AP Physics C Mechanics Quiz: Resistive Forces

20 questions · exam conditions

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Resistive ForcesQuestion 1 of 20

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?

v_0 k / m

The object never stops, so the distance is infinite.

v_0 m / k

v_0 m / (2k)

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Resistive Forces

Practice Resistive Forces in AP Physics C Mechanics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Resistive Forces, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics C Mechanics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$v_0 k / m$
The object never stops, so the distance is infinite.
$v_0 m / k$ (correct answer)
$v_0 m / (2k)$

Explanation: The total distance is the integral of the velocity function from

t=0

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}

Question 2

A boat of mass $m$ starts from rest at $t=0$ . A constant force $F_{app}$ is applied, and it experiences a resistive force $F_r = -kv$ . The boat's speed $v$ as a function of time $t$ is given by which expression?

$v(t) = \frac{F_{app}}{k}(1 - e^{-kt/m})$ (correct answer)
$v(t) = \frac{F_{app}}{k} e^{-kt/m}$
$v(t) = \frac{F_{app}}{m}t - \frac{k}{2m}t^2$
$v(t) = \frac{F_{app}}{m}(1 - e^{-kt/m})$

Explanation: The differential equation for the boat's motion is

m \frac{dv}{dt} = F_{app} - kv

. This equation is mathematically analogous to that of a falling object, with the constant gravitational force

mg

replaced by the constant applied force

F_{app}

. The solution, with the initial condition

v(0)=0

, is therefore

v(t) = \frac{F_{app}}{k}(1 - e^{-kt/m})

, where

F_{app}/k

is the terminal velocity.

Question 3

A sphere of mass $m$ is released from rest in a viscous fluid. The fluid exerts a resistive force with magnitude $F_r = kv$ , where $v$ is the speed of the sphere and $k$ is a positive constant. What is the terminal speed of the sphere?

$k/mg$
$mg/k$ (correct answer)
$\sqrt{mg/k}$
$mgk$

Explanation: Terminal speed is reached when the net force on the sphere is zero. The forces acting on the sphere are gravity (

mg

downward) and the resistive force (

kv

upward). Setting the net force to zero:

\sum F = mg - kv_T = 0

. Solving for the terminal speed

v_T

gives

v_T = mg/k

Question 4

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?

$m \frac{dv}{dt} = mg - kv$ (correct answer)
$m \frac{dv}{dt} = mg + kv$
$m \frac{dv}{dt} = -kv$
$m \frac{dv}{dt} = kv - mg$

Explanation: 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

Question 5

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?

$v_{TB} = v_{TA} / 2$
$v_{TB} = v_{TA}$
$v_{TB} = 2v_{TA}$ (correct answer)
$v_{TB} = \sqrt{2}v_{TA}$

Explanation: 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.

Question 6

A motorboat of mass $m$ is propelled by a constant force $F_{app}$ . The water exerts a resistive force of magnitude $kv$ , where $v$ is the speed of the boat. Which differential equation describes the motion of the boat?

$m \frac{dv}{dt} = F_{app} + kv$
$m \frac{dv}{dt} = F_{app} - kv$ (correct answer)
$m \frac{dv}{dt} = kv - F_{app}$
$m \frac{dv}{dt} = -F_{app} - kv$

Explanation: Using Newton's second law,

F_{net} = ma

. The net force is the sum of the applied force

F_{app}

(in the direction of motion) and the resistive force

kv

(opposite to the direction of motion). Thus,

F_{net} = F_{app} - kv

. Setting this equal to

ma = m \frac{dv}{dt}

gives the correct differential equation.

Question 7

Based on the scenario, a $600\,\text{kg}$ boat has drag $F_d=kv^2$ with $k=100\,\text{N\,s}^2/\text{m}^2$ ; calculate terminal speed if engine thrust is constant $1600\,\text{N}$ .

$4.0\,\text{m/s}$ (correct answer)
$16\,\text{m/s}$
$40\,\text{m/s}$
$2.7\,\text{m/s}$

Explanation: This question tests understanding of resistive forces in translational dynamics, specifically finding terminal velocity when thrust balances drag force. Resistive forces, such as water drag, oppose motion and at terminal speed, the propulsive thrust exactly balances the drag force for zero acceleration. In this scenario, the boat reaches terminal speed when thrust equals drag: 1600 = kv², so v = √(1600/k) = √(1600/100) = √16 = 4.0 m/s. Choice A is correct because it accurately calculates terminal speed using the force balance equation, demonstrating understanding of equilibrium conditions. Choice C is incorrect because it appears to use v = √(F×k) instead of v = √(F/k), a common algebraic manipulation error. To help students: Emphasize setting up equilibrium equations correctly, practice algebraic manipulation of quadratic relationships, and verify units throughout the calculation process.

Question 8

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?

$v(t) = v_0 - \frac{k}{m}t$
$v(t) = v_0(1 - e^{-kt/m})$
$v(t) = v_0 e^{-kt/m}$ (correct answer)
$v(t) = v_0 + \frac{k}{m}t$

Explanation: 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

v

and from 0 to

t

yields

\ln(v/v_0) = -kt/m

. Exponentiating both sides gives

v(t) = v_0 e^{-kt/m}

Question 9

A skydiver jumps from a plane and falls vertically. The resistive force of the air is proportional to the skydiver's speed. Which statement best describes the skydiver's motion before the parachute is opened?

Speed increases at a constant rate, and acceleration remains constant at $g$ .
Speed increases and approaches a constant value, while acceleration decreases from $g$ to zero. (correct answer)
Speed increases exponentially without bound, while acceleration remains constant.
Speed remains constant throughout the entire fall, and acceleration is always zero.

Explanation: Initially, the only force is gravity, so acceleration is

g

. As speed increases, the upward resistive force increases, which reduces the net downward force and thus reduces the acceleration. The speed continues to increase, but at a decreasing rate, until the resistive force equals the force of gravity. At this point, the net force is zero, acceleration is zero, and the speed becomes constant (terminal velocity).

Question 10

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?

$mg/c$
$\sqrt{c/mg}$
$c/(mg)^2$
$\sqrt{mg/c}$ (correct answer)

Explanation: 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}

Question 11

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 $t=0$ ?

0
$g$ (correct answer)
$v_T$
It cannot be determined without the value of $v_T$ .

Explanation: 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

Question 12

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 $\tau$ ?

$\tau' = \tau / 2$
$\tau' = 2\tau$ (correct answer)
$\tau' = \tau$
$\tau' = 4\tau$

Explanation: 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.

Question 13

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?

0
$g - k/m$
$g$ (correct answer)
$mg/k$

Explanation: 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

Question 14

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?

$F_{app}k$
$F_{app}/(mk)$
$\sqrt{F_{app}/k}$
$F_{app}/k$ (correct answer)

Explanation: 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

Question 15

An object of mass $m$ falls through a fluid with a resistive force $F_r = kv$ . After a long time, the object reaches a terminal speed $v_T$ . At this speed, what is the rate at which mechanical energy is being dissipated by the resistive force?

$mgv_T / 2$
$mgv_T$ (correct answer)
$k(v_T)^2/m$
Zero, because the kinetic energy is constant.

Explanation: The rate of energy dissipation by a force is power, given by

P = F \cdot v

. For the resistive force,

P_{dissipated} = F_r v

. At terminal speed

v_T

, the resistive force magnitude is equal to the gravitational force,

F_r = mg

. Therefore, the power dissipated is

P = (mg)v_T

Question 16

The motion of a falling object with linear drag is described by $m \frac{dv}{dt} = mg - kv$ . To solve for $v(t)$ using separation of variables, the equation must be rearranged for integration. Which of the following shows the correctly separated equation ready for integration from an initial velocity $v_0$ ?

$\int_{v_0}^v \frac{m}{mg - kv} dv = \int_0^t dt$
$\int_{v_0}^v \frac{1}{mg - kv} dv = \int_0^t \frac{1}{m} dt$ (correct answer)
$\int_{v_0}^v (mg - kv) dv = \int_0^t m dt$
$\int_{v_0}^v dv = \int_0^t (g - \frac{kv}{m}) dt$

Explanation: To separate variables, all terms involving

v

and

dv

must be on one side, and all terms involving

t

and

dt

on the other. Starting with

m \frac{dv}{dt} = mg - kv

, we can divide by

m

and multiply by

dt

to get

dv = (g - \frac{k}{m}v) dt

. Then, we divide by the term in parentheses:

\frac{dv}{g - \frac{k}{m}v} = dt

, which is equivalent to

\frac{dv}{mg - kv} = \frac{dt}{m}

. Integrating both sides gives the expression in choice B.

Question 17

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?

The object's speed increases until it reaches a new, higher terminal speed.
The object's speed decreases until it reaches its terminal speed. (correct answer)
The object's speed remains constant at its initial value because it is already moving fast.
The object immediately reverses direction and moves upward toward the surface.

Explanation: 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.

Question 18

Two identical blocks slide across horizontal surfaces with different coefficients of kinetic friction. Block A experiences $\mu_k = 0.25$ and block B experiences $\mu_k = 0.40$ . If both blocks start with the same initial speed and come to rest, what is the ratio of the distance traveled by block A to the distance traveled by block B?

0.63, because block A experiences less friction throughout its motion
1.6, because block A experiences proportionally less resistive force
0.63, because block A decelerates more slowly than block B
1.6, because block A decelerates more slowly than block B (correct answer)

Explanation: Using

v^2 = v_0^2 - 2ad

with

a = \mu_k g

, we get

d = v_0^2/(2\mu_k g)

. The ratio is

d_A/d_B = \mu_{kB}/\mu_{kA} = 0.40/0.25 = 1.6

. Block A travels farther because it decelerates more slowly due to lower friction. Choices A and C incorrectly invert the ratio. Choice B has correct ratio but wrong reasoning about resistive force.

Question 19

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?

45 m/s², directed radially inward toward the center of curvature (correct answer)
44 m/s², directed radially inward accounting for frictional effects
45 m/s², directed tangentially along the path of motion
44 m/s², directed tangentially due to the resistive friction force

Explanation: 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.

Question 20

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?

The terminal velocity increases by a factor of $\sqrt{100/70} = 1.20$ (correct answer)
The terminal velocity increases by a factor of $\sqrt{70/100} = 0.84$
The terminal velocity decreases by a factor of $\sqrt{100/70} = 1.20$
The terminal velocity decreases by a factor of $\sqrt{70/100} = 0.84$

Explanation: 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.