacement, velocity, acceleration, and time when acceleration is constant:

v = v₀ + at — Final velocity from initial velocity and acceleration
d = v₀t + ½at² — Displacement from initial velocity and acceleration
v² = v₀² + 2ad — Final velocity without time
d = ½(v₀ + v)t — Displacement using average velocity

Where: v₀ = initial velocity, v = final velocity, a = acceleration, t = time, d = displacement

Free Fall

Free fall is motion under the influence of gravity alone. Near Earth's surface:

Acceleration due to gravity: g = 9.8 m/s² (downward)
Air resistance is ignored in ideal free fall problems
Use kinematic equations with a = g (or a = -g if up is positive)

Newton's Laws of Motion

Dynamics explains why objects move the way they do—it's about forces.

Law	Statement	Implication
First Law (Inertia)	An object at rest stays at rest, and an object in motion stays in motion at constant velocity, unless acted upon by a net force.	Objects resist changes to their motion
Second Law	F = ma (Force equals mass times acceleration)	Larger forces cause larger accelerations; more massive objects need more force to accelerate
Third Law	For every action, there is an equal and opposite reaction.	Forces always come in pairs; if A pushes B, B pushes A with equal force

Common Forces

Force	Symbol	Description	Formula
Weight (Gravity)	Fg or W	Force due to gravity	W = mg
Normal Force	FN or N	Perpendicular contact force from a surface	Depends on situation
Friction	Ff or f	Force opposing sliding between surfaces	f = μN
Tension	T	Force transmitted through a rope/string	Depends on situation
Applied Force	Fa	Push or pull applied to an object	Given in problem

Friction

Friction opposes motion between surfaces in contact.

Static friction (fs): Prevents motion from starting. Maximum: fs(max) = μsN
Kinetic friction (fk): Acts on sliding objects. fk = μkN
μs > μk (it's harder to start motion than to maintain it)
μ (mu) = coefficient of friction (depends on surfaces)

Free-Body Diagrams

A free-body diagram shows all forces acting on a single object:

Draw the object as a dot or simple shape
Draw all forces as arrows pointing away from the object
Label each force
Show only forces acting ON the object, not forces the object exerts

Solving Force Problems

Draw a free-body diagram
Choose a coordinate system (usually: +x in direction of motion, +y perpendicular)
Break forces into components if needed
Apply Newton's Second Law: ΣF = ma (separately for x and y)
Solve for unknowns

💡 Examples

Work through these mechanics problems step by step.

Example 1: Kinematics

Problem: A car accelerates from rest at 3 m/s² for 8 seconds. (a) What is its final velocity? (b) How far does it travel?

Solution

Given: v₀ = 0 m/s (starts from rest), a = 3 m/s², t = 8 s

(a) Final velocity:

v = v₀ + at = 0 + (3)(8) = 24 m/s

(b) Displacement:

d = v₀t + ½at² = 0(8) + ½(3)(8)² = ½(3)(64) = 96 m

Example 2: Free Fall

Problem: A ball is dropped from a height of 45 m. How long does it take to hit the ground? What is its velocity just before impact? (Use g = 10 m/s²)

Solution

Given: v₀ = 0, d = 45 m (down), g = 10 m/s² (down)

Time to fall:

d = v₀t + ½gt² → 45 = 0 + ½(10)t² → 45 = 5t² → t² = 9 → t = 3 s

Final velocity:

v = v₀ + gt = 0 + (10)(3) = 30 m/s (downward)

Example 3: Newton's Second Law

Problem: A 5 kg box is pushed with a force of 40 N on a frictionless surface. What is its acceleration?

Solution

Given: m = 5 kg, F = 40 N, frictionless (f = 0)

Apply F = ma:

40 = (5)a → a = 40/5 = 8 m/s²

Example 4: Friction

Problem: A 10 kg box sits on a surface with μs = 0.4 and μk = 0.3. (a) What is the maximum static friction force? (b) If pushed with 50 N, does it move? (c) If it's sliding, what is the kinetic friction force?

Solution

Given: m = 10 kg, g = 10 m/s², μs = 0.4, μk = 0.3

Normal force: N = mg = (10)(10) = 100 N

(a) Maximum static friction:

fs(max) = μsN = (0.4)(100) = 40 N

(b) With 50 N push:

Since 50 N > 40 N (max static friction), yes, it moves

(c) Kinetic friction:

fk = μkN = (0.3)(100) = 30 N

Example 5: Net Force with Friction

Problem: A 20 kg box is pushed with 80 N on a surface where μk = 0.2. What is the acceleration?

Solution

Given: m = 20 kg, Fa = 80 N, μk = 0.2, g = 10 m/s²

Normal force: N = mg = (20)(10) = 200 N

Kinetic friction: fk = μkN = (0.2)(200) = 40 N

Net force: ΣF = Fa - fk = 80 - 40 = 40 N

Acceleration: a = ΣF/m = 40/20 = 2 m/s²

✏️ Practice

Solve these mechanics problems on your own. Use g = 10 m/s² where needed.

1. A car traveling at 20 m/s brakes with an acceleration of -4 m/s². How far does it travel before stopping?

2. A ball is thrown straight up with an initial velocity of 25 m/s. How high does it go? How long until it returns to the starting point?

3. What net force is required to accelerate a 1500 kg car at 2.5 m/s²?

4. A 8 kg box is on a frictionless surface. If pushed with 24 N, what is its acceleration?

5. A 50 kg person stands on a scale in an elevator. What does the scale read when the elevator: (a) is at rest? (b) accelerates upward at 2 m/s²? (c) accelerates downward at 2 m/s²?

6. A 15 kg box rests on a surface with μs = 0.5. What is the minimum force needed to start it moving?

7. A 30 kg box is pushed with 100 N on a surface where μk = 0.2. What is its acceleration?

8. An object falls from rest. After 4 seconds, what is its velocity and how far has it fallen?

9. A 2000 kg car travels at 30 m/s. What braking force is needed to stop it in 50 m?

10. Two forces act on a 4 kg object: 20 N east and 12 N west. What is the object's acceleration?

Answer Key

Using v² = v₀² + 2ad: 0 = (20)² + 2(-4)d → d = 400/8 = 50 m
At highest point v = 0: v² = v₀² - 2gh → 0 = 625 - 20h → h = 31.25 m. Time up: t = v₀/g = 25/10 = 2.5 s. Total time = 2 × 2.5 = 5 s
F = ma = (1500)(2.5) = 3750 N
a = F/m = 24/8 = 3 m/s²
(a) At rest: Scale reads mg = (50)(10) = 500 N. (b) Accelerating up: N = m(g+a) = 50(12) = 600 N. (c) Accelerating down: N = m(g-a) = 50(8) = 400 N
fs(max) = μsN = (0.5)(15)(10) = 75 N
N = 300 N, fk = (0.2)(300) = 60 N. ΣF = 100 - 60 = 40 N. a = 40/30 = 1.33 m/s²
v = gt = (10)(4) = 40 m/s. d = ½gt² = ½(10)(16) = 80 m
v² = v₀² + 2ad → 0 = 900 + 2a(50) → a = -9 m/s². F = ma = (2000)(-9) = -18,000 N (braking force)
ΣF = 20 - 12 = 8 N east. a = 8/4 = 2 m/s² east

✅ Check Your Understanding

1. What is the difference between speed and velocity?

Show Answer

Speed is a scalar quantity—it tells you how fast an object is moving (magnitude only). Velocity is a vector quantity—it includes both how fast (magnitude) and in what direction. A car going 50 km/h has speed; a car going 50 km/h north has velocity.

2. Explain Newton's Third Law with an example.

Show Answer

Newton's Third Law states that for every action force, there is an equal and opposite reaction force. For example, when you push against a wall (action), the wall pushes back against you with equal force (reaction). When you walk, your foot pushes backward on the ground, and the ground pushes forward on your foot, propelling you forward.

3. Why is static friction typically greater than kinetic friction?

Show Answer

Static friction is greater because when surfaces are stationary relative to each other, microscopic irregularities (bumps and valleys) interlock more completely. Once motion begins, surfaces skim over each other more quickly, with less time for interlocking. This is why it's harder to start pushing a heavy box than to keep it moving.

4. An astronaut in space pushes a heavy object. What happens to the astronaut? Why?

Show Answer

The astronaut moves backward (in the opposite direction of the push). By Newton's Third Law, when the astronaut pushes the object, the object pushes back on the astronaut with equal force. In space, without friction or gravity, this reaction force causes the astronaut to accelerate backward. Since the astronaut has less mass than a heavy object, the astronaut will accelerate more (F = ma → a = F/m).

🚀 Next Steps

Review any concepts that felt challenging
Move on to the next lesson when ready
Return to practice problems periodically for review