Resistive Forces - AP Physics C Mechanics

An object is thrown downward with an initial speed greater than its terminal speed in a fluid with a resistive force proportional to speed. Which statement correctly describes the subsequent motion?
Answer: The object's speed decreases until it reaches its terminal speed. If the initial downward speed $v_0$ is greater than the terminal speed $v_T$ , the initial upward resistive force $kv_0$ is greater than the downward gravitational force $mg$ . This results in a net upward force, causing an upward acceleration (i.e., a deceleration). The object slows down, the resistive force decreases, and the speed approaches the terminal speed from above.
An object of mass $m$ is released from rest in a fluid that exerts a resistive force $F_r = kv$ , where $v$ is the object's speed. What is the initial acceleration of the object immediately after it is released?
Answer: $g$ Immediately after release, at $t=0$ , the object's speed is $v=0$ . Therefore, the resistive force $F_r = kv = 0$ . The only force acting on the object is gravity, $mg$ . By Newton's second law, $F_{net} = mg = ma$ , so the initial acceleration is $a = g$ .
An object is falling through a fluid and experiences a resistive force proportional to its velocity. The object reaches terminal velocity when which of the following conditions is met?
Answer: the magnitude of the resistive force equals the magnitude of the gravitational force. Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling equals the force of gravity. At this point, the net force on the object is zero, and its acceleration becomes zero. Therefore, the magnitude of the resistive force must be equal to the magnitude of the gravitational force, $mg$ .
An object of mass $m$ is dropped from rest and experiences a drag force of magnitude $F_r = kv$ , where $v$ is its speed and $k$ is a constant. Taking the downward direction as positive, which of the following differential equations correctly describes the object's motion?
Answer: $m \frac{dv}{dt} = mg - kv$ According to Newton's second law, $F_{net} = ma$ . The net force is the sum of the gravitational force ( $mg$ , acting downward, which is the positive direction) and the drag force ( $kv$ , acting upward, which is the negative direction). Therefore, $F_{net} = mg - kv$ . Since acceleration $a = dv/dt$ , the equation of motion is $m \frac{dv}{dt} = mg - kv$ .
A particle of mass $m$ moves in a straight line with velocity $v(t) = v_0 e^{-kt/m}$ . Its position at $t=0$ is $x=0$ . What is its position $x$ as a function of time $t$ ?
Answer: $x(t) = \frac{mv_0}{k}(1 - e^{-kt/m})$ Position is the integral of velocity with respect to time. $x(t) = \int_0^t v(\tau) d\tau = \int_0^t v_0 e^{-k\tau/m} d\tau$ . Evaluating the integral gives $x(t) = v_0 [-\frac{m}{k}e^{-k\tau/m}]_0^t = -\frac{mv_0}{k}(e^{-kt/m} - e^0) = \frac{mv_0}{k}(1 - e^{-kt/m})$ .
A ball of mass $m$ is falling through the air, where the drag force is $kv$ . At an instant when its speed is half of its terminal speed, what is the magnitude of its acceleration?
Answer: $g/2$ The terminal speed is $v_T = mg/k$ . The speed is given as $v = v_T/2 = mg/(2k)$ . The net force on the ball is $F_{net} = mg - kv = mg - k(mg/(2k)) = mg - mg/2 = mg/2$ . From Newton's second law, $F_{net} = ma$ , so $ma = mg/2$ , which gives an acceleration of $a = g/2$ .
The velocity of an object falling from rest through a fluid is described by the equation $v(t) = v_T(1 - e^{-t/\tau})$ , where $v_T$ is the terminal velocity and $\tau$ is the time constant. What is the object's acceleration at time $t=0$ ?
Answer: $g$ Acceleration is the derivative of velocity with respect to time: $a(t) = \frac{dv}{dt} = \frac{d}{dt}[v_T(1 - e^{-t/\tau})] = v_T(-e^{-t/\tau})(-\frac{1}{\tau}) = \frac{v_T}{\tau}e^{-t/\tau}$ . At $t=0$ , $a(0) = \frac{v_T}{\tau}e^0 = \frac{v_T}{\tau}$ . For linear drag, $v_T = mg/k$ and $\tau = m/k$ , so $a(0) = (mg/k)/(m/k) = g$ .
Two small spheres, A and B, have the same radius and shape, but sphere A has mass $M$ and sphere B has mass $2M$ . They are dropped from rest in air, and each experiences a drag force proportional to its speed, with the same proportionality constant $k$ . Let $v_{TA}$ and $v_{TB}$ be their respective terminal speeds. Which of the following is true?
Answer: $v_{TB} = 2v_{TA}$ The terminal speed $v_T$ for an object with linear drag is given by $v_T = mg/k$ . For sphere A, $v_{TA} = Mg/k$ . For sphere B, the mass is $2M$ , so its terminal speed is $v_{TB} = (2M)g/k = 2(Mg/k) = 2v_{TA}$ . The more massive object has a higher terminal speed.
An object of mass $m$ is moving horizontally with an initial speed $v_0$ . It is subject only to a resistive force of magnitude $F_r = kv$ , where $v$ is its speed and $k$ is a constant. The speed $v$ of the object as a function of time $t$ is given by which expression?
Answer: $v(t) = v_0 e^{-kt/m}$ The only horizontal force is the resistive force, so Newton's second law gives $F_{net} = -kv = m \frac{dv}{dt}$ . Separating variables gives $\frac{dv}{v} = -\frac{k}{m}dt$ . Integrating from $v_0$ to $v$ and from 0 to $t$ yields $\ln(v/v_0) = -kt/m$ . Exponentiating both sides gives $v(t) = v_0 e^{-kt/m}$ .
Based on the scenario, a $1200\,\text{kg}$ car travels at constant $25\,\text{m/s}$ on level road with rolling resistance $F_r=\mu mg$ , $\mu=0.020$ ; find $F_r$ .
Answer: $240\,\text{N}$ This question tests understanding of resistive forces in translational dynamics, specifically calculating rolling resistance force for a vehicle on a level road. Resistive forces, such as rolling resistance, oppose the motion of objects and are proportional to the normal force, which equals weight on level ground. In this scenario, the car experiences rolling resistance calculated using $F_r$ = μmg, where μ = 0.020, m = 1200 kg, and g = 9.8 m/s². Choice A is correct because it accurately calculates the resistive force: $F_r$ = 0.020 × 1200 × 9.8 = 235.2 N, which rounds to 240 N. Choice D is incorrect because it has the wrong units (kg instead of N), showing confusion between mass and force concepts. To help students: Emphasize the difference between mass and force, practice identifying when normal force equals weight, and reinforce proper unit analysis throughout calculations.
A motorboat engine provides a constant propulsive force $F_{app}$ . The boat experiences a resistive drag force from the water given by $F_r = kv$ , where $v$ is the boat's speed and $k$ is a constant. What is the maximum speed the boat can achieve?
Answer: $F_{app}/k$ The maximum speed, or terminal speed, is achieved when the net force on the boat is zero. The forces are the propulsive force $F_{app}$ and the drag force $kv$ . Setting the net force to zero gives $F_{net} = F_{app} - kv_{max} = 0$ . Solving for $v_{max}$ yields $v_{max} = F_{app}/k$ .
An object of mass $m$ is dropped from rest into a fluid with a resistive force $F_r = kv$ . Its velocity as a function of time $t$ is given by which of the following expressions?
Answer: $v(t) = \frac{mg}{k} (1 - e^{-kt/m})$ The differential equation is $m \frac{dv}{dt} = mg - kv$ . Solving this equation with the initial condition $v(0)=0$ yields the solution $v(t) = \frac{mg}{k}(1 - e^{-kt/m})$ . This function correctly shows the velocity starting at 0 and asymptotically approaching the terminal velocity $v_T = mg/k$ .
A small bead slides down a curved wire track. At one point, the bead moves in a circular path of radius 0.80 m with speed 6.0 m/s. If the coefficient of kinetic friction between bead and wire is 0.15, what is the magnitude of the net centripetal acceleration at this point?
Answer: 45 m/s², directed radially inward toward the center of curvature Centripetal acceleration depends only on speed and radius: $a_c = v^2/r = (6.0)^2/0.80 = 45$ m/s², directed radially inward. Friction affects tangential motion and energy loss, but doesn't change the geometric requirement for centripetal acceleration. The net centripetal force must provide exactly this acceleration regardless of friction. Choices B incorrectly subtract friction effects. Choices C and D have wrong direction for centripetal acceleration.
Considering the conditions described, a $75\,\text{kg}$ skydiver has drag $F_d=kv^2$ with $k=0.80\,\text{N\,s}^2/\text{m}^2$ ; calculate terminal velocity.
Answer: $30\,\text{m/s}$ This question tests understanding of resistive forces in translational dynamics, specifically determining terminal velocity when drag force equals gravitational force. Resistive forces, such as drag or friction, oppose the motion of objects and at terminal velocity, the net force becomes zero as drag balances weight. In this scenario, the skydiver reaches terminal velocity when $F_d$ = mg, so kv² = mg, giving v = √(mg/k) = √(75 × 9.8 / 0.80). Choice A is correct because it accurately calculates terminal velocity: v = √(735/0.80) = √(918.75) ≈ 30.3 m/s, which rounds to 30 m/s. Choice B is incorrect because it appears to use an incorrect formula or calculation method, possibly confusing the relationship between forces at terminal velocity. To help students: Emphasize the equilibrium condition at terminal velocity, practice setting up force balance equations, and ensure proper algebraic manipulation when solving for velocity from quadratic relationships.
Based on the scenario, a $70\,\text{kg}$ skydiver falls at $v=50\,\text{m/s}$ with drag $F_d=kv^2$ , $k=0.60\,\text{N\,s}^2/\text{m}^2$ ; what is $F_d$ ?
Answer: $1500\,\text{N}$ This question tests understanding of resistive forces in translational dynamics, specifically calculating drag force using the quadratic drag equation. Resistive forces, such as drag or friction, oppose the motion of objects and are often proportional to velocity or velocity squared, affecting acceleration and energy. In this scenario, the skydiver experiences a drag force calculated using $F_d$ = kv², where k = 0.60 N·s²/m² and v = 50 m/s. Choice B is correct because it accurately calculates the resistive force: $F_d$ = 0.60 × (50)² = 0.60 × 2500 = 1500 N, demonstrating the student's understanding of quadratic drag relationships. Choice C is incorrect because it appears to divide k by v² instead of multiplying, a common error when students misunderstand the drag equation. To help students: Emphasize the quadratic nature of air resistance at high speeds, practice substituting values carefully with proper units, and reinforce that drag force increases dramatically with velocity squared.
A parachutist of mass 70 kg reaches terminal velocity when the drag force equals 686 N. If the drag coefficient and air density remain constant, what would be the terminal velocity of the same parachutist carrying an additional 30 kg of equipment?
Answer: The terminal velocity increases by a factor of $\sqrt{100/70} = 1.20$ At terminal velocity, $mg = F_d = \frac{1}{2}\rho C_d A v^2$ . For constant air density, drag coefficient, and area: $v \propto \sqrt{mg}$ . The ratio is $v_2/v_1 = \sqrt{m_2/m_1} = \sqrt{100/70} = 1.20$ . Terminal velocity increases because more weight requires higher speed to generate equal drag. Choice B inverts the mass ratio. Choices C and D incorrectly suggest velocity decreases with increased mass.
An object with mass $m$ is projected horizontally with initial velocity $v_0$ on a surface where the only horizontal force is a resistive force given by $F = -kv$ . The velocity of the object as a function of time is $v(t) = v_0 e^{-kt/m}$ . What is the total distance the object travels as it slows to a stop?
Answer: $v_0 m / k$ The total distance is the integral of the velocity function from $t=0$ to $t=\infty$ . $x_{total} = \int_0^\infty v(t) dt = \int_0^\infty v_0 e^{-kt/m} dt$ . The integral evaluates to $v_0 [-\frac{m}{k}e^{-kt/m}]_0^\infty = -\frac{mv_0}{k}(0 - 1) = \frac{mv_0}{k}$ .
Considering the conditions described, a $90\,\text{kg}$ cyclist rides at $v=8.0\,\text{m/s}$ with air drag $F_d=kv^2$ , $k=0.25$ ; what is the resistive force magnitude?
Answer: $16\,\text{N}$ This question tests understanding of resistive forces in translational dynamics, specifically calculating air drag on a cyclist using the quadratic drag equation. Resistive forces, such as air drag, oppose the motion of cyclists and are proportional to velocity squared, becoming significant factors in cycling performance. In this scenario, the cyclist experiences drag force calculated using $F_d$ = kv², where k = 0.25 N·s²/m² and v = 8.0 m/s. Choice B is correct because it accurately calculates the resistive force: $F_d$ = 0.25 × (8.0)² = 0.25 × 64 = 16 N, demonstrating proper application of the drag equation. Choice C is incorrect because it appears to divide k by v² instead of multiplying, showing confusion about the mathematical relationship. To help students: Emphasize the multiplicative nature of the drag coefficient, practice substituting values systematically, and relate calculations to real-world cycling experiences where doubling speed quadruples air resistance.
A small stone of mass $m$ is dropped into a deep lake. The stone experiences a drag force of magnitude $F_r = cv^2$ , where $v$ is the speed of the stone and $c$ is a positive constant. What is the terminal speed of the stone?
Answer: $\sqrt{mg/c}$ Terminal speed is reached when the net force is zero. The forces are gravity ( $mg$ downward) and drag ( $cv^2$ upward). At terminal speed $v_T$ , we have $mg - cv_T^2 = 0$ . Solving for $v_T$ gives $cv_T^2 = mg$ , so $v_T^2 = mg/c$ , and $v_T = \sqrt{mg/c}$ .
A particle of mass $m$ is subject to a resistive force $F_r = -kv$ . The characteristic time constant for its motion is $\tau = m/k$ , which indicates how quickly the particle's velocity changes. If the mass of the particle is doubled while $k$ remains the same, how does the new time constant $\tau'$ compare to the original $\tau$ ?
Answer: $\tau' = 2\tau$ The time constant is given by the formula $\tau = m/k$ . If the new mass is $m' = 2m$ , the new time constant will be $\tau' = m'/k = (2m)/k = 2(m/k) = 2\tau$ . A larger mass means more inertia, so the object resists changes in velocity more, taking longer to approach its terminal velocity.

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