FULL REVIEW
Full Review: Newton's Laws and Free-Body Diagrams
Express Newton's laws as vector differential equations and use integration or differential-equation models when forces vary with time, position, or velocity.
TIME
45–60 minutes
BEST FOR
A complete topic review
FINISH WITH
A readiness check
After this full review, you'll be able to...
move fluently between free-body diagrams, vector differential equations, impulse, and variable-force motion models while preserving the physical meaning of every force.
Choose how you want to review
Unit Alignment
This bundle is aligned to the approved Physics Sensei unit specification below. Use it to recover the unit structure, reinforce key decisions, and confirm readiness for the next study task.
ARCHITECTURE: Physics Sensei Independent Mechanics
UNIT: MEC-U05 — Newton's Laws and Free-Body Diagrams
SCOPE: Unit Review
PHYSICS LEVEL: Calculus-Based
BEST USED
✓ Before calculus-based dynamics homework
✓ When forces vary with time or velocity
✓ When you need to connect FBDs to differential equations
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 45–60 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Recall Activity
State Newton second law in differential form.
For constant mass, write Newton's second law using velocity and using momentum.
Reveal Answers
ΣF = m dv/dt and, more generally, ΣF = dp/dt.
Why it works: Force determines the time rate of change of momentum. For constant mass, dp/dt = m dv/dt.
ACTIVITY 2
Recall Activity
Recognize what an ODE solution needs.
After a free-body diagram gives m dv/dt = F(t), what additional information is needed to determine a unique velocity function?
Reveal Answers
An initial velocity (or another equivalent velocity condition).
Why it works: Integrating acceleration introduces a constant of integration that is fixed by the initial condition.
ACTIVITY 3
Recall Activity
Connect steady motion to zero net force.
For linear drag with downward positive, m dv/dt = mg - bv. What condition defines terminal speed?
Reveal Answers
dv/dt = 0, so mg - bv_t = 0.
Why it works: Terminal speed is the steady state of the differential equation: acceleration vanishes even though gravity and drag remain nonzero and balanced.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Let's rebuild the key ideas one step at a time. Focus on understanding the relationships before worrying about solving problems.
KEY CONCEPT 1
The free-body diagram becomes a vector differential equation
Choose a system and write ΣF_ext = dp/dt. For constant mass, each component gives m d²r/dt² = ΣF(r,v,t). The FBD supplies the physics; calculus evolves the state.
EXAMPLE If F(t) = (6t i + 4 j) N on a 2 kg particle, then dvₓ/dt = 3t and dvᵧ/dt = 2.
SENSEI NOTE Do not begin with integration before the force model and coordinate directions are correct.
KEY CONCEPT 2
Integrals connect force, momentum, velocity, and position
Integrating net force over time gives impulse and momentum change. Integrating acceleration gives velocity; integrating velocity gives position, with constants fixed by initial conditions.
EXAMPLE J = ∫ΣF dt = Δp. For constant mass, Δv = (1/m)∫ΣF dt.
SENSEI NOTE An integral is not decorative calculus; it is required when the force or acceleration varies over the interval.
KEY CONCEPT 3
Velocity- and position-dependent forces create differential-equation models
Drag, springs, and other variable forces require equations such as m dv/dt = mg - bv or m x¨ = -kx. Characteristic time or frequency emerges from the force law.
EXAMPLE Linear drag gives v_t = mg/b and time constant τ = m/b. A spring gives ω = √(k/m).
SENSEI NOTE State the sign convention and force direction before solving; most ODE sign errors begin in the FBD.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Now it's time to apply what you've reviewed.
Work through each activity in order. The examples become gradually more challenging, and each one prepares you for the final readiness check.
PRACTICE 1
Worked Example
Integrate a time-dependent vector force twice.
A 2.0 kg particle starts from rest at the origin under F(t) = (6t i + 4 j) N. Find v(2 s) and r(2 s).
Reveal Answers
v(2) = (6 i + 4 j) m/s; r(2) = (4 i + 4 j) m.
Why it works: a = F/m = (3t i + 2 j) m/s². Integrating with v(0)=0 gives v = (1.5t² i + 2t j). Integrating again with r(0)=0 gives r = (0.5t³ i + t² j). At t=2 s these are (6,4) m/s and (4,4) m.
PRACTICE 2
Guided Problem
Solve a first-order drag equation and identify its time scale.
A 1.5 kg object falls from rest with linear drag b = 3.0 N·s/m. Take downward positive and use g = 9.8 m/s². Find terminal speed and speed at t = 1.0 s.
Reveal Answers
v_t = 4.90 m/s; v(1.0 s) = 4.24 m/s downward.
Why it works: m dv/dt = mg - bv gives v(t)=v_t[1-e^{-t/τ}], where v_t=mg/b=4.90 m/s and τ=m/b=0.50 s. Thus v(1)=4.90(1-e^{-2})=4.24 m/s.
PRACTICE 3
Independent Problem
Solve a position-dependent force model.
A 2.0 kg mass experiences Fₓ = -8.0x N. If x(0)=0.30 m and v(0)=0, find x(t) and the velocity at t = π/4 s.
Reveal Answers
x(t)=0.30 cos(2t) m; v(π/4) = -0.60 m/s.
Why it works: The equation is 2x¨=-8x, so x¨+4x=0 and ω=2 rad/s. The initial conditions give x=0.30cos(2t), v=-0.60sin(2t). At t=π/4, sin(π/2)=1, so v=-0.60 m/s.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Momentum form
Identify the general law.
What is the most general compact form of Newton's second law for a particle system with momentum p?
Reveal Answers
ΣF_ext = dp/dt.
Why it works: The momentum form remains the fundamental statement; m dv/dt follows for constant mass.
QUICK CHECK 2
Impulse from variable force
Integrate force over time.
A 2.0 kg particle initially at rest experiences Fₓ = 4t N from t=0 to 3.0 s. What is its final x-velocity?
Reveal Answers
9.0 m/s.
Why it works: Impulse J = ∫₀³4t dt = 18 N·s = Δp. With m=2.0 kg and initial velocity zero, v = 18/2 = 9.0 m/s.
QUICK CHECK 3
Terminal condition
Read the steady state from the differential equation.
For m dv/dt = mg - bv with downward positive, what equality holds at terminal speed?
Reveal Answers
bv_t = mg.
Why it works: At terminal speed dv/dt=0, so the drag force exactly balances weight while the object continues at constant velocity.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
FBD first, differential equation second
Calculus evolves a correct physical model; it cannot repair a missing or misdirected force.
KEY TAKEAWAY 2
Integrals encode accumulated force effects
Impulse is the time integral of net force and determines momentum change.
KEY TAKEAWAY 3
Variable force laws set natural scales
Drag introduces τ=m/b and springs introduce ω=√(k/m), both derived directly from Newton second law.
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Next Step
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