AP Physics C Mechanics- 2.9 Resistive Forces- Study Notes- New Syllabus
AP Physics C Mechanics- 2.9 Resistive Forces – Study Notes
AP Physics C Mechanics- 2.9 Resistive Forces – Study Notes – per latest Syllabus.
Key Concepts:
- Resistive Forces
- Newton’s Second Law with a Resistive Force
- Terminal Velocity
Resistive Forces
A resistive force (also known as a damping force or drag force) is a force that acts opposite to the direction of an object’s velocity and depends on the object’s speed. It always acts to reduce the object’s motion by converting kinetic energy into thermal energy or dissipating it through the surrounding medium (such as air or fluid).
Mathematical Expression:
\( \mathrm{\vec{F}_r = -k\vec{v}} \)
- \( \mathrm{\vec{F}_r} \): resistive (drag) force (N)
- \( \mathrm{k} \): proportionality constant (N·s/m), depends on properties of the medium and object’s shape
- \( \mathrm{\vec{v}} \): velocity vector of the object (m/s)
Key Idea:
- The negative sign indicates the force acts in the opposite direction to the velocity.
- At low speeds (e.g., motion through viscous fluids), resistive force is often proportional to velocity (\( \mathrm{F \propto v} \)).
- At higher speeds (e.g., air resistance), resistive force can depend on the square of velocity (\( \mathrm{F \propto v^2} \)).
Physical Meaning:
Resistive forces oppose motion — they slow down moving objects and prevent indefinite acceleration under constant driving forces. This concept is essential for understanding terminal velocity, damping, and real-world motion in fluids or air.
Example
A small object moves through a viscous fluid such that the resistive force on it is given by \( \mathrm{F_r = -k v} \), where \( \mathrm{k = 0.8\,N·s/m.} \) If the object is moving at \( \mathrm{v = 3.5\,m/s,} \) determine the magnitude and direction of the resistive force acting on it.
▶️ Answer / Explanation
Step 1: Use the resistive force formula:
\( \mathrm{F_r = -k v} \)
Step 2: Substitute the given values:
\( \mathrm{F_r = -(0.8)(3.5) = -2.8\,N.} \)
Step 3: Interpret the result:
- The negative sign indicates that the resistive force acts opposite to the object’s direction of motion.
- The magnitude of the resistive force is \( \mathrm{2.8\,N.} \)
Result: The object experiences a resistive force of \( \mathrm{2.8\,N} \) acting in the direction opposite to its velocity.
Newton’s Second Law with a Resistive Force
When an object moves through a medium (such as air or water), it experiences a resistive force that depends on its velocity. Applying Newton’s Second Law to such motion leads to a differential equation for the object’s velocity.
Starting Point (Newton’s Second Law):
\( \mathrm{\sum F = m\dfrac{dv}{dt}} \)
For a downward-moving object with gravity and a resistive force \( \mathrm{F_r = -kv} \):
\( \mathrm{mg – kv = m\dfrac{dv}{dt}} \)
This is a first-order linear differential equation describing how velocity changes over time.
Separation of Variables
Method:
Rearranging terms allows separation of variables for integration:
\( \mathrm{\dfrac{dv}{(mg – kv)} = \dfrac{dt}{m}} \)
Integrating both sides over the appropriate limits yields the velocity as a function of time.
Integration Result:
\( \mathrm{v(t) = \dfrac{mg}{k}\left(1 – e^{-\tfrac{k}{m}t}\right)} \)
Key Idea: This shows that velocity increases exponentially toward a limiting value (terminal velocity) determined by \( \mathrm{mg/k.} \)
Determining Acceleration and Position
Once the velocity function \( \mathrm{v(t)} \) is known, both acceleration and position can be derived using calculus.
- Acceleration: \( \mathrm{a(t) = \dfrac{dv}{dt} = \dfrac{gk}{m} e^{-\tfrac{k}{m}t}} \)
- Position: \( \mathrm{x(t) = \int v(t)\,dt = \dfrac{mg}{k}t – \dfrac{m^2g}{k^2}\left(1 – e^{-\tfrac{k}{m}t}\right)} \)
Interpretation:
- Initially, acceleration is maximum (equal to \( \mathrm{g} \)) but decreases exponentially over time.
- Velocity increases asymptotically to a constant terminal velocity \( \mathrm{v_t = \dfrac{mg}{k}} \).
- Position increases nonlinearly, approaching a linear relationship at large times when velocity becomes constant.
Exponential Motion and Asymptotic Behavior
The velocity, acceleration, and position functions for an object moving under a velocity-dependent resistive force (\( \mathrm{F_r = -kv} \)) are all exponential in time and approach asymptotic limits determined by the object’s initial conditions and external forces.
Functional Forms:
- Velocity: \( \mathrm{v(t) = v_t(1 – e^{-\tfrac{k}{m}t})} \)
- Acceleration: \( \mathrm{a(t) = g e^{-\tfrac{k}{m}t}} \)
- Position: \( \mathrm{x(t) = v_t t – \dfrac{m v_t}{k}(1 – e^{-\tfrac{k}{m}t})} \)
Asymptotic Behavior:
- As \( \mathrm{t \to \infty} \): \( \mathrm{v(t) \to v_t, \, a(t) \to 0, \, x(t) \to v_t t.} \)
- The object eventually moves at a constant terminal velocity where drag balances weight.
Example
A raindrop falls through air, experiencing both gravity and air resistance. Initially, the drop accelerates due to gravity, but as its velocity increases, air resistance grows. Over time, the resistive force equals the drop’s weight, and the acceleration becomes zero — the raindrop continues to fall at a constant terminal speed. Explain this.
▶️ Answer / Explanation
Step 1: Apply Newton’s Second Law:
\( \mathrm{mg – kv = m\dfrac{dv}{dt}} \)
Step 2: Solve the differential equation using separation of variables:
\( \mathrm{v(t) = \dfrac{mg}{k}\left(1 – e^{-\tfrac{k}{m}t}\right)} \)
Step 3: Analyze behavior:
- At \( \mathrm{t = 0} \): \( \mathrm{v = 0} \) (starts from rest)
- As \( \mathrm{t \to \infty} \): \( \mathrm{v \to \dfrac{mg}{k}} \) — terminal velocity reached
Step 4: Conclusion:
The velocity follows an exponential increase toward a maximum, while acceleration decreases exponentially toward zero. The position function also approaches a linear asymptote. These behaviors illustrate how resistive forces produce motion that stabilizes over time instead of increasing indefinitely.
Example
A \( \mathrm{0.5\,kg} \) ball is dropped from rest and experiences a resistive force proportional to its velocity: \( \mathrm{F_r = -0.2v.} \) Calculate its velocity after \( \mathrm{5\,s} \) of falling. Take \( \mathrm{g = 9.8\,m/s^2.} \)
▶️ Answer / Explanation
Step 1: Write Newton’s second law with resistive force:
\( \mathrm{m\dfrac{dv}{dt} = mg – kv.} \)
Step 2: Substitute constants and solve the differential equation solution form:
\( \mathrm{v(t) = \dfrac{mg}{k}\left(1 – e^{-\tfrac{k}{m}t}\right)} \)
Step 3: Substitute values:
\( \mathrm{v(t) = \dfrac{(0.5)(9.8)}{0.2}\left(1 – e^{-\tfrac{0.2}{0.5}(5)}\right)} \)
\( \mathrm{v(t) = 24.5(1 – e^{-2})} \)
\( \mathrm{e^{-2} \approx 0.135, \; \therefore v(t) = 24.5(1 – 0.135) = 24.5(0.865) = 21.17\,m/s.} \)
Step 4: Interpretation:
- The velocity approaches the terminal value \( \mathrm{v_t = \dfrac{mg}{k} = 24.5\,m/s.} \)
- After \( \mathrm{5\,s,} \) the ball’s velocity is \( \mathrm{21.2\,m/s,} \) still below terminal speed.
Terminal Velocity
Terminal velocity is the maximum constant speed that an object reaches while moving through a fluid (like air or water) when the resistive force acting opposite to the motion exactly balances the driving force (usually gravity).
At terminal velocity:
\( \mathrm{F_{net} = 0} \)
\( \mathrm{F_{gravity} = F_{resistive}} \)
Derivation:
If an object of mass \( \mathrm{m} \) falls under gravity and experiences a resistive force \( \mathrm{F_r = k v} \):
\( \mathrm{mg = kv_t} \)
Terminal Velocity Expression:
\( \mathrm{v_t = \dfrac{mg}{k}} \)
- \( \mathrm{v_t} \): terminal velocity (m/s)
- \( \mathrm{m} \): mass of the object (kg)
- \( \mathrm{k} \): resistive constant (N·s/m)
- \( \mathrm{g} \): acceleration due to gravity (m/s²)
Key Idea: At terminal velocity, the downward gravitational force and the upward resistive force are equal in magnitude and opposite in direction, so acceleration becomes zero.
Example
A raindrop of mass \( \mathrm{3.0\times10^{-5}\,kg} \) falls through air and experiences a resistive force given by \( \mathrm{F_r = k v,} \) where \( \mathrm{k = 1.2\times10^{-4}\,N·s/m.} \) Determine the terminal velocity of the raindrop.
▶️ Answer / Explanation
Step 1: At terminal velocity, \( \mathrm{mg = kv_t.} \)
\( \mathrm{v_t = \dfrac{mg}{k}} \)
Step 2: Substitute values:
\( \mathrm{v_t = \dfrac{(3.0\times10^{-5})(9.8)}{1.2\times10^{-4}}} \)
\( \mathrm{v_t = \dfrac{2.94\times10^{-4}}{1.2\times10^{-4}} = 2.45\,m/s.} \)
Result: The raindrop reaches a terminal velocity of approximately \( \mathrm{2.45\,m/s,} \) falling at a steady speed when the drag force balances its weight.