Statics and Dynamics Quiz: Friction Drag And Spring Forces

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Friction Drag And Spring ForcesQuestion 1 of 9

A particle of mass $m = 2 \text{ kg}$ is launched vertically upward with initial speed $v_0 = 20 \text{ m/s}$ through air that exerts a drag force $F_D = bv$ with $b = 0.4 \text{ N}\cdot\text{s/m}$ . Take $g = 9.81 \text{ m/s}^2$ .

Compare the time to reach maximum height $t_{up}$ to the time to fall back to the launch point $t_{down}$ . Which statement is correct?

t_{up} < t_{down}

, and the return speed at the launch point equals the launch speed

v_0 = 20 \text{ m/s}

, because energy conservation requires the same kinetic energy at the same height regardless of the drag force encountered along the path.

t_{up} = t_{down}

, because for linear drag the equations of motion are symmetric about the apex: the speed-versus-time profile during ascent mirrors the descent profile, and the total time up equals the total time down for any initial speed.

t_{up} > t_{down}

, because the particle must overcome both gravity and drag during ascent, spending more time decelerating on the way up than it does accelerating on the way down, where only gravity drives the motion.

t_{up} < t_{down}

, because during ascent both gravity and drag act downward, decelerating the particle rapidly, while during descent gravity and drag oppose each other, producing a smaller net acceleration and a longer return journey.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Friction Drag And Spring Forces

Practice Friction Drag And Spring Forces in Statics and Dynamics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Friction Drag And Spring Forces, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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Question 1

Compare the time to reach maximum height $t_{up}$ to the time to fall back to the launch point $t_{down}$ . Which statement is correct?

$t_{up} < t_{down}$ , and the return speed at the launch point equals the launch speed $v_0 = 20 \text{ m/s}$ , because energy conservation requires the same kinetic energy at the same height regardless of the drag force encountered along the path.
$t_{up} = t_{down}$ , because for linear drag the equations of motion are symmetric about the apex: the speed-versus-time profile during ascent mirrors the descent profile, and the total time up equals the total time down for any initial speed.
$t_{up} > t_{down}$ , because the particle must overcome both gravity and drag during ascent, spending more time decelerating on the way up than it does accelerating on the way down, where only gravity drives the motion.
$t_{up} < t_{down}$ , because during ascent both gravity and drag act downward, decelerating the particle rapidly, while during descent gravity and drag oppose each other, producing a smaller net acceleration and a longer return journey. (correct answer)

Explanation: When a drag force acts on a moving object, you must always ask: in which direction does drag point relative to gravity, and how does that change the net force at each phase of motion? During ascent, both gravity (

mg

) and drag (

bv

, opposing upward motion) act downward, so the net decelerating force is

mg + bv

. This large combined force brings the particle to rest quickly — meaning

t_{up}

is relatively short. During descent, gravity pulls down while drag now acts upward (opposing downward motion), so the net accelerating force is only

mg - bv

. This reduced net force means the particle accelerates slowly back down, taking longer to cover the same distance. Therefore

t_{up} < t_{down}

, confirming D is correct. A is wrong on two counts: drag is a non-conservative force, so mechanical energy is not conserved — the return speed must be less than

v_0

. Energy conservation arguments simply don't apply when friction or drag is present. B is wrong because the equations of motion are not symmetric. Ascent has net force

mg + bv

while descent has

mg - bv

; these are fundamentally different differential equations with different time scales. C gets the direction of inequality backwards. Yes, ascent involves both gravity and drag decelerating the particle — but that larger combined force means the particle stops sooner, not later. It's descent, with the smaller net force, that takes more time. Study tip: Whenever you see linear drag, sketch a free-body diagram for each phase separately — drag always flips direction with velocity, so the net force changes completely between ascent and descent.

Question 2

A particle of mass $m$ falls vertically through a fluid that exerts a drag force of the form $F_D = bv^2$ , where $b$ is a positive drag coefficient and $v$ is the particle's speed. The particle is released from rest.

Which of the following correctly describes the time required for the particle to reach half of its terminal velocity, $v_t/2$ , where $v_t = \sqrt{mg/b}$ ?

$t_{1/2} = \frac{m}{b v_t} \tanh^{-1}\!\left(\frac{1}{2}\right)$ , because the equation of motion with quadratic drag integrates to an inverse hyperbolic tangent in time, and evaluating at $v = v_t/2$ gives a finite, well-defined result. (correct answer)
$t_{1/2} = \frac{m}{2 b v_t}$ , because at $v = v_t/2$ the drag force is one-quarter of its terminal value, and the remaining net force divided into the momentum change gives a linear time estimate.
$t_{1/2} = \frac{m}{b v_t} \ln 2$ , because the velocity profile for quadratic drag is a natural logarithm in time, and evaluating at $v = v_t/2$ yields $\ln 2$ as the dimensionless time constant.
$t_{1/2} = \frac{\pi m}{4 b v_t}$ , because the velocity solution for quadratic drag involves a tangent function whose argument at half terminal velocity equals $\pi/4$ , introducing a factor of $\pi$ into the result.

Explanation: When you encounter drag-force problems, your first move should always be to write Newton's second law and actually solve the differential equation — don't let the answer choices anchor your reasoning before you've done the physics. For a particle falling with quadratic drag, Newton's second law gives

m\frac{dv}{dt} = mg - bv^2

. Substituting

v_t = \sqrt{mg/b}

, this becomes

\frac{dv}{dt} = \frac{b v_t^2}{m}\left(1 - \frac{v^2}{v_t^2}\right)

. Separating variables and integrating, you get

\int_0^v \frac{dv'}{1-(v'/v_t)^2} = \frac{bv_t}{m}t

. The left side integrates to

v_t \tanh^{-1}(v/v_t)

, giving

v(t) = v_t \tanh\!\left(\frac{bv_t}{m}t\right)

. Inverting for

t

v = v_t/2

t_{1/2} = \frac{m}{bv_t}\tanh^{-1}\!\left(\frac{1}{2}\right)

. This confirms A is correct. Choice B is wrong because it treats the problem as a simple impulse calculation using a "linear estimate" of net force — this ignores the nonlinear dynamics entirely and produces an incorrect dimensional shortcut. Choice C is the classic trap:

\ln 2

appears in the linear drag case (

F_D = bv

), where the solution involves an exponential. Students who confuse linear and quadratic drag land here. Choice D is subtler — the velocity solution does involve

\tanh

, not

\tan

, so no factor of

\pi

enters. The tangent function appears in the upward-motion quadratic drag problem, not free fall from rest. Your key study tip: memorize the two drag solutions separately — linear drag gives exponential/

\ln

behavior, quadratic drag gives hyperbolic tangent behavior. Mixing them up is the most common error on these problems.

Question 3

A block of mass $m = 6 \text{ kg}$ is pushed against a vertical wall by a horizontal force $P$ . The block is also attached to a vertical spring ( $k = 300 \text{ N/m}$ ) anchored to the floor directly below, with the spring currently stretched by $\delta = 0.08 \text{ m}$ (pulling the block downward). The coefficient of static friction between the block and the wall is $\mu_s = 0.6$ . Take $g = 9.81 \text{ m/s}^2$ .

What is the minimum horizontal force $P_{min}$ required to keep the block from sliding downward along the wall?

$P_{min} = \frac{mg + k\delta}{\mu_s} - k\delta = \frac{82.86}{0.6} - 24 = 114.1 \text{ N}$ , because the spring force must be subtracted from the friction force requirement after computing the gross normal force needed, accounting for the spring's direct vertical load path to the floor.
$P_{min} = \frac{mg - k\delta}{\mu_s} = \frac{58.86 - 24}{0.6} = 57.8 \text{ N}$ , because the spring is stretched and pulls downward, so the spring partially supports the block against gravity, reducing the net downward load that friction must resist.
$P_{min} = \frac{mg}{\mu_s} = \frac{58.86}{0.6} = 98.1 \text{ N}$ , because the spring force acts vertically and therefore does not affect the horizontal normal force $P$ ; only gravity must be balanced by wall friction, and the spring contribution cancels in the vertical equilibrium equation.
$P_{min} = \frac{mg + k\delta}{\mu_s} = \frac{58.86 + 24}{0.6} = 138.1 \text{ N}$ , because the spring pulls the block downward, increasing the total downward force that friction (acting upward) must balance, and the normal force is provided entirely by $P$ . (correct answer)

Explanation: When a block is pressed against a wall by a horizontal force, you're dealing with a classic friction equilibrium problem. The key framework: identify all forces in each direction separately. The normal force on the wall equals

P

(horizontal equilibrium), and friction — which acts upward when the block tends to slide down — must balance every downward force in vertical equilibrium. Here, the block has two downward forces: gravity

mg = (6)(9.81) = 58.86 \text{ N}

and the stretched spring pulling down with

k\delta = (300)(0.08) = 24 \text{ N}

. Vertical equilibrium requires friction to resist both:

f = mg + k\delta = 82.86 \text{ N}

. Since maximum static friction is

f_{max} = \mu_s P

, setting

\mu_s P_{min} = 82.86

gives

P_{min} = \frac{82.86}{0.6} \approx 138.1 \text{ N}

. That's answer D. Answer B reverses the spring's effect — a stretched spring pulling the block downward is an additional load, not a helpful one. Only a compressed spring pushing upward would reduce the friction requirement. Answer C ignores the spring entirely, treating the vertical equilibrium as if only gravity matters; every force in the free-body diagram must be accounted for. Answer A performs a convoluted two-step that incorrectly subtracts the spring force after computing the normal force, which has no physical basis — the spring force goes straight into vertical equilibrium, not into a correction on

P

. Your strategy: always draw a complete free-body diagram before writing equilibrium equations. Label every force's direction carefully — a stretched spring pulls, a compressed spring pushes. Getting that sign wrong is the most common trap on friction-against-wall problems.

Question 4

A particle of mass $m$ moves horizontally and is subject to a velocity-dependent drag force of the form $F_D = -cv^n$ , where $c > 0$ and $n$ is a positive integer. No other horizontal forces act on the particle. The particle has initial velocity $v_0 > 0$ .

For which value of $n$ does the particle theoretically require infinite time to come to rest, yet travel only a finite total distance before stopping?

$n = 1$ (linear drag), because the velocity decays exponentially as $v(t) = v_0 e^{-ct/m}$ , which never reaches zero in finite time, while the total distance $\int_0^\infty v\,dt = mv_0/c$ converges to a finite value. (correct answer)
$n = 2$ (quadratic drag), because the velocity decays as $v(t) = v_0/(1 + cv_0 t/m)$ , which never reaches zero in finite time, but the total distance $\int_0^\infty v\,dt$ diverges logarithmically, yielding infinite travel distance.
$n = 3$ (cubic drag), because the velocity decays as $v(t) = v_0/\sqrt{1 + 2cv_0^2 t/m}$ , which asymptotically approaches zero, and the total distance integral converges to a finite value unlike the quadratic case.
Both $n = 1$ and $n = 3$ simultaneously satisfy the condition of infinite stopping time with finite stopping distance, because any odd power of $v$ in the drag law produces exponential-like decay with convergent distance integrals.

Explanation: When a drag force

F_D = -cv^n

acts on a particle, you need to ask two separate questions: does the velocity reach zero in finite time, and does the total distance

\int_0^\infty v\,dt

converge? These two conditions are independent, and the exponent

n

controls both. For linear drag (

n = 1

), Newton's second law gives

m\dot{v} = -cv

, which yields

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

. This exponential decay never reaches zero — the particle asymptotically approaches rest, requiring infinite time. Yet the total distance is

\int_0^\infty v_0 e^{-ct/m}\,dt = \frac{mv_0}{c}

, which is perfectly finite. Both conditions are satisfied simultaneously, confirming A is correct. Choice B is factually accurate about quadratic drag's velocity formula

v(t) = v_0/(1 + cv_0 t/m)

, and it's true that stopping takes infinite time — but the distance integral

\int_0^\infty \frac{v_0}{1+cv_0 t/m}\,dt

diverges logarithmically. So

n=2

fails the finite-distance requirement. B correctly identifies the flaw but presents it as disqualifying when the question asks which case does satisfy both conditions. Choice C describes cubic drag (

n=3

) with

v(t) = v_0/\sqrt{1+2cv_0^2 t/m}

. While stopping time is indeed infinite, the distance integral

\int_0^\infty (1+2cv_0^2 t/m)^{-1/2}\,dt

also diverges (it grows like

\sqrt{t}

), so

n=3

actually fails the finite-distance condition. Choice D is wrong because odd powers do not universally produce convergent distance integrals —

n=3

is a direct counterexample. Study tip: When analyzing drag problems, treat stopping time and stopping distance as separate convergence questions. Exponential decay (

n=1

) is the unique case where the faster-than-power-law decay makes both integrals behave simultaneously.

Question 5

A 4 kg block sits on a horizontal surface ( $\mu_s = 0.4$ , $\mu_k = 0.3$ ). A spring ( $k = 500 \text{ N/m}$ ) is attached horizontally to the block at one end and to a wall at the other. The block is displaced $x_1 = 0.06 \text{ m}$ from the spring's natural length position (stretching the spring) and released from rest. Take $g = 9.81 \text{ m/s}^2$ .

Immediately after release, the block does not move. What is the minimum spring displacement $x_{min}$ at which the block would just begin to slide, and does the block eventually come to rest at the spring's natural length position after one or more oscillations?

$x_{min} = \mu_s mg/k = 0.031 \text{ m}$ ; the block does not return to the natural length position but comes to rest at a displaced position where the remaining spring force is insufficient to overcome static friction. (correct answer)
$x_{min} = \mu_k mg/k = 0.024 \text{ m}$ ; the block does not return to the natural length because once kinetic friction dissipates enough energy, the block stops wherever the kinetic friction force equals the spring restoring force.
$x_{min} = \mu_s mg/k = 0.031 \text{ m}$ ; the block does eventually come to rest exactly at the natural length position because the work done by friction equals the initial spring potential energy over complete oscillations.
$x_{min} = \mu_s mg/k = 0.031 \text{ m}$ ; whether the block moves at all depends on whether $x_1 > x_{min}$ , and since $0.06 > 0.031$ , the block slides; it eventually stops at the natural length position only if the number of half-cycles is exactly an integer.

Explanation: When a spring-mass system involves friction, the key question isn't just "will it oscillate?" but "where will it finally stop?" The block begins to slide only when the spring force exceeds maximum static friction:

f_s^{max} = \mu_s mg = 0.4 \times 4 \times 9.81 = 15.7 \text{ N}

. Dividing by

k

gives the threshold displacement

x_{min} = \mu_s mg / k = 15.7/500 \approx 0.031 \text{ m}

. Since the initial displacement

x_1 = 0.06 \text{ m} > 0.031 \text{ m}

, the block does slide — confirming the first part of answer A. For the final resting position, recognize that during each half-cycle, kinetic friction dissipates energy and shifts the block's equilibrium. The block stops whenever its velocity reaches zero and the remaining spring force is less than

f_s^{max}

. This doesn't have to occur at the natural length — it occurs wherever the spring compression or extension falls within the "dead zone"

|x| \leq x_{min}

. The block lands somewhere inside that band, not necessarily at

x = 0

. Answer A captures this correctly. Answer B is wrong on two counts: it uses

\mu_k

instead of

\mu_s

for the sliding threshold (kinetic friction only applies during motion, not at the onset), and the stopping condition it describes is also imprecise. Answer C incorrectly assumes the block returns to exactly

x = 0

; energy balance doesn't guarantee the natural length is the stopping point. Answer D introduces an unnecessary and incorrect condition about integer half-cycles. Your study tip: always distinguish between the static friction threshold (governs whether motion starts) and the kinetic friction effect (governs where motion ends). These two roles are never interchangeable.

Question 6

A block of mass $m = 5 \text{ kg}$ is placed on a horizontal surface with coefficients of static and kinetic friction $\mu_s = 0.45$ and $\mu_k = 0.35$ , respectively. A horizontal spring with stiffness $k = 200 \text{ N/m}$ is attached to the block. The spring is slowly stretched until the block begins to slide, and at that instant the spring extension is $x_0$ . Immediately after the block starts sliding, the spring continues to pull the block and the block accelerates.

What is the magnitude of the block's acceleration immediately after it begins to slide, expressed in terms of the given quantities? Use $g = 9.81 \text{ m/s}^2$ .

$a = \frac{k x_0 - \mu_k m g}{m}$ , which evaluates to approximately $0.98 \text{ m/s}^2$ when $x_0 = \mu_s m g / k$
$a = \frac{k x_0 - \mu_s m g}{m}$ , which evaluates to zero at the instant of slip since the applied spring force exactly equals the maximum static friction force at that moment
$a = \frac{\mu_s m g - \mu_k m g}{m} = (\mu_s - \mu_k)g$ , which evaluates to approximately $0.98 \text{ m/s}^2$ independent of spring stiffness (correct answer)
$a = \frac{k x_0}{m}$ , which evaluates to approximately $4.42 \text{ m/s}^2$ because friction vanishes the instant kinetic sliding begins and no longer contributes to the net force

Explanation: When a block transitions from static to kinetic friction, the critical insight is that the resisting friction force drops the moment sliding begins — and that drop is what produces the initial acceleration. Here's the key sequence: the spring is stretched just enough to overcome maximum static friction, so at the instant of slip,

k x_0 = \mu_s m g

. This tells you

x_0 = \frac{\mu_s m g}{k}

. Once sliding begins, friction doesn't vanish — it switches from static to kinetic, meaning the resistive force drops from

\mu_s m g

\mu_k m g

. Applying Newton's second law immediately after slip:

ma = k x_0 - \mu_k m g

. Substituting

k x_0 = \mu_s m g

gives

a = \frac{\mu_s m g - \mu_k m g}{m} = (\mu_s - \mu_k)g = (0.45 - 0.35)(9.81) \approx 0.98 \text{ m/s}^2

. This is answer C, and notice it's independent of spring stiffness because

k

cancels out entirely. Answer A is actually algebraically equivalent to C before substituting

x_0

, but its numerical evaluation is correct only if you properly substitute — the issue is that A presents the formula without recognizing it simplifies to C's cleaner form, making C the more complete and instructive answer. Answer B uses

\mu_s

in the net force equation after sliding begins, which is wrong — once kinetic friction applies, you must use

\mu_k

, giving zero acceleration, which contradicts the physical reality. Answer D ignores kinetic friction entirely after slip, which is a common misconception; friction doesn't disappear just because motion starts. Remember: static friction prevents motion; kinetic friction opposes motion but persists. The acceleration at slip onset comes entirely from the difference between the two friction coefficients.

Question 7

A small sphere of mass $m$ and radius $r$ falls through a viscous fluid. At low Reynolds number, Stokes drag applies: $F_D = 6\pi\mu r v$ , where $\mu$ is the dynamic viscosity and $v$ is the speed. The sphere also experiences a buoyancy force $F_B = \rho_f V g$ , where $\rho_f$ is the fluid density and $V = \frac{4}{3}\pi r^3$ is the sphere's volume. The sphere's density is $\rho_s$ .

Starting from rest, the sphere reaches terminal velocity $v_t$ . If the sphere's radius is doubled while keeping all other properties ( $\rho_s$ , $\rho_f$ , $\mu$ ) constant, by what factor does the terminal velocity change?

The terminal velocity increases by a factor of 2, because terminal velocity is directly proportional to radius through the balance of gravitational and drag forces, with mass scaling as $r^3$ and drag scaling as $r$ .
The terminal velocity increases by a factor of 4, because at terminal velocity the net gravitational force (scaling as $r^3$ ) balances Stokes drag (scaling as $r$ ), making $v_t \propto r^2$ , and doubling $r$ gives $2^2 = 4$ . (correct answer)
The terminal velocity increases by a factor of 8, because the weight scales as $r^3$ , buoyancy scales as $r^3$ , and drag scales as $r$ , so the net force scales as $r^3$ and dividing by the drag coefficient (scaling as $r$ ) gives $v_t \propto r^2$ ; then doubling $r$ while accounting for the mass increase gives an eightfold increase.
The terminal velocity increases by a factor of 2, because both the net downward force and the Stokes drag coefficient scale identically with radius, leaving only a linear dependence on $r$ from the velocity term in the drag expression.

Explanation: When a sphere falls at terminal velocity, the net force is zero — drag exactly balances the net downward force (weight minus buoyancy). Setting up this balance is the key move for any Stokes-drag problem. At terminal velocity:

6\pi\mu r v_t = (\rho_s - \rho_f)\frac{4}{3}\pi r^3 g

. Solving for

v_t

v_t = \frac{2(\rho_s - \rho_f)g}{9\mu} r^2

Everything outside

r^2

is held constant, so

v_t \propto r^2

. Doubling the radius gives

v_t \propto (2r)^2 = 4r^2

, a factor of 4 increase — confirming B is correct. The key insight is that the net driving force scales as

r^3

(both weight and buoyancy grow cubically, and their difference preserves that scaling), while Stokes drag scales as only

r^1

, leaving an

r^2

dependence for velocity. A is wrong because it claims

v_t \propto r

, incorrectly treating the problem as if only mass (not buoyancy-corrected mass) enters, or misapplying the force balance to get a linear relationship. C reaches the right scaling (

r^2

) but then incorrectly applies an additional factor of

r

, confusing itself by double-counting the mass increase separately — the

r^2

scaling already accounts for everything. D claims the net force and drag scale identically, which would cancel radius entirely; this ignores that drag scales as

r^1

while the net gravitational force scales as

r^3

. As a study habit: whenever asked "by what factor does X change," derive the proportionality first (

v_t \propto r^2

), then plug in the scaling factor. This two-step method prevents the errors seen in A, C, and D.

Question 8

Two blocks, A (mass $m_A = 3 \text{ kg}$ ) and B (mass $m_B = 5 \text{ kg}$ ), are connected by a massless spring ( $k = 400 \text{ N/m}$ ) and rest on a horizontal surface. The coefficient of kinetic friction between each block and the surface is $\mu_k = 0.25$ . Block A is given an initial velocity of $v_0 = 4 \text{ m/s}$ toward block B while block B is initially at rest. At the instant shown, the spring is compressed by $\delta = 0.05 \text{ m}$ and block A has velocity $v_A = 1.5 \text{ m/s}$ (toward B) while block B has velocity $v_B = 0.8 \text{ m/s}$ (away from A). Take $g = 9.81 \text{ m/s}^2$ .

At the described instant, what is the net force on block B, and in which direction does it act?

$F_{net,B} = 12.64 \text{ N}$ directed away from A, because the compressed spring pushes B away from A with 20 N, and the kinetic friction force on B is computed using block A's mass, giving $\mu_k m_A g = 0.25(3)(9.81) = 7.36 \text{ N}$ opposing motion, for a net of $20 - 7.36 = 12.64 \text{ N}$ .
$F_{net,B} = 7.74 \text{ N}$ directed toward A, because the compressed spring force (20 N away from A) is partially offset by kinetic friction (12.26 N away from A, acting in the same direction as B's motion), leaving a net force toward A.
$F_{net,B} = 7.74 \text{ N}$ directed away from A, because the spring pushes B with $k\delta = 400(0.05) = 20 \text{ N}$ away from A, and kinetic friction opposes B's motion (toward A) with $\mu_k m_B g = 0.25(5)(9.81) = 12.26 \text{ N}$ , giving a net force of $20 - 12.26 = 7.74 \text{ N}$ away from A. (correct answer)
$F_{net,B} = 20 \text{ N}$ directed away from A, because kinetic friction acts parallel to the surface but perpendicular to the direction of the spring force, so only the spring force contributes to block B's net horizontal acceleration.

Explanation: When analyzing forces on a single block in a spring-friction system, always isolate that block and identify every force acting on it specifically — don't mix in properties of other blocks. For block B at this instant, two horizontal forces act: the spring force and kinetic friction. The compressed spring exerts

F_{spring} = k\delta = 400(0.05) = 20 \text{ N}

pushing B away from A. Since B is already moving away from A, kinetic friction opposes that motion — meaning friction acts toward A on block B. Using block B's own weight:

f_k = \mu_k m_B g = 0.25(5)(9.81) = 12.26 \text{ N}

. The net force is therefore

20 - 12.26 = 7.74 \text{ N}

directed away from A, confirming answer C is correct. Answer A contains a critical conceptual error: it uses block A's mass to compute friction on block B. Each block's friction force depends on its own normal force (

m_B g

, not

m_A g

). Never substitute one block's mass into another block's friction calculation. Answer B correctly identifies the spring force magnitude and direction but then claims friction acts away from A — the same direction as B's motion. That directly contradicts the definition of kinetic friction, which always opposes the direction of motion. Answer D ignores friction entirely, incorrectly claiming the forces are perpendicular. Both the spring force and friction act along the same horizontal axis; they absolutely combine to determine net force. Your strategy tip: when you see a multi-body problem, draw a separate free-body diagram for each object, label the direction of motion, and immediately draw friction opposing that specific direction. This one habit eliminates the most common errors on force-and-motion questions.

Question 9

A particle of mass $m = 0.5 \text{ kg}$ is connected to a spring ( $k = 80 \text{ N/m}$ , natural length $L_0 = 0.3 \text{ m}$ ) and moves along a straight horizontal track. The particle also experiences a linear drag force $F_D = c\dot{x}$ with $c = 4 \text{ N}\cdot\text{s/m}$ . At time $t = 0$ , the particle is at the natural length position with velocity $v_0 = 2 \text{ m/s}$ in the positive direction.

Classify the system's damping and determine the damped natural frequency $\omega_d$ (in rad/s) of the oscillation.

The system is underdamped with $\omega_d = \sqrt{\omega_n^2 - \zeta^2} \approx 11.3 \text{ rad/s}$ , where $\omega_n = \sqrt{k/m} \approx 12.65 \text{ rad/s}$ and $\zeta = c/(2m) = 4 \text{ rad/s}$ is the exponential decay rate.
The system is underdamped with $\omega_d = \omega_n\sqrt{1 - \xi^2} \approx 12.0 \text{ rad/s}$ , where $\omega_n = \sqrt{k/m} \approx 12.65 \text{ rad/s}$ and the damping ratio $\xi = c/(2m\omega_n) \approx 0.316$ . (correct answer)
The system is critically damped because $c = 4 \text{ N·s/m}$ equals the critical damping coefficient $c_{cr} = 2\sqrt{km} = 2\sqrt{80 \times 0.5} \approx 12.65 \text{ N·s/m}$ , so no oscillatory frequency exists and the particle returns to equilibrium without oscillating.
The system is overdamped with no oscillatory frequency because the damping ratio $\xi = c/(2\sqrt{km}) = 4/(2\sqrt{40}) \approx 0.316$ , which exceeds the threshold $\xi > 0.25$ required for overdamped behavior in spring-mass-damper systems.

Explanation: When analyzing a spring-mass-damper system, your first step is always to classify the damping by comparing the actual damping coefficient to the critical value, then compute the appropriate frequency from there. Start by finding the undamped natural frequency:

\omega_n = \sqrt{k/m} = \sqrt{80/0.5} = \sqrt{160} \approx 12.65 \text{ rad/s}

. Next, compute the critical damping coefficient:

c_{cr} = 2m\omega_n = 2(0.5)(12.65) \approx 12.65 \text{ N·s/m}

. Since

c = 4 \text{ N·s/m} < c_{cr}

, the system is underdamped — it will oscillate with a decaying amplitude. The damping ratio is

\xi = c/c_{cr} = 4/12.65 \approx 0.316

, and the damped natural frequency is

\omega_d = \omega_n\sqrt{1 - \xi^2} = 12.65\sqrt{1 - 0.316^2} \approx 12.0 \text{ rad/s}

. This confirms B. Choice A gets the classification right but uses the wrong formula. The quantity

\zeta = c/(2m) = 4 \text{ rad/s}

is the exponential decay rate (sometimes called

\sigma

), not the damping ratio, and

\omega_d \neq \sqrt{\omega_n^2 - \zeta^2}

— that expression conflates two different quantities with incompatible roles in the formula. Choice C incorrectly claims

c = c_{cr}

. It confuses the numerical coincidence that

c_{cr} \approx 12.65

N·s/m with the given

c = 4

N·s/m — always recompute

c_{cr}

explicitly rather than assuming. Choice D fabricates a rule: there is no "

\xi > 0.25

overdamped threshold." The actual threshold is

\xi > 1

. A reliable memory anchor: under/over/critical damping depends solely on whether $\xi$ is less than, greater than, or equal to 1 — no other thresholds exist.