Statics and Dynamics Quiz: Newtons Second Law N T

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Newtons Second Law N TQuestion 1 of 10

A small bead of mass $m$ slides without friction along the inside of a vertical circular loop of radius $R$ . At the top of the loop the bead has speed $v_0$ . Take $g$ as the gravitational acceleration.

Which expression correctly gives the normal force $N$ that the track exerts on the bead at the top of the loop?

N = m\left(\dfrac{v_0^2}{R} - g\right)

, because at the top of the inside of the loop the track is above the bead, so

N

points downward (toward the center) and gravity also points downward (toward the center); their sum equals the centripetal force, giving

N + mg = mv_0^2/R

N = m\left(\dfrac{v_0^2}{R} + g\right)

, because the normal force points toward the center while gravity points away from the center at the top, so the centripetal force equals

N - mg

N = mg - \dfrac{mv_0^2}{R}

, because gravity points toward the center (downward) and the normal force points away from the center (upward) at the top, so the net inward force is

mg - N

N = \dfrac{mv_0^2}{R}

, because at the top of the loop the normal force alone supplies the entire centripetal acceleration while gravity acts perpendicular to the motion.

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Statics and Dynamics Quiz

Statics and Dynamics Quiz: Newtons Second Law N T

Practice Newtons Second Law N T 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 Newtons Second Law N T, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

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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.

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Question 1

Which expression correctly gives the normal force $N$ that the track exerts on the bead at the top of the loop?

$N = m\left(\dfrac{v_0^2}{R} - g\right)$ , because at the top of the inside of the loop the track is above the bead, so $N$ points downward (toward the center) and gravity also points downward (toward the center); their sum equals the centripetal force, giving $N + mg = mv_0^2/R$ . (correct answer)
$N = m\left(\dfrac{v_0^2}{R} + g\right)$ , because the normal force points toward the center while gravity points away from the center at the top, so the centripetal force equals $N - mg$ .
$N = mg - \dfrac{mv_0^2}{R}$ , because gravity points toward the center (downward) and the normal force points away from the center (upward) at the top, so the net inward force is $mg - N$ .
$N = \dfrac{mv_0^2}{R}$ , because at the top of the loop the normal force alone supplies the entire centripetal acceleration while gravity acts perpendicular to the motion.

Explanation: When analyzing circular motion problems, your first job is always to identify the direction of the centripetal acceleration (toward the center) and then apply Newton's second law along that direction:

\sum F_{\text{toward center}} = \frac{mv^2}{R}

. At the top of a vertical circular loop, the center is below the bead. For a bead on the inside of the loop, the track sits above the bead, so the normal force pushes downward — toward the center. Gravity also pulls downward — also toward the center. Both forces point inward, so they both contribute to centripetal acceleration. Setting up Newton's second law:

N + mg = \frac{mv_0^2}{R} \implies N = m\left(\frac{v_0^2}{R} - g\right)

This is choice A, the correct answer. Note that if

v_0

is too small (specifically,

v_0 < \sqrt{gR}

N

would be negative, meaning the bead has lost contact with the track — a physically meaningful check. Choice B claims gravity points away from the center at the top, which is wrong — gravity always points downward, and at the top, downward is toward the center. Choice C reverses both force directions, placing the normal force upward (away from center) and solving accordingly — this would describe a bead on the outside of the loop, not the inside. Choice D ignores gravity entirely, as if it acts perpendicular to motion; at the top of a vertical loop, gravity is radial, not tangential. Study tip: Always draw a free-body diagram and explicitly label "toward center" as your positive direction before writing any equation — this single habit eliminates most circular motion errors.

Question 2

A 2 kg collar slides without friction along a curved rod in the vertical plane. At the instant shown, the collar is moving along a section of the rod that has a radius of curvature $\rho = 1.5 \text{ m}$ . The tangent to the rod at this point makes an angle of $45°$ with the horizontal. The collar's speed is $v = 6 \text{ m/s}$ and decreasing at a rate of $2 \text{ m/s}^2$ . The center of curvature lies above and to the left of the collar's current position. Take $g = 9.81 \text{ m/s}^2$ .

What is the magnitude of the rod's normal force on the collar at this instant? (The rod is frictionless, so it can exert a force only in the n-direction — perpendicular to the path.)

$N = m(v^2/\rho - g\cos45°) = 2(24 - 6.937) \approx 34.13 \text{ N}$ , because gravity has a component $mg\cos45°$ directed toward the center, so it assists the normal force in providing centripetal acceleration.
$N = m(v^2/\rho + g\cos45°) = 2(24 + 6.937) \approx 61.87 \text{ N}$ , because the n-direction points toward the center (up-left at $45°$ from vertical), gravity has a component $mg\cos45°$ directed away from the center, and $\sum F_n = ma_n$ gives $N - mg\cos45° = mv^2/\rho$ . (correct answer)
$N = mv^2/\rho = 2(36/1.5) = 48 \text{ N}$ , because the frictionless rod's normal force alone provides the centripetal acceleration, and gravity acts entirely in the tangential direction at this position.
$N = m\sqrt{(v^2/\rho)^2 + a_t^2} = 2\sqrt{576 + 4} \approx 48.08 \text{ N}$ , because the rod's reaction force must balance the entire acceleration vector, so both the centripetal and tangential decelerations contribute to the rod's normal force.

Explanation: Whenever you see a curved-path dynamics problem, your first move should be to set up normal-tangential (n-t) coordinates and carefully identify which direction the n-axis points — it always points toward the center of curvature. Then apply Newton's second law separately in each direction:

\sum F_n = ma_n = mv^2/\rho

and

\sum F_t = ma_t

. Here, the center of curvature is above and to the left, so the n-axis points up-left at 45° from vertical (equivalently, 45° from horizontal). Now ask: what component does gravity contribute along this n-axis? Gravity pulls straight down with magnitude

mg

. The angle between the downward direction and the negative n-axis (pointing away from center, i.e., down-right) is 45°, so gravity has a component

mg\cos45°

pointing away from the center — in the negative n-direction. Writing Newton's second law in the n-direction (positive toward center):

N - mg\cos45° = mv^2/\rho

, which gives

N = m(v^2/\rho + g\cos45°) = 2(24 + 6.937) \approx 61.87 \text{ N}

. That's answer B. A is wrong because it subtracts

mg\cos45°

, which would be correct only if gravity's component pointed toward the center — it doesn't here, it points away. C is wrong because it assumes gravity is entirely tangential at this position. That would only be true if the n-axis were perfectly horizontal; here, the n-axis has a vertical component, so gravity absolutely contributes to the n-equation. D is wrong because the rod is frictionless — it can only push perpendicular to the path (n-direction). The tangential deceleration is irrelevant to calculating the normal force; it would matter for a friction or tension force acting along the path. Your study tip: always sketch the n-axis, then project every force (including gravity) onto it. The direction gravity's component points relative to the center determines whether you add or subtract it from

mv^2/\rho

Question 3

A particle of mass $m$ moves along a curve such that its speed varies as $v = ct$ , where $c$ is a positive constant and $t$ is time. At time $t = t_1$ the radius of curvature is $\rho_1$ .

At time $t = t_1$ , what is the angle $\phi$ that the net force vector makes with the normal (n) direction?

$\phi = \arctan\!\left(\dfrac{a_t}{a_n}\right) = \arctan\!\left(\dfrac{ct_1}{c^2t_1^2/\rho_1}\right) = \arctan\!\left(\dfrac{\rho_1}{ct_1}\right)$ , because $a_t = v = ct_1$ at time $t_1$ , and $a_n = c^2t_1^2/\rho_1$ , giving a ratio that depends on $ct_1$ rather than $c$ .
$\phi = \arctan\!\left(\dfrac{a_n}{a_t}\right) = \arctan\!\left(\dfrac{c^2t_1^2/\rho_1}{c}\right) = \arctan\!\left(\dfrac{ct_1^2}{\rho_1}\right)$ , because the angle from the n-direction is the arctangent of the normal-to-tangential acceleration ratio.
$\phi = \arctan\!\left(\dfrac{a_t}{a_n}\right) = \arctan\!\left(\dfrac{c}{c^2t_1^2/\rho_1}\right) = \arctan\!\left(\dfrac{\rho_1}{ct_1^2}\right)$ , because $a_t = dv/dt = c$ and $a_n = v^2/\rho_1 = c^2t_1^2/\rho_1$ , and the angle from the n-direction is the arctangent of the tangential-to-normal acceleration ratio. (correct answer)
$\phi = \arctan\!\left(\dfrac{a_n}{a_t}\right) = \arctan\!\left(\dfrac{\rho_1}{ct_1^2}\right)$ , which equals the same numerical value as the angle from the n-direction but is derived by incorrectly labeling the t-direction angle as the n-direction angle, so the formula is the same as option A.

Explanation: When a particle moves along a curved path, its acceleration has two components: a tangential component

a_t = dv/dt

(along the path) and a normal component

a_n = v^2/\rho

(pointing toward the center of curvature). The net force — and therefore the net acceleration vector — lives in the plane spanned by these two directions. The angle

\phi

from the normal direction to the net force vector is found by asking: "how far does the vector lean away from n, toward t?" That lean is captured by

\tan\phi = a_t / a_n

. For this problem, since

v = ct

, the tangential acceleration is

a_t = dv/dt = c

. At

t = t_1

, the speed is

v = ct_1

, so the normal acceleration is

a_n = v^2/\rho_1 = c^2t_1^2/\rho_1

. Plugging into the angle formula gives

\phi = \arctan\!\left(\dfrac{c}{c^2t_1^2/\rho_1}\right) = \arctan\!\left(\dfrac{\rho_1}{ct_1^2}\right)

, which is exactly C. Choice A commits a critical error by setting

a_t = v = ct_1

instead of

a_t = dv/dt = c

— confusing speed with tangential acceleration. Choice B flips the ratio, computing

a_n/a_t

instead of

a_t/a_n

; that formula gives the angle from the tangential direction, not the normal. Choice D arrives at the same numerical expression as C but claims it comes from an inverted ratio, so the reasoning is internally contradictory and wrong. A reliable memory aid:

\phi

from n means "tangential over normal" — just as

\phi

from t would mean "normal over tangential." Always differentiate

v

to get

a_t

; never substitute speed directly.

Question 4

A car of mass $M$ travels at constant speed $v$ along a banked curve of radius $R$ and bank angle $\theta$ . There is no friction between the tires and the road. The normal force on the car from the road surface is $N$ .

Applying Newton's second law in n–t coordinates to this frictionless banked curve, which pair of equations is correct for the n-direction (horizontal, toward center) and the vertical direction (upward positive)?

n-direction: $N + Mg\sin\theta = Mv^2/R$ ; vertical: $N\cos\theta - Mg = 0$ , because both the normal force and the inward component of gravity contribute to the centripetal acceleration on a banked surface.
n-direction: $N\cos\theta = Mv^2/R$ ; vertical: $N\sin\theta - Mg = 0$ , because the component of the normal force perpendicular to the bank surface provides centripetal force while the component along the bank balances gravity.
n-direction: $Mg\sin\theta = Mv^2/R$ ; vertical: $N - Mg\cos\theta = 0$ , because the component of gravity along the bank surface drives the centripetal motion and the normal force balances the perpendicular component of gravity.
n-direction: $N\sin\theta = Mv^2/R$ ; vertical: $N\cos\theta - Mg = 0$ , because the normal force has a horizontal centripetal component and a vertical component that balances gravity, with no tangential acceleration since speed is constant. (correct answer)

Explanation: When analyzing a banked curve in n–t coordinates, your job is to decompose every force into its horizontal (n-direction, toward center) and vertical components — not along or perpendicular to the bank surface. Since speed is constant, there is zero tangential acceleration, so you only need two equations: centripetal (n-direction) and vertical equilibrium. On a frictionless banked curve, the only contact force is the normal force

N

, which acts perpendicular to the road surface. If the bank angle is

\theta

from horizontal, then

N

points at angle

\theta

from vertical. Decomposing: the horizontal component is

N\sin\theta

(directed toward the center) and the vertical component is

N\cos\theta

(directed upward). Gravity

Mg

acts purely downward. Applying Newton's second law gives exactly what D states:

N\sin\theta = Mv^2/R

in the n-direction and

N\cos\theta - Mg = 0

vertically. This is the correct answer. A is wrong because it adds

Mg\sin\theta

to the centripetal equation — gravity has no horizontal component on its own; it acts straight down. B swaps the sine and cosine terms, incorrectly assigning

N\cos\theta

as the centripetal component and

N\sin\theta

as the vertical component — this reverses the geometry of the normal force decomposition. C eliminates

N

from the centripetal equation entirely, claiming gravity drives centripetal motion, which ignores that gravity is vertical and cannot directly provide a horizontal net force. Study tip: Always decompose forces into true horizontal and vertical components first, not along-the-surface components. Mixing coordinate systems (tilted vs. Cartesian) within the same equation is the most common mistake on banked-curve problems.

Question 5

A pilot pulls an aircraft out of a dive along a circular arc of radius $\rho = 500 \text{ m}$ . At the bottom of the pull-out, the aircraft's speed is $v = 100 \text{ m/s}$ and is constant. The pilot has mass $m = 80 \text{ kg}$ and $g = 9.81 \text{ m/s}^2$ .

What is the normal force (seat force) that the seat exerts on the pilot at the bottom of the pull-out, and what apparent weight multiplier (g-load) does the pilot experience?

$N = m(v^2/\rho + g) = 80(20 + 9.81) \approx 2385 \text{ N}$ , giving a g-load of $N/(mg) \approx 3.04$ , because at the bottom the seat force and gravity oppose each other while the net upward force provides centripetal acceleration. (correct answer)
$N = m(v^2/\rho - g) = 80(20 - 9.81) \approx 815 \text{ N}$ , giving a g-load of $N/(mg) \approx 1.04$ , because the centripetal acceleration exceeds gravity so only the difference is felt by the pilot.
$N = mv^2/\rho = 80 \times 20 = 1600 \text{ N}$ , giving a g-load of $N/(mg) \approx 2.04$ , because the seat force alone provides the centripetal acceleration and gravity cancels with the pilot's inertia at the bottom of the arc.
$N = mg = 80 \times 9.81 \approx 785 \text{ N}$ , giving a g-load of exactly 1.0, because at the bottom of the arc the speed is constant so there is no net force and the pilot experiences only normal weight.

Explanation: Whenever you see a circular motion problem involving a pilot, car, or roller coaster at the bottom of a curved path, your first instinct should be to apply Newton's second law in the radial direction, where the net upward force must equal the centripetal force

mv^2/\rho

. At the bottom of the pull-out, the seat pushes the pilot upward with normal force

N

, while gravity pulls downward with

mg

. The net upward force provides centripetal acceleration toward the center of the arc (which is directly above the pilot at this point). Setting up the equation:

N - mg = \frac{mv^2}{\rho}

Solving:

N = m\left(\frac{v^2}{\rho} + g\right) = 80\left(\frac{10000}{500} + 9.81\right) = 80(20 + 9.81) \approx 2385 \text{ N}

The g-load is

N/(mg) = 2385/785 \approx 3.04

. This is answer A, the correct choice. Answer B subtracts gravity instead of adding it — this would apply at the top of a loop, where both

N

and gravity point toward the center, not at the bottom. Answer C ignores gravity entirely, treating the seat force as if it only needs to produce centripetal acceleration without also supporting the pilot's weight against gravity — a classic omission error. Answer D incorrectly assumes that constant speed means zero net force; constant speed only means zero tangential force, but a net radial (centripetal) force is still required to maintain circular motion. Study tip: Always draw a free-body diagram and identify which direction is "toward the center." At the bottom of any arc, the normal force and gravity oppose each other — so they add in the Newton's second law equation.

Question 6

A 3 kg block slides along the inside of a smooth cylindrical bowl. At a certain instant the block is at a position where the tangent to the path makes an angle of $30°$ with the horizontal. The block's speed at this instant is $v = 5 \text{ m/s}$ and the radius of curvature is $\rho = 2 \text{ m}$ . The only forces acting are the normal force $N$ (perpendicular to the path, directed toward the center of curvature) and gravity (downward). Take $g = 9.81 \text{ m/s}^2$ .

What is the magnitude of the normal force $N$ on the block at this instant?

$N = mv^2/\rho = 3 \times 25/2 = 37.5 \text{ N}$ , because the normal force alone provides the centripetal acceleration and gravity acts entirely in the tangential direction at this position.
$N = m(v^2/\rho + g\sin30°) = 3(12.5 + 4.905) \approx 52.22 \text{ N}$ , because the component of gravity in the n-direction is $mg\sin30°$ when the tangent makes $30°$ with the horizontal.
$N = m(v^2/\rho + g\cos30°) = 3(12.5 + 8.496) \approx 62.99 \text{ N}$ , because when the tangent makes $30°$ with the horizontal the n-direction makes $30°$ with the vertical, so the component of gravity opposing the normal force (directed away from center) is $mg\cos30°$ . (correct answer)
$N = m(v^2/\rho - g\cos30°) = 3(12.5 - 8.496) \approx 12.01 \text{ N}$ , because both the normal force and the inward component of gravity contribute to the centripetal acceleration, so the gravity component is subtracted from the centripetal requirement.

Explanation: When a block moves along a curved path, you apply Newton's second law in the normal (n) direction — perpendicular to the path, pointing toward the center of curvature — and the tangential (t) direction separately. The key geometric challenge here is finding how gravity projects onto each direction. When the tangent makes 30° with the horizontal, the normal direction (perpendicular to the tangent) makes 30° with the vertical. Gravity points straight down, so its component in the normal direction is

mg\cos30°

. Writing Newton's second law in the n-direction (toward the center):

N - mg\cos30° = \frac{mv^2}{\rho}

N = m\!\left(\frac{v^2}{\rho} + g\cos30°\right) = 3(12.5 + 8.496) \approx 62.99\text{ N}

This is answer C — the normal force must overcome the inward gravity component and supply the centripetal acceleration. A is wrong because it ignores gravity's contribution in the n-direction entirely, treating gravity as purely tangential. That would only be true if the tangent were vertical (90° from horizontal). B uses

\sin30°

instead of

\cos30°

. This is the classic geometry trap: students instinctively pair "30° angle" with

\sin30°

, but the relevant angle between the n-direction and gravity is the complement — so cosine applies. D flips the sign, incorrectly assuming gravity assists the normal force toward the center. In this bowl geometry, the center of curvature is above the block, so gravity pulls away from the center, meaning it must be overcome, not subtracted. Study tip: Always draw a free-body diagram and explicitly identify the angle between gravity and your n-axis — don't assume sin or cos based on the given angle alone.

Question 7

A 1500 kg car travels over a hill whose crest has a radius of curvature $\rho = 60 \text{ m}$ . The car maintains a constant speed as it crests the hill. Gravity acts downward with $g = 9.81 \text{ m/s}^2$ .

What is the minimum speed at which the car's tires lose contact with the road at the crest of the hill?

$v_{\min} = \sqrt{\rho g} = \sqrt{60 \times 9.81} \approx 24.3 \text{ m/s}$ , found by setting the normal force to zero so that gravity alone provides the centripetal acceleration directed toward the center of curvature. (correct answer)
$v_{\min} = \sqrt{2\rho g} = \sqrt{2 \times 60 \times 9.81} \approx 34.3 \text{ m/s}$ , found by requiring that the centripetal force equal twice the gravitational force at the crest.
$v_{\min} = \rho g / v = 60 \times 9.81 / v$ , which is indeterminate without additional information about the normal force distribution along the path.
$v_{\min} = \sqrt{\rho g / 2} = \sqrt{60 \times 9.81 / 2} \approx 17.2 \text{ m/s}$ , found by setting the centripetal acceleration equal to half of $g$ at the instant contact is lost.

Explanation: When a car crests a hill, you're dealing with circular motion in the vertical plane. The key insight is identifying which direction "toward the center of curvature" points — at the top of a hill, the center lies below the road, meaning the net downward force must supply the centripetal acceleration. Applying Newton's second law in the radial direction (positive toward center, i.e., downward):

mg - N = \frac{mv^2}{\rho}

The tires lose contact when the normal force

N = 0

, meaning the road no longer pushes on the car. Setting

N = 0

and solving for

v

mg = \frac{mv^2}{\rho} \implies v_{\min} = \sqrt{\rho g} = \sqrt{60 \times 9.81} \approx 24.3 \text{ m/s}

This confirms A is correct — gravity alone provides exactly the centripetal force needed, and any faster, the required centripetal force exceeds gravity, making contact impossible. B is wrong because there's no physical reason to set centripetal force equal to twice gravity. The factor of 2 has no basis in the free-body diagram — it's an invented multiplier. C is wrong because it's circular reasoning: the expression

\rho g / v

still contains

v

, making it unsolvable and physically meaningless as written. D is wrong because setting centripetal acceleration to

g/2

is arbitrary. Contact is lost when

N = 0

, which corresponds to the full

g

, not half. Study tip: For any curved-path problem, always draw your free-body diagram first, then write

\sum F = mv^2/\rho

toward the center. The "loss of contact" condition always means setting

N = 0

— this is a reliable pattern across many exam problems.

Question 8

A 0.5 kg ball on a string moves in a horizontal circular path of radius $r = 0.8 \text{ m}$ on a frictionless table. The string makes no angle with the horizontal. At a certain instant the ball's speed is $v = 3 \text{ m/s}$ and the string's tension is $T$ .

If the string is suddenly cut at this instant, which statement correctly describes the ball's subsequent motion and the value of $T$ just before cutting?

$T = m(v^2/r + g) \approx 6.51 \text{ N}$ ; after cutting, the ball moves tangentially because the normal force from the table cancels gravity and only the centripetal component of tension remains relevant to subsequent motion.
$T = mv^2/r = 5.625 \text{ N}$ ; after cutting, the ball spirals outward from the release point because the centripetal force is removed and the ball's inertia carries it in an expanding curve away from the center.
$T = mv/r = 0.5 \times 3/0.8 = 1.875 \text{ N}$ ; after cutting, the ball moves radially outward because the centripetal acceleration was directed inward and its removal causes motion in the opposite direction.
$T = mv^2/r = 0.5 \times 9/0.8 = 5.625 \text{ N}$ ; after cutting, the ball moves in a straight line tangent to the circle at the point of release, because the tension was the sole source of centripetal force and no tangential force existed. (correct answer)

Explanation: When a ball moves in uniform circular motion on a frictionless horizontal surface, two separate force analyses are happening simultaneously: vertical equilibrium (normal force balances gravity) and horizontal centripetal acceleration (tension provides the inward force). The string lies flat, so tension acts purely horizontally — gravity never enters the tension calculation. The centripetal force formula gives you

T = \frac{mv^2}{r} = \frac{0.5 \times (3)^2}{0.8} = \frac{4.5}{0.8} = 5.625 \text{ N}

. This tension is the only horizontal force on the ball, and it points radially inward — it has no tangential component at all. When the string is cut, that inward force vanishes instantly. With no net force remaining (the table still supports the ball vertically), Newton's First Law takes over: the ball continues at constant velocity in whatever direction it was already moving — tangent to the circle at the release point. This confirms D is correct. A is wrong because it adds gravity into the tension formula. Since the string is horizontal on a flat table, gravity is already balanced by the normal force and contributes nothing to string tension. B correctly calculates

T

but misidentifies post-cut motion as a spiral. Spiraling would require a continuous force — once tension disappears, no force remains to curve the path. C uses

mv/r

instead of

mv^2/r

, mixing up momentum with centripetal force, and also incorrectly predicts radial outward motion. A useful pattern to remember: the direction of motion at any instant is always tangential, while centripetal force points radially inward. Removing the centripetal force doesn't reverse or redirect existing velocity — it simply lets that tangential velocity persist unchanged.

Question 9

A particle moves along a curved path. Its position along the path is described by the arc-length parameter $s$ . The speed of the particle is given by $v = \sqrt{2as}$ , where $a$ is a positive constant and $s$ is measured from rest ( $v=0$ at $s=0$ ). At a location where the radius of curvature is $\rho$ , a net force F acts on the particle of mass $m$ .

What is the angle that the net force F makes with the tangential direction at this location?

$\phi = \arctan\!\left(\dfrac{\rho}{2s}\right)$ , because the angle from the tangential direction is the arctangent of the tangential-to-normal acceleration ratio, with the tangential component in the numerator.
$\phi = \arctan\!\left(\dfrac{2s}{\rho}\right)$ , because the tangential acceleration is $a_t = a$ (constant) and the normal acceleration is $a_n = v^2/\rho = 2as/\rho$ , so the angle from the tangential direction is $\arctan(a_n/a_t) = \arctan(2s/\rho)$ . (correct answer)
$\phi = \arctan\!\left(\dfrac{2as}{\rho}\right)$ , because $a_t = v = \sqrt{2as}$ and $a_n = v^2/\rho = 2as/\rho$ , giving a ratio $a_n/a_t = 2as/\rho \div \sqrt{2as} = \sqrt{2as}/\rho$ , so the angle depends on the square root of $s$ .
$\phi = \arctan\!\left(\dfrac{\sqrt{2as}}{\rho}\right)$ , because the tangential acceleration is $a_t = dv/dt$ and the normal acceleration is $a_n = v/\rho = \sqrt{2as}/\rho$ , and the angle from the tangential direction is $\arctan(a_n/a_t)$ .

Explanation: When a particle moves along a curved path, the net force has two perpendicular components: a tangential component (

F_t = ma_t

) driving the change in speed, and a normal component (

F_n = ma_n

) steering the particle around the curve. The angle the net force makes with the tangential direction is therefore

\phi = \arctan(a_n / a_t)

, where

a_n

goes in the numerator because it's the component perpendicular to the reference direction. For this problem, start by finding

a_t

. Since

v = \sqrt{2as}

, differentiate with respect to time:

a_t = dv/dt = \frac{a}{\sqrt{2as}} \cdot \dot{s} = \frac{a}{\sqrt{2as}} \cdot v = \frac{a \cdot \sqrt{2as}}{\sqrt{2as}} = a

. So the tangential acceleration is simply the constant

a

. Next,

a_n = v^2/\rho = 2as/\rho

. The angle is

\phi = \arctan\!\left(\frac{2as/\rho}{a}\right) = \arctan\!\left(\frac{2s}{\rho}\right)

, confirming B. A is wrong because it flips the ratio — putting the tangential component in the numerator gives the complement of the correct angle, not the angle from the tangential direction. C is wrong due to a computational error: the student mistakenly uses

v

instead of

dv/dt

for the tangential acceleration, which is a fundamental confusion between speed and acceleration. D shares the same flaw as C — using

a_n = v/\rho

instead of the correct

v^2/\rho

for normal acceleration, which has units of velocity, not acceleration. Study tip: Always derive

a_t = dv/dt

explicitly — don't substitute speed for acceleration. And remember:

a_n = v^2/\rho

, never

v/\rho

Question 10

A 2 kg particle moves along a curved path. At a particular instant, the particle's speed is 4 m/s and is increasing at a rate of 3 m/s². The radius of curvature of the path at that instant is 8 m. A single force F acts on the particle at this instant.

What is the magnitude of the net force F acting on the particle at this instant?

$|\mathbf{F}| = \sqrt{(2 \times 3)^2 + (2 \times 4^2/8)^2} = \sqrt{36 + 16} \approx 7.21 \text{ N}$ , combining the tangential and normal components in quadrature. (correct answer)
$|\mathbf{F}| = 2 \times (3 + 4^2/8) = 2 \times 5 = 10 \text{ N}$ , adding the tangential acceleration and centripetal acceleration before multiplying by mass.
$|\mathbf{F}| = 2 \times 4^2/8 = 4 \text{ N}$ , because only the centripetal (normal) component of acceleration arises from the net force when the path curves.
$|\mathbf{F}| = 2 \times 3 = 6 \text{ N}$ , because only the tangential component of acceleration contributes to the net force magnitude when speed is changing.

Explanation: When a particle moves along a curved path with changing speed, its acceleration has two perpendicular components: a tangential component

a_t

(responsible for changing speed) and a normal (centripetal) component

a_n

(responsible for changing direction). Because these components are perpendicular, you must combine them using the Pythagorean theorem — not by simple addition. Here,

a_t = 3 \text{ m/s}^2

and

a_n = v^2/\rho = 4^2/8 = 2 \text{ m/s}^2

. The net acceleration magnitude is

|\mathbf{a}| = \sqrt{a_t^2 + a_n^2} = \sqrt{9 + 4} = \sqrt{13} \text{ m/s}^2

. Applying Newton's second law,

|\mathbf{F}| = m|\mathbf{a}| = 2\sqrt{13} \approx 7.21 \text{ N}

, confirming answer A is correct. Answer B adds the two accelerations arithmetically before multiplying by mass — this would only be valid if the components pointed in the same direction, but they're perpendicular, so this overcounts the magnitude. Answer C ignores the tangential acceleration entirely, treating the net force as if only the direction change matters. But Newton's second law demands you account for all components of acceleration. Answer D makes the opposite mistake — it ignores the centripetal acceleration and considers only the speed change. A particle curving through space requires a normal force component even if its speed were constant. Your study tip: whenever you see curved-path motion with changing speed, immediately sketch the two acceleration components as perpendicular vectors. The net force always requires

\sqrt{F_t^2 + F_n^2}

— never a simple sum.