Part 1: Static vs Kinetic Friction Static vs Kinetic Friction
Part 1 of 7 — Friction & Inclines
Friction is one of the most common forces you'll encounter on the AP Physics C exam. Understanding the distinction between static and kinetic friction — and when each applies — is essential.
The Nature of Friction
Friction arises from microscopic interactions between surfaces in contact. At the AP Physics C level, we model friction with two simple laws:
Type Symbol Condition Formula Static f s f_s f s Object at rest f s ≤ μ s N f_s \leq \mu_s N f s ≤ μ s N Kinetic f k f_k f k Object sliding f k = μ k N f_k = \mu_k N f k = μ k N
Key Differences
Static friction is a variable force: it adjusts to match the applied force, up to a maximum f s , max = μ s N f_{s,\max} = \mu_s N f s , m a x = μ s N Kinetic friction is a constant force (for a given normal force): f k = μ k N f_k = \mu_k N f k = μ k N Typically μ s > μ k \mu_s > \mu_k μ s > μ k start motion than to maintain it.
The Normal Force
The normal force N N N
N = m g − F sin θ N = mg - F\sin\theta N = m g − F sin θ
where F F F θ \theta θ θ \theta θ
N = m g + F sin θ N = mg + F\sin\theta N = m g + F sin θ
Free Body Diagrams with Friction
When drawing FBDs involving friction, always:
Identify the contact surface — friction acts along it.Determine the normal force — perpendicular to the surface.Check if the object moves — this determines static vs kinetic.Direction of friction — opposes motion (kinetic) or opposes tendency of motion (static).
Worked Example
A 5 kg block is pushed along a floor by a force F = 40 F = 40 F = 40 30 ° 30° 30° μ k = 0.25 \mu_k = 0.25 μ k = 0.25
Step 1: Normal force (vertical equilibrium):
N = m g + F sin 30 ° = 5 ( 10 ) + 40 ( 0.5 ) = 70 N N = mg + F\sin 30° = 5(10) + 40(0.5) = 70 \text{ N} N = m g + F sin 30° = 5 ( 10 ) + 40 ( 0.5 ) = 70 N
Step 2: Kinetic friction:
f k = μ k N = 0.25 × 70 = 17.5 N f_k = \mu_k N = 0.25 \times 70 = 17.5 \text{ N} f k = μ k N = 0.25 × 70 = 17.5 N
Step 3: Horizontal acceleration:
F cos 30 ° − f k = m a F\cos 30° - f_k = ma F cos 30° − f k = ma 40 ( 0.866 ) − 17.5 = 5 a 40(0.866) - 17.5 = 5a 40 ( 0.866 ) − 17.5 = 5 a a = 34.64 − 17.5 5 = 3.43 m/s 2 a = \frac{34.64 - 17.5}{5} = 3.43 \text{ m/s}^2 a = 5 34.64 − 17.5 = 3.43 m/s 2
Key Insight: Pushing downward increases the normal force and thus the friction force. Pulling upward decreases it.
The Threshold of Motion
A critical AP Physics C skill is determining whether a block moves under a given force. The procedure:
Assume static: Calculate the force needed to keep the block stationary.Compare to f s , max f_{s,\max} f s , m a x If the required friction ≤ μ s N \leq \mu_s N ≤ μ s N If exceeded: The block accelerates. Switch to kinetic friction for the dynamics.
The Transition: Calculus Perspective
At the instant motion begins, the friction force drops from f s , max = μ s N f_{s,\max} = \mu_s N f s , m a x = μ s N f k = μ k N f_k = \mu_k N f k = μ k N
a ( t = 0 + ) = F applied − μ k N m a(t = 0^+) = \frac{F_{\text{applied}} - \mu_k N}{m} a ( t = 0 + ) = m F applied − μ k N
This discontinuity in friction is why objects often "jerk" when they start sliding.
Optimal Angle for Pulling
To minimize the pulling force F F F θ \theta θ
F = μ s m g cos θ + μ s sin θ F = \frac{\mu_s mg}{\cos\theta + \mu_s \sin\theta} F = c o s θ + μ s s i n θ μ s m g
Setting d F d θ = 0 \frac{dF}{d\theta} = 0 d θ d F = 0
sin θ − μ s cos θ = 0 ⟹ tan θ = μ s \sin\theta - \mu_s \cos\theta = 0 \implies \tan\theta = \mu_s sin θ − μ s cos θ = 0 ⟹ tan θ = μ s
θ opt = arctan ( μ s ) \theta_{\text{opt}} = \arctan(\mu_s) θ opt = arctan ( μ s )
Part 1 Summary
Concept Key Formula Static friction f s ≤ μ s N f_s \leq \mu_s N f s ≤ μ s N Kinetic friction f k = μ k N f_k = \mu_k N f k = μ k N Normal force (pull up) N = m g − F sin θ N = mg - F\sin\theta N = m g − F sin θ Normal force (push down) N = m g + F sin θ N = mg + F\sin\theta N = m g + F sin θ Optimal pull angle θ = arctan ( μ s ) \theta = \arctan(\mu_s) θ = arctan ( μ s )
Next up: Part 2 — Inclined Planes (No Friction), where we analyze motion on smooth ramps using calculus.
Part 2: Frictionless Inclines Inclined Planes (No Friction)
Part 2 of 7 — Friction & Inclines
Before adding friction, let's master the frictionless incline — the foundation for all ramp problems.
Setting Up Coordinates
For an incline at angle θ \theta θ
Direction Axis Forces Along the incline x x x m g sin θ mg\sin\theta m g sin θ Perpendicular to incline y y y N − m g cos θ = 0 N - mg\cos\theta = 0 N − m g cos θ = 0
Newton's Second Law on an Incline
Along the incline (taking down-the-ramp as positive):
m a = m g sin θ ma = mg\sin\theta ma = m g sin θ a = g sin θ a = g\sin\theta a = g sin θ
This is constant acceleration , so all kinematics equations apply:
v ( t ) = v 0 + ( g sin θ ) t v(t) = v_0 + (g\sin\theta)\,t v ( t ) = v 0 + ( g sin θ ) t x ( t ) = x 0 + v 0 t + 1 2 ( g sin θ ) t 2 x(t) = x_0 + v_0 t + \frac{1}{2}(g\sin\theta)\,t^2 x ( t ) = x 0 + v 0 t + 2 1 ( g sin θ ) t 2 v 2 = v 0 2 + 2 ( g sin θ ) Δ x v^2 = v_0^2 + 2(g\sin\theta)\Delta x v 2 = v 0 2 + 2 ( g sin θ ) Δ x
Energy Methods on Inclines
For a frictionless incline, mechanical energy is conserved:
1 2 m v 0 2 + m g h 0 = 1 2 m v 2 + m g h \frac{1}{2}mv_0^2 + mgh_0 = \frac{1}{2}mv^2 + mgh 2 1 m v 0 2 + m g h 0 = 2 1 m v 2 + m g h
The height change on a ramp of length L L L θ \theta θ
Δ h = L sin θ \Delta h = L\sin\theta Δ h = L sin θ
Worked Example: Launched Up a Ramp
A block is launched up a frictionless 45 ° 45° 45° v 0 = 10 v_0 = 10 v 0 = 10
How far up the ramp does it travel?
Using energy conservation:
1 2 m v 0 2 = m g ( L sin θ ) \frac{1}{2}mv_0^2 = mg(L\sin\theta) 2 1 m v 0 2 = m g ( L sin θ ) L = v 0 2 2 g sin θ = 100 2 ( 10 ) ( sin 45 ° ) = 100 14.14 = 7.07 m L = \frac{v_0^2}{2g\sin\theta} = \frac{100}{2(10)(\sin 45°)} = \frac{100}{14.14} = 7.07 \text{ m} L = 2 g s i n θ v 0 2 = 2 ( 10 ) ( s i n 45° ) 100 = 14.14 100 = 7.07 m
Time to reach the top:
Decelerating at a = − g sin 45 ° = − 7.07 a = -g\sin 45° = -7.07 a = − g sin 45° = − 7.07 m / s 2 m/s^{2} m / s 2 v = v 0 − ( g sin θ ) t = 0 v = v_0 - (g\sin\theta)t = 0 v = v 0 − ( g sin θ ) t = 0 t = v 0 g sin θ = 10 7.07 = 1.41 s t = \frac{v_0}{g\sin\theta} = \frac{10}{7.07} = 1.41 \text{ s} t = g s i n θ v 0 = 7.07 10 = 1.41 s
Velocity as a Function of Position
Using v 2 = v 0 2 − 2 g sin θ ⋅ x v^2 = v_0^2 - 2g\sin\theta \cdot x v 2 = v 0 2 − 2 g sin θ ⋅ x
v ( x ) = v 0 2 − 2 g x sin θ v(x) = \sqrt{v_0^2 - 2gx\sin\theta} v ( x ) = v 0 2 − 2 g x sin θ
This is valid until v = 0 v = 0 v = 0 x max = v 0 2 2 g sin θ x_{\max} = \frac{v_0^2}{2g\sin\theta} x m a x = 2 g s i n θ v 0 2
Calculus Approach: Variable Angle Ramps
What if the ramp angle changes? Consider a curved ramp where θ = θ ( x ) \theta = \theta(x) θ = θ ( x )
For a small displacement d x dx d x d h = sin θ ( x ) d x dh = \sin\theta(x)\,dx d h = sin θ ( x ) d x
The equation of motion becomes:
m d v d t = m g sin θ ( x ) m\frac{dv}{dt} = mg\sin\theta(x) m d t d v = m g sin θ ( x )
Using the chain rule: d v d t = v d v d x \frac{dv}{dt} = v\frac{dv}{dx} d t d v = v d x d v
m v d v d x = m g sin θ ( x ) mv\frac{dv}{dx} = mg\sin\theta(x) m v d x d v = m g sin θ ( x )
∫ v 0 v v ′ d v ′ = g ∫ 0 x sin θ ( x ′ ) d x ′ \int_{v_0}^{v} v'\,dv' = g\int_0^x \sin\theta(x')\,dx' ∫ v 0 v v ′ d v ′ = g ∫ 0 x sin θ ( x ′ ) d x ′
v 2 − v 0 2 2 = g ∫ 0 x sin θ ( x ′ ) d x ′ \frac{v^2 - v_0^2}{2} = g\int_0^x \sin\theta(x')\,dx' 2 v 2 − v 0 2 = g ∫ 0 x sin θ ( x ′ ) d x ′
Example: Parabolic Ramp
For a ramp shaped like y = x 2 2 L y = \frac{x^2}{2L} y = 2 L x 2 x x x
tan θ = d y d x = x L \tan\theta = \frac{dy}{dx} = \frac{x}{L} tan θ = d x d y = L x
For small angles: sin θ ≈ tan θ = x / L \sin\theta \approx \tan\theta = x/L sin θ ≈ tan θ = x / L
v 2 2 = g ∫ 0 x x ′ L d x ′ = g x 2 2 L \frac{v^2}{2} = g\int_0^x \frac{x'}{L}\,dx' = \frac{gx^2}{2L} 2 v 2 = g ∫ 0 x L x ′ d x ′ = 2 L g x 2
v ( x ) = x g L v(x) = x\sqrt{\frac{g}{L}} v ( x ) = x L g
The speed increases linearly with position — a result unique to this geometry.
Part 2 Summary
Concept Formula Acceleration on incline a = g sin θ a = g\sin\theta a = g sin θ Normal force N = m g cos θ N = mg\cos\theta N = m g cos θ Speed at bottom v = 2 g L sin θ = 2 g h v = \sqrt{2gL\sin\theta} = \sqrt{2gh} v = 2 g L sin θ = 2 g h Height change Δ h = L sin θ \Delta h = L\sin\theta Δ h = L sin θ Chain rule substitution a = v d v d x a = v\frac{dv}{dx} a = v d x d v
Next up: Part 3 — Inclined Planes with Friction, combining everything from Parts 1 and 2.
Part 3: Inclines with Friction Inclined Planes with Friction
Part 3 of 7 — Friction & Inclines
Now we combine friction with inclined planes — the bread and butter of AP Physics C mechanics problems.
Forces on a Rough Incline
For a block on an incline of angle θ \theta θ
Perpendicular to incline (y y y
N = m g cos θ N = mg\cos\theta N = m g cos θ
Along the incline (x x x
Scenario Net Force (down ramp +) Sliding down m g sin θ − μ k m g cos θ = m a mg\sin\theta - \mu_k mg\cos\theta = ma m g sin θ − μ k m g cos θ = ma Sliding up − m g sin θ − μ k m g cos θ = m a -mg\sin\theta - \mu_k mg\cos\theta = ma − m g sin θ − μ k m g cos θ = ma At rest (checking) m g sin θ ≤ μ s m g cos θ mg\sin\theta \leq \mu_s mg\cos\theta m g sin θ ≤ μ s m g cos θ
Critical Angle
A block starts sliding when the component of gravity along the ramp exceeds maximum static friction:
m g sin θ c = μ s m g cos θ c mg\sin\theta_c = \mu_s mg\cos\theta_c m g sin θ c = μ s m g cos θ c tan θ c = μ s \tan\theta_c = \mu_s tan θ c = μ s θ c = arctan ( μ s ) \theta_c = \arctan(\mu_s) θ c = arctan ( μ s )
This gives a clean way to measure μ s \mu_s μ s
Dynamics on Rough Inclines
Sliding Down
Acceleration (taking down-ramp as positive):
a down = g ( sin θ − μ k cos θ ) a_{\text{down}} = g(\sin\theta - \mu_k \cos\theta) a down = g ( sin θ − μ k cos θ )
The block slides down only if tan θ > μ s \tan\theta > \mu_s tan θ > μ s θ > θ c \theta > \theta_c θ > θ c
Sliding Up (Launched Upward)
When a block is launched up a rough incline, both gravity and friction decelerate it:
a up = − g ( sin θ + μ k cos θ ) a_{\text{up}} = -g(\sin\theta + \mu_k \cos\theta) a up = − g ( sin θ + μ k cos θ )
Caution: After the block stops, it may or may not slide back down. It stays at rest if tan θ ≤ μ s \tan\theta \leq \mu_s tan θ ≤ μ s
Worked Example
A block is launched up a 37 ° 37° 37° μ k = 0.3 \mu_k = 0.3 μ k = 0.3 v 0 = 12 v_0 = 12 v 0 = 12
Deceleration going up:
a up = − g ( sin 37 ° + 0.3 cos 37 ° ) = − 10 ( 0.6 + 0.24 ) = − 8.4 m/s 2 a_{\text{up}} = -g(\sin 37° + 0.3\cos 37°) = -10(0.6 + 0.24) = -8.4 \text{ m/s}^2 a up = − g ( sin 37° + 0.3 cos 37° ) = − 10 ( 0.6 + 0.24 ) = − 8.4 m/s 2
Distance traveled up:
v 2 = v 0 2 + 2 a Δ x = 0 v^2 = v_0^2 + 2a\Delta x = 0 v 2 = v 0 2 + 2 a Δ x = 0 Δ x = v 0 2 2 ∣ a ∣ = 144 16.8 = 8.57 m \Delta x = \frac{v_0^2}{2|a|} = \frac{144}{16.8} = 8.57 \text{ m} Δ x = 2∣ a ∣ v 0 2 = 16.8 144 = 8.57 m
Does it slide back? tan 37 ° = 0.75 > μ s \tan 37° = 0.75 > \mu_s tan 37° = 0.75 > μ s μ s ≈ 0.35 \mu_s \approx 0.35 μ s ≈ 0.35
Acceleration sliding back down:
a down = g ( sin 37 ° − 0.3 cos 37 ° ) = 10 ( 0.6 − 0.24 ) = 3.6 m/s 2 a_{\text{down}} = g(\sin 37° - 0.3\cos 37°) = 10(0.6 - 0.24) = 3.6 \text{ m/s}^2 a down = g ( sin 37° − 0.3 cos 37° ) = 10 ( 0.6 − 0.24 ) = 3.6 m/s 2
Speed at the bottom (returning):
v = 2 ( 3.6 ) ( 8.57 ) = 61.7 = 7.85 m/s v = \sqrt{2(3.6)(8.57)} = \sqrt{61.7} = 7.85 \text{ m/s} v = 2 ( 3.6 ) ( 8.57 ) = 61.7 = 7.85 m/s
Notice: v return < v 0 v_{\text{return}} < v_0 v return < v 0
Energy Methods with Friction
When friction is present, mechanical energy is not conserved. The work-energy theorem gives:
1 2 m v 2 − 1 2 m v 0 2 = W net = − m g h − μ k N ⋅ d \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 = W_{\text{net}} = -mgh - \mu_k N \cdot d 2 1 m v 2 − 2 1 m v 0 2 = W net = − m g h − μ k N ⋅ d
For a ramp of length d d d θ \theta θ
1 2 m v 2 = 1 2 m v 0 2 + m g d sin θ − μ k m g d cos θ \frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 + mgd\sin\theta - \mu_k mgd\cos\theta 2 1 m v 2 = 2 1 m v 0 2 + m g d sin θ − μ k m g d cos θ
The energy dissipated by friction:
Δ E thermal = μ k m g d cos θ = f k ⋅ d \Delta E_{\text{thermal}} = \mu_k mgd\cos\theta = f_k \cdot d Δ E thermal = μ k m g d cos θ = f k ⋅ d
Round Trip Energy Loss
For a block launched up and sliding back to the start:
Δ E total = 2 μ k m g d cos θ \Delta E_{\text{total}} = 2\mu_k mgd\cos\theta Δ E total = 2 μ k m g d cos θ
Since 1 2 m v return 2 = 1 2 m v 0 2 − 2 μ k m g d cos θ \frac{1}{2}mv_{\text{return}}^2 = \frac{1}{2}mv_0^2 - 2\mu_k mgd\cos\theta 2 1 m v return 2 = 2 1 m v 0 2 − 2 μ k m g d cos θ
v return = v 0 2 − 4 μ k g d cos θ v_{\text{return}} = \sqrt{v_0^2 - 4\mu_k gd\cos\theta} v return = v 0 2 − 4 μ k g d cos θ
where d = v 0 2 2 g ( sin θ + μ k cos θ ) d = \frac{v_0^2}{2g(\sin\theta + \mu_k\cos\theta)} d = 2 g ( s i n θ + μ k c o s θ ) v 0 2
Part 3 Summary
Concept Formula Critical angle θ c = arctan ( μ s ) \theta_c = \arctan(\mu_s) θ c = arctan ( μ s ) Accel. sliding down a = g ( sin θ − μ k cos θ ) a = g(\sin\theta - \mu_k\cos\theta) a = g ( sin θ − μ k cos θ ) Decel. sliding up a = − g ( sin θ + μ k cos θ ) a = -g(\sin\theta + \mu_k\cos\theta) a = − g ( sin θ + μ k cos θ ) Energy loss (one way) Δ E = μ k m g d cos θ \Delta E = \mu_k mgd\cos\theta Δ E = μ k m g d cos θ
Next up: Part 4 — Velocity-Dependent Friction, where we use differential equations to handle drag and other calculus-based friction models.
Part 4: Velocity-Dependent Friction Friction with Calculus (Velocity-Dependent)
Part 4 of 7 — Friction & Inclines
In many real-world situations, the resistive force depends on velocity. This is the hallmark of AP Physics C — using differential equations to solve dynamics problems.
Linear Drag: f = b v f = bv f = b v
A common model for low-speed drag:
m d v d t = F applied − b v m\frac{dv}{dt} = F_{\text{applied}} - bv m d t d v = F applied − b v
Terminal Velocity
At terminal velocity, a = 0 a = 0 a = 0
v T = F applied b v_T = \frac{F_{\text{applied}}}{b} v T = b F applied
For an object falling under gravity with linear drag:
v T = m g b v_T = \frac{mg}{b} v T = b m g
Solving the ODE
m d v d t = m g − b v m\frac{dv}{dt} = mg - bv m d t d v = m g − b v
Separation of variables:
d v m g − b v = d t m \frac{dv}{mg - bv} = \frac{dt}{m} m g − b v d v = m d t
− 1 b ln ( m g − b v m g ) = t m -\frac{1}{b}\ln\left(\frac{mg - bv}{mg}\right) = \frac{t}{m} − b 1 ln ( m g m g − b v ) = m t
v ( t ) = m g b ( 1 − e − b t / m ) = v T ( 1 − e − t / τ ) v(t) = \frac{mg}{b}\left(1 - e^{-bt/m}\right) = v_T\left(1 - e^{-t/\tau}\right) v ( t ) = b m g ( 1 − e − b t / m ) = v T ( 1 − e − t / τ )
where τ = m / b \tau = m/b τ = m / b time constant .
Quadratic Drag: f = c v 2 f = cv^2 f = c v 2
At higher speeds, drag is proportional to v 2 v^2 v 2
m d v d t = m g − c v 2 m\frac{dv}{dt} = mg - cv^2 m d t d v = m g − c v 2
Terminal Velocity
v T = m g c v_T = \sqrt{\frac{mg}{c}} v T = c m g
Solving the ODE
d v g − ( c / m ) v 2 = d t \frac{dv}{g - (c/m)v^2} = dt g − ( c / m ) v 2 d v = d t
Let α = c / m \alpha = c/m α = c / m
d v g − α v 2 = d t \frac{dv}{g - \alpha v^2} = dt g − α v 2 d v = d t
Using partial fractions or the substitution v = v T tanh ( u ) v = v_T \tanh(u) v = v T tanh ( u )
v ( t ) = v T tanh ( g t v T ) v(t) = v_T \tanh\left(\frac{gt}{v_T}\right) v ( t ) = v T tanh ( v T g t )
where v T = g / α = m g / c v_T = \sqrt{g/\alpha} = \sqrt{mg/c} v T = g / α = m g / c
Position by Integration
x ( t ) = ∫ 0 t v T tanh ( g t ′ v T ) d t ′ = v T 2 g ln cosh ( g t v T ) x(t) = \int_0^t v_T \tanh\left(\frac{gt'}{v_T}\right)dt' = \frac{v_T^2}{g}\ln\cosh\left(\frac{gt}{v_T}\right) x ( t ) = ∫ 0 t v T tanh ( v T g t ′ ) d t ′ = g v T 2 ln cosh ( v T g t )
Key Behavior
Time Speed Acceleration t = 0 t = 0 t = 0 0 0 0 g g g t → ∞ t \to \infty t → ∞ v T v_T v T 0 0 0 t = v T / g t = v_T/g t = v T / g ≈ 0.76 v T \approx 0.76\,v_T ≈ 0.76 v T ≈ 0.42 g \approx 0.42\,g ≈ 0.42 g
Velocity-Dependent Friction on Inclines
Consider a block sliding down an incline with velocity-dependent friction f = b v f = bv f = b v
m d v d t = m g sin θ − b v m\frac{dv}{dt} = mg\sin\theta - bv m d t d v = m g sin θ − b v
This has the same form as free fall with linear drag. The solution:
v ( t ) = m g sin θ b ( 1 − e − b t / m ) v(t) = \frac{mg\sin\theta}{b}\left(1 - e^{-bt/m}\right) v ( t ) = b m g s i n θ ( 1 − e − b t / m )
Terminal velocity on the incline:
v T = m g sin θ b v_T = \frac{mg\sin\theta}{b} v T = b m g s i n θ
Worked Example
A 0.5 kg block slides down a 30 ° 30° 30° f = 2 v f = 2v f = 2 v
Terminal velocity:
v T = m g sin θ b = 0.5 × 10 × 0.5 2 = 1.25 m/s v_T = \frac{mg\sin\theta}{b} = \frac{0.5 \times 10 \times 0.5}{2} = 1.25 \text{ m/s} v T = b m g s i n θ = 2 0.5 × 10 × 0.5 = 1.25 m/s
Time constant:
τ = m b = 0.5 2 = 0.25 s \tau = \frac{m}{b} = \frac{0.5}{2} = 0.25 \text{ s} τ = b m = 2 0.5 = 0.25 s
Speed at t = 0.5 t = 0.5 t = 0.5
v ( 0.5 ) = 1.25 ( 1 − e − 0.5 / 0.25 ) = 1.25 ( 1 − e − 2 ) = 1.25 ( 0.865 ) = 1.08 m/s v(0.5) = 1.25\left(1 - e^{-0.5/0.25}\right) = 1.25(1 - e^{-2}) = 1.25(0.865) = 1.08 \text{ m/s} v ( 0.5 ) = 1.25 ( 1 − e − 0.5/0.25 ) = 1.25 ( 1 − e − 2 ) = 1.25 ( 0.865 ) = 1.08 m/s
Acceleration at t = 0.5 t = 0.5 t = 0.5
a ( t ) = g sin θ ⋅ e − b t / m = 5 ⋅ e − 2 = 0.677 m/s 2 a(t) = g\sin\theta \cdot e^{-bt/m} = 5 \cdot e^{-2} = 0.677 \text{ m/s}^2 a ( t ) = g sin θ ⋅ e − b t / m = 5 ⋅ e − 2 = 0.677 m/s 2
Part 4 Summary
Model ODE Solution v T v_T v T Linear drag (f = b v f=bv f = b v m v ˙ = m g − b v m\dot{v} = mg - bv m v ˙ = m g − b v v T ( 1 − e − t / τ ) v_T(1-e^{-t/\tau}) v T ( 1 − e − t / τ ) m g / b mg/b m g / b Quadratic drag (f = c v 2 f=cv^2 f = c v 2 m v ˙ = m g − c v 2 m\dot{v} = mg - cv^2 m v ˙ = m g − c v 2 v T tanh ( g t / v T ) v_T\tanh(gt/v_T) v T tanh ( g t / v T ) m g / c \sqrt{mg/c} m g / c Linear drag, no driving m v ˙ = − b v m\dot{v} = -bv m v ˙ = − b v v 0 e − t / τ v_0 e^{-t/\tau} v 0 e − t / τ 0 0 0
AP Tip: The AP Physics C exam frequently tests your ability to set up and solve first-order ODEs with separation of variables.
Next up: Part 5 — Systems on Inclines (Atwood machines, connected blocks).
Part 5: Systems on Inclines Systems on Inclines
Part 5 of 7 — Friction & Inclines
Many AP Physics C problems involve multiple objects connected by strings, often on inclines. The key: apply Newton's second law to each object separately , then combine.
Atwood Machine on an Incline
A classic setup: mass m 1 m_1 m 1 θ \theta θ μ k \mu_k μ k m 2 m_2 m 2
Free Body Diagrams
Block on incline (m 1 m_1 m 1
T − m 1 g sin θ − μ k m 1 g cos θ = m 1 a T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a T − m 1 g sin θ − μ k m 1 g cos θ = m 1 a
Hanging block (m 2 m_2 m 2
m 2 g − T = m 2 a m_2 g - T = m_2 a m 2 g − T = m 2 a
Solving for a a a T T T
Adding the equations:
m 2 g − m 1 g sin θ − μ k m 1 g cos θ = ( m 1 + m 2 ) a m_2 g - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = (m_1 + m_2)a m 2 g − m 1 g sin θ − μ k m 1 g cos θ = ( m 1 + m 2 ) a
a = m 2 g − m 1 g ( sin θ + μ k cos θ ) m 1 + m 2 a = \frac{m_2 g - m_1 g(\sin\theta + \mu_k\cos\theta)}{m_1 + m_2} a = m 1 + m 2 m 2 g − m 1 g ( s i n θ + μ k c o s θ )
T = m 2 ( g − a ) = m 1 m 2 g ( 1 + sin θ + μ k cos θ ) m 1 + m 2 T = m_2(g - a) = \frac{m_1 m_2 g(1 + \sin\theta + \mu_k\cos\theta)}{m_1 + m_2} T = m 2 ( g − a ) = m 1 + m 2 m 1 m 2 g ( 1 + s i n θ + μ k c o s θ )
Two Blocks on Different Inclines
Consider masses m 1 m_1 m 1 m 2 m_2 m 2 θ 1 \theta_1 θ 1 θ 2 \theta_2 θ 2
Assuming m 1 m_1 m 1 m 2 m_2 m 2
For m 1 m_1 m 1
T − m 1 g sin θ 1 − μ 1 m 1 g cos θ 1 = m 1 a T - m_1 g\sin\theta_1 - \mu_1 m_1 g\cos\theta_1 = m_1 a T − m 1 g sin θ 1 − μ 1 m 1 g cos θ 1 = m 1 a
For m 2 m_2 m 2
m 2 g sin θ 2 − T − μ 2 m 2 g cos θ 2 = m 2 a m_2 g\sin\theta_2 - T - \mu_2 m_2 g\cos\theta_2 = m_2 a m 2 g sin θ 2 − T − μ 2 m 2 g cos θ 2 = m 2 a
Adding:
a = m 2 g ( sin θ 2 − μ 2 cos θ 2 ) − m 1 g ( sin θ 1 + μ 1 cos θ 1 ) m 1 + m 2 a = \frac{m_2 g(\sin\theta_2 - \mu_2\cos\theta_2) - m_1 g(\sin\theta_1 + \mu_1\cos\theta_1)}{m_1 + m_2} a = m 1 + m 2 m 2 g ( s i n θ 2 − μ 2 c o s θ 2 ) − m 1 g ( s i n θ 1 + μ 1 c o s θ 1 )
Checking Direction
Before solving, determine which way the system tends to move:
Compare m 1 g sin θ 1 m_1 g\sin\theta_1 m 1 g sin θ 1 m 2 g sin θ 2 m_2 g\sin\theta_2 m 2 g sin θ 2
The side with greater gravitational component tends to slide down.
If a < 0 a < 0 a < 0
Stacked Blocks on an Incline
For a block of mass m 1 m_1 m 1 m 2 m_2 m 2
The friction between blocks is what accelerates/decelerates m 1 m_1 m 1
If the blocks move together: treat as one system to find a a a
If they slide relative to each other: apply kinetic friction at the interface.
Constraint Equations
When blocks are connected by strings through pulleys, the constraint equation relates their accelerations.
Simple Constraint (Single String)
If a single string connects two blocks, their speeds are equal:
v 1 = v 2 ⟹ a 1 = a 2 = a v_1 = v_2 \implies a_1 = a_2 = a v 1 = v 2 ⟹ a 1 = a 2 = a
Pulley Ratio Constraint
If a string wraps around a movable pulley attached to block 2:
Δ x 2 = Δ x 1 2 ⟹ a 2 = a 1 2 \Delta x_2 = \frac{\Delta x_1}{2} \implies a_2 = \frac{a_1}{2} Δ x 2 = 2 Δ x 1 ⟹ a 2 = 2 a 1
And the tensions relate as:
T 2 = 2 T 1 T_2 = 2T_1 T 2 = 2 T 1
Worked Example: Pulley System
A 5 kg block on a 37 ° 37° 37° μ k = 0.25 \mu_k = 0.25 μ k = 0.25
With constraint a hang = 2 a incline a_{\text{hang}} = 2a_{\text{incline}} a hang = 2 a incline
Let a a a
Incline block: 2 T − m 1 g sin θ − μ k m 1 g cos θ = m 1 a 2T - m_1 g\sin\theta - \mu_k m_1 g\cos\theta = m_1 a 2 T − m 1 g sin θ − μ k m 1 g cos θ = m 1 a
Hanging block: m 2 g − T = m 2 ( 2 a ) m_2 g - T = m_2(2a) m 2 g − T = m 2 ( 2 a )
From the hanging block: T = m 2 g − 2 m 2 a = 3 ( 10 ) − 6 a = 30 − 6 a T = m_2 g - 2m_2 a = 3(10) - 6a = 30 - 6a T = m 2 g − 2 m 2 a = 3 ( 10 ) − 6 a = 30 − 6 a
Substituting: 2 ( 30 − 6 a ) − 50 sin 37 ° − 0.25 ( 50 ) cos 37 ° = 5 a 2(30 - 6a) - 50\sin 37° - 0.25(50)\cos 37° = 5a 2 ( 30 − 6 a ) − 50 sin 37° − 0.25 ( 50 ) cos 37° = 5 a
60 − 12 a − 30 − 10 = 5 a 60 - 12a - 30 - 10 = 5a 60 − 12 a − 30 − 10 = 5 a 20 = 17 a ⟹ a = 1.18 m/s 2 20 = 17a \implies a = 1.18 \text{ m/s}^2 20 = 17 a ⟹ a = 1.18 m/s 2
Part 5 Summary
System Type Key Approach Atwood on incline Separate FBDs, same a a a T T T Two inclines Compare gravitational components to find direction Stacked blocks Check if friction is sufficient; if not, blocks separate Pulley constraints Relate a 1 a_1 a 1 a 2 a_2 a 2
Next up: Part 6 — Problem-Solving Workshop with multi-step AP-style problems.
Part 6: Problem-Solving Workshop Problem-Solving Workshop
Part 6 of 7 — Friction & Inclines
This workshop presents AP Physics C–style problems that integrate concepts from Parts 1–5. Practice the systematic approach:
Problem-Solving Framework
Step Action 1 Draw a diagram and label all forces 2 Choose a coordinate system (tilted for inclines) 3 Write Newton's 2nd law for each object 4 Identify constraints (strings, contact) 5 Solve the system of equations 6 Check units and limiting cases
Problem 2: Block Launched Up a Ramp
A 3 kg block is launched up a 37 ° 37° 37° μ k = 0.3 \mu_k = 0.3 μ k = 0.3 v 0 = 10 v_0 = 10 v 0 = 10
Solution
Phase 1: Going up
a up = − g ( sin 37 ° + μ k cos 37 ° ) = − 10 ( 0.6 + 0.24 ) = − 8.4 m/s 2 a_{\text{up}} = -g(\sin 37° + \mu_k \cos 37°) = -10(0.6 + 0.24) = -8.4 \text{ m/s}^2 a up = − g ( sin 37° + μ k cos 37° ) = − 10 ( 0.6 + 0.24 ) = − 8.4 m/s 2
Distance traveled up:
0 = v 0 2 + 2 a up d ⟹ d = v 0 2 2 × 8.4 = 100 16.8 = 5.95 m 0 = v_0^2 + 2a_{\text{up}}d \implies d = \frac{v_0^2}{2 \times 8.4} = \frac{100}{16.8} = 5.95 \text{ m} 0 = v 0 2 + 2 a up d ⟹ d = 2 × 8.4 v 0 2 = 16.8 100 = 5.95 m
Time to reach top:
t 1 = v 0 ∣ a up ∣ = 10 8.4 = 1.19 s t_1 = \frac{v_0}{|a_{\text{up}}|} = \frac{10}{8.4} = 1.19 \text{ s} t 1 = ∣ a up ∣ v 0 = 8.4 10 = 1.19 s
Phase 2: Check if it slides back
tan 37 ° = 0.75 \tan 37° = 0.75 tan 37° = 0.75 μ s ≈ 0.35 < 0.75 \mu_s \approx 0.35 < 0.75 μ s ≈ 0.35 < 0.75
a down = g ( sin 37 ° − μ k cos 37 ° ) = 10 ( 0.6 − 0.24 ) = 3.6 m/s 2 a_{\text{down}} = g(\sin 37° - \mu_k \cos 37°) = 10(0.6 - 0.24) = 3.6 \text{ m/s}^2 a down = g ( sin 37° − μ k cos 37° ) = 10 ( 0.6 − 0.24 ) = 3.6 m/s 2
Speed at bottom:
v = 2 ( 3.6 ) ( 5.95 ) = 42.84 = 6.55 m/s v = \sqrt{2(3.6)(5.95)} = \sqrt{42.84} = 6.55 \text{ m/s} v = 2 ( 3.6 ) ( 5.95 ) = 42.84 = 6.55 m/s
Energy check: Energy dissipated = μ k m g cos θ × 2 d = 0.3 ( 30 ) ( 0.8 ) ( 11.9 ) = 85.7 \mu_k mg\cos\theta \times 2d = 0.3(30)(0.8)(11.9) = 85.7 μ k m g cos θ × 2 d = 0.3 ( 30 ) ( 0.8 ) ( 11.9 ) = 85.7
Initial KE = 1 2 ( 3 ) ( 100 ) = 150 \frac{1}{2}(3)(100) = 150 2 1 ( 3 ) ( 100 ) = 150 150 − 85.7 = 64.3 150 - 85.7 = 64.3 150 − 85.7 = 64.3 v = 2 ( 64.3 ) / 3 = 6.55 v = \sqrt{2(64.3)/3} = 6.55 v = 2 ( 64.3 ) /3 = 6.55
Problem 3: Velocity-Dependent Friction on a Ramp
A 1 kg block starts from rest at the top of a 45 ° 45° 45° f = 4 v f = 4v f = 4 v
(a) Terminal velocity:
v T = m g sin θ b = 1 ( 10 ) ( 2 2 ) 4 = 7.07 4 = 1.77 m/s v_T = \frac{mg\sin\theta}{b} = \frac{1(10)(\frac{\sqrt{2}}{2})}{4} = \frac{7.07}{4} = 1.77 \text{ m/s} v T = b m g s i n θ = 4 1 ( 10 ) ( 2 2 ) = 4 7.07 = 1.77 m/s
(b) Velocity as a function of time:
v ( t ) = v T ( 1 − e − b t / m ) = 1.77 ( 1 − e − 4 t ) v(t) = v_T(1 - e^{-bt/m}) = 1.77(1 - e^{-4t}) v ( t ) = v T ( 1 − e − b t / m ) = 1.77 ( 1 − e − 4 t )
(c) Distance traveled in 1 second:
x ( t ) = ∫ 0 t v ( t ′ ) d t ′ = v T [ t + m b e − b t / m ] 0 t = v T ( t + 1 4 e − 4 t − 1 4 ) x(t) = \int_0^t v(t')\,dt' = v_T\left[t + \frac{m}{b}e^{-bt/m}\right]_0^t = v_T\left(t + \frac{1}{4}e^{-4t} - \frac{1}{4}\right) x ( t ) = ∫ 0 t v ( t ′ ) d t ′ = v T [ t + b m e − b t / m ] 0 t = v T ( t + 4 1 e − 4 t − 4 1 )
x ( 1 ) = 1.77 ( 1 + 0.25 e − 4 − 0.25 ) = 1.77 ( 1 + 0.00458 − 0.25 ) = 1.77 ( 0.755 ) = 1.34 m x(1) = 1.77\left(1 + 0.25e^{-4} - 0.25\right) = 1.77(1 + 0.00458 - 0.25) = 1.77(0.755) = 1.34 \text{ m} x ( 1 ) = 1.77 ( 1 + 0.25 e − 4 − 0.25 ) = 1.77 ( 1 + 0.00458 − 0.25 ) = 1.77 ( 0.755 ) = 1.34 m
(d) Acceleration at t = 0.5 t = 0.5 t = 0.5
a ( t ) = g sin θ ⋅ e − b t / m = 7.07 e − 2 = 0.957 m/s 2 a(t) = g\sin\theta \cdot e^{-bt/m} = 7.07 e^{-2} = 0.957 \text{ m/s}^2 a ( t ) = g sin θ ⋅ e − b t / m = 7.07 e − 2 = 0.957 m/s 2
Workshop Summary
Common AP Pitfalls
Mistake Correction Wrong friction direction Always opposes motion (or tendency of motion) Forgetting N ≠ m g N \neq mg N = m g N = m g cos θ N = mg\cos\theta N = m g cos θ Using μ s \mu_s μ s Use μ k \mu_k μ k Ignoring the "does it move?" check Always compare applied force to f s , max f_{s,\max} f s , m a x Wrong sign on a a a Friction always opposes velocity direction
Next up: Part 7 — Review & Applications, consolidating everything with real-world contexts.
Part 7: Review & Applications Review & Applications
Part 7 of 7 — Friction & Inclines
Complete Topic Reference
Concept Formula Part Static friction f s ≤ μ s N f_s \leq \mu_s N f s ≤ μ s N 1 Kinetic friction f k = μ k N f_k = \mu_k N f k = μ k N 1 Incline acceleration (no friction) a = g sin θ a = g\sin\theta a = g sin θ 2 Incline acceleration (friction, down) a = g ( sin θ − μ k cos θ ) a = g(\sin\theta - \mu_k\cos\theta) a = g ( sin θ − μ k cos θ ) 3 Critical angle θ c = arctan ( μ s ) \theta_c = \arctan(\mu_s) θ c = arctan ( μ s ) 3 Linear drag terminal velocity v T = m g / b v_T = mg/b v T = m g / b 4 Quadratic drag terminal velocity v T = m g / c v_T = \sqrt{mg/c} v T = m g / c 4 Falling with linear drag v = v T ( 1 − e − b t / m ) v = v_T(1-e^{-bt/m}) v = v T ( 1 − e − b t / m ) 4 Atwood system on incline a = m 2 g − m 1 g ( sin θ + μ k cos θ ) m 1 + m 2 a = \frac{m_2 g - m_1 g(\sin\theta + \mu_k\cos\theta)}{m_1+m_2} a = m 1 + m 2 m 2 g − m 1 g ( s i n θ + μ k c o s θ ) 5
Application: Braking on a Hill
A car of mass m m m v 0 v_0 v 0 θ \theta θ F b F_b F b
Equation of motion:
m a = m g sin θ − F b ma = mg\sin\theta - F_b ma = m g sin θ − F b
Stopping distance:
d = m v 0 2 2 ( F b − m g sin θ ) d = \frac{mv_0^2}{2(F_b - mg\sin\theta)} d = 2 ( F b − m g s i n θ ) m v 0 2
This is valid only if F b > m g sin θ F_b > mg\sin\theta F b > m g sin θ
Critical Braking Condition
The minimum braking force to stop on a hill:
F b > m g sin θ F_b > mg\sin\theta F b > m g sin θ
If the tires provide braking through friction: F b = μ k N = μ k m g cos θ F_b = \mu_k N = \mu_k mg\cos\theta F b = μ k N = μ k m g cos θ
The car can stop only if:
μ k m g cos θ > m g sin θ ⟹ μ k > tan θ \mu_k mg\cos\theta > mg\sin\theta \implies \mu_k > \tan\theta μ k m g cos θ > m g sin θ ⟹ μ k > tan θ
This is why steep icy hills (μ k \mu_k μ k θ \theta θ
With ABS (Maintaining Static Friction)
Modern ABS brakes prevent wheel lock, using μ s \mu_s μ s μ k \mu_k μ k
d ABS = v 0 2 2 g ( μ s cos θ − sin θ ) d_{\text{ABS}} = \frac{v_0^2}{2g(\mu_s\cos\theta - \sin\theta)} d ABS = 2 g ( μ s c o s θ − s i n θ ) v 0 2
Since μ s > μ k \mu_s > \mu_k μ s > μ k
Application: Drag on a Falling Object
A skydiver (m = 80 m = 80 m = 80 f = 0.25 v 2 f = 0.25v^2 f = 0.25 v 2
Terminal velocity:
v T = m g c = 800 0.25 = 3200 = 56.6 m/s ≈ 127 mph v_T = \sqrt{\frac{mg}{c}} = \sqrt{\frac{800}{0.25}} = \sqrt{3200} = 56.6 \text{ m/s} \approx 127 \text{ mph} v T = c m g = 0.25 800 = 3200 = 56.6 m/s ≈ 127 mph
Velocity as a function of time:
v ( t ) = 56.6 tanh ( 10 t 56.6 ) = 56.6 tanh ( 0.177 t ) v(t) = 56.6\tanh\left(\frac{10t}{56.6}\right) = 56.6\tanh(0.177t) v ( t ) = 56.6 tanh ( 56.6 10 t ) = 56.6 tanh ( 0.177 t )
At t = 5 t = 5 t = 5
v ( 5 ) = 56.6 tanh ( 0.883 ) = 56.6 ( 0.708 ) = 40.1 m/s v(5) = 56.6\tanh(0.883) = 56.6(0.708) = 40.1 \text{ m/s} v ( 5 ) = 56.6 tanh ( 0.883 ) = 56.6 ( 0.708 ) = 40.1 m/s
At t = 15 t = 15 t = 15
v ( 15 ) = 56.6 tanh ( 2.65 ) = 56.6 ( 0.990 ) = 56.0 m/s v(15) = 56.6\tanh(2.65) = 56.6(0.990) = 56.0 \text{ m/s} v ( 15 ) = 56.6 tanh ( 2.65 ) = 56.6 ( 0.990 ) = 56.0 m/s
The skydiver is essentially at terminal velocity after about 15 seconds.
🎉 Topic Complete: Friction & Inclines
You've mastered the full AP Physics C treatment of friction and inclines:
Part Topic Status 1 Static vs kinetic friction ✅ 2 Inclined planes (no friction) ✅ 3 Inclined planes with friction ✅ 4 Velocity-dependent friction (calculus) ✅ 5 Systems on inclines ✅ 6 Problem-solving workshop ✅ 7 Review & applications ✅
Key Takeaway: On the AP exam, friction problems test your ability to (1) set up correct FBDs, (2) handle the static/kinetic transition, and (3) solve differential equations for velocity-dependent forces.