AP Physics C Mechanics Quiz: Forces And Free Body Diagrams

20 questions · exam conditions

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Forces And Free Body DiagramsQuestion 1 of 20

Two blocks, $M_1$ and $M_2$ , are in contact on a frictionless horizontal surface. A horizontal force $F$ is applied to $M_1$ , causing both blocks to accelerate. What forces should be shown on a free-body diagram for block $M_2$ ?

Only the applied force

F

, gravity, and the normal force from the surface.

Gravity, the normal force from the surface, and a contact force exerted by

M_1

M_2

Gravity, the normal force from the surface, the applied force

F

, and a contact force from

M_1

M_2

Gravity and the normal force from the surface, as the horizontal forces are internal to the system.

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AP Physics C Mechanics Quiz

AP Physics C Mechanics Quiz: Forces And Free Body Diagrams

Practice Forces And Free Body Diagrams 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 Forces And Free Body Diagrams, 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

Only the applied force $F$ , gravity, and the normal force from the surface.
Gravity, the normal force from the surface, and a contact force exerted by $M_1$ on $M_2$ . (correct answer)
Gravity, the normal force from the surface, the applied force $F$ , and a contact force from $M_1$ on $M_2$ .
Gravity and the normal force from the surface, as the horizontal forces are internal to the system.

Explanation: A free-body diagram shows forces exerted on a specific object. For block

M_2

, the forces are: gravity (down), the normal force from the horizontal surface (up), and the contact force from block

M_1

pushing it (horizontally). The applied force

F

acts directly on

M_1

, not

M_2

Question 2

When constructing a free-body diagram for an object, forces are represented as vectors originating from a single point representing the object. What physical location does this point typically represent?

The geometric center of the object.
The point of application of the largest force.
The center of mass of the object. (correct answer)
An arbitrary point chosen for convenience.

Explanation: For the purpose of analyzing translational motion, a rigid body can be treated as a point mass with all its mass concentrated at its center of mass. Forces are drawn originating from this point because the net force on the object determines the acceleration of its center of mass, as described by Newton's second law (

\Sigma F_{ext} = Ma_{cm}

Question 3

A uniform ladder leans against a smooth, vertical wall while resting on a rough, horizontal floor. The ladder is in static equilibrium. The free-body diagram for the ladder must include which combination of forces?

Gravity, a normal force from the wall, and a normal force from the floor.
Gravity, a normal force from the wall, a normal force from the floor, and a static friction force from the wall.
Gravity, a normal force from the wall, a normal force from the floor, and a static friction force from the floor. (correct answer)
Gravity and a normal force from the floor only.

Explanation: The forces acting on the ladder are: its weight (gravity) acting at its center of mass, a normal force from the floor directed upward, a normal force from the wall directed horizontally away from the wall, and a static friction force from the floor directed horizontally toward the wall. The wall is smooth, so it exerts no friction. The friction from the floor is necessary to prevent the base of the ladder from slipping away from the wall.

Question 4

A $12.0\,\text{kg}$ box is pulled up a $30^\circ$ incline at constant speed by a rope parallel to the plane; $\mu_k=0.15$ . Consider the scenario described above, with kinetic friction opposing the motion and $g=9.8\,\text{m/s}^2$ . The normal force is perpendicular to the incline and weight is vertical. No air resistance acts. Calculate the frictional force given the conditions.

$15.3\,\text{N}$ up the plane
$15.3\,\text{N}$ down the plane (correct answer)
$17.6\,\text{N}$ down the plane
$14.7\,\text{N}$ down the plane
$18\,\text{N}$ down the plane

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams help visualize all forces acting on an object, including weight, normal force, friction, and applied forces, which is essential for analyzing motion on inclined planes. In this scenario, a 12.0 kg box moves up a 30° incline at constant speed, meaning the net force is zero and all forces balance. The problem asks specifically for the frictional force, which opposes motion (pointing down the plane). Choice B is correct because the normal force equals mg cos(30°) = 101.8 N, and the kinetic friction force equals μₖN = 0.15 × 101.8 = 15.3 N down the plane. Choice A is incorrect because it has the right magnitude but wrong direction - friction must oppose motion, so it points down the plane, not up. To help students: Emphasize that friction always opposes relative motion between surfaces, and at constant velocity, the net force must be zero. Practice identifying the direction of friction based on the direction of motion or impending motion. Watch for: students confusing the direction of friction, mixing up static and kinetic friction, or forgetting that constant speed means zero acceleration.

Question 5

Two masses $m_1=2.0\,\text{kg}$ and $m_2=2.5\,\text{kg}$ are connected by a light rope over a frictionless pulley. Consider the scenario described above, with $m_2$ moving downward and the system accelerating. Take $g=9.8\,\text{m/s}^2$ and assume the rope is massless. Tension is the same on both sides. Determine the tension in the rope.

$22.1\,\text{N}$
$4.9\,\text{N}$
$24.5\,\text{N}$
$21.8\,\text{N}$ (correct answer)
$19.6\,\text{N}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams for Atwood machines show that tension is the same throughout a massless rope, and both masses experience the same magnitude of acceleration in opposite directions. In this scenario, masses of 2.0 kg and 2.5 kg are connected over a frictionless pulley, with the heavier mass accelerating downward. Choice D is correct because the system acceleration is a = (m₂ - m₁)g/(m₁ + m₂) = (0.5)(9.8)/4.5 = 1.09 m/s², and the tension T = m₁(g + a) = 2.0(9.8 + 1.09) = 21.8 N, which also equals m₂(g - a) as a check. Choice E is incorrect because it represents the weight of the lighter mass (19.6 N), showing the common error of assuming tension equals weight when the system accelerates. To help students: Set up separate free-body diagrams and equations for each mass, use the constraint that both accelerations have the same magnitude, and solve the system of equations. Verify the answer by checking that the same tension satisfies both force equations. Watch for: students assuming tension equals one of the weights, using different accelerations for each mass, or making sign errors in the force equations.

Question 6

Two masses $m_1=4.0\,\text{kg}$ and $m_2=6.0\,\text{kg}$ hang on either side of a light rope over a frictionless pulley. Consider the scenario described above, with $m_2$ accelerating downward and the rope remaining taut. Take $g=9.8\,\text{m/s}^2$ and neglect pulley mass. Tension is uniform throughout the rope. Determine the tension in the rope.

$47.0\,\text{N}$ (correct answer)
$39.2\,\text{N}$
$58.8\,\text{N}$
$78.4\,\text{N}$
$23.5\,\text{N}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams for connected objects must show tension forces that obey Newton's third law, with the rope providing equal tension on both masses in an Atwood machine setup. In this scenario, masses of 4.0 kg and 6.0 kg hang on either side of a frictionless pulley, with the heavier mass accelerating downward. Choice A is correct because applying Newton's second law to the system: the net force is (m₂ - m₁)g = 19.6 N, giving acceleration a = 19.6/10 = 1.96 m/s², and the tension T = m₁(g + a) = 4.0(9.8 + 1.96) = 47.0 N. Choice B is incorrect because it represents the weight of the lighter mass (39.2 N), which students often confuse with tension when they don't account for the acceleration of the system. To help students: Draw separate free-body diagrams for each mass, write Newton's second law equations for each object, and solve the system of equations simultaneously. Emphasize that tension is not simply equal to the weight of either mass when the system accelerates. Watch for: students assuming tension equals weight, forgetting to use the same acceleration for both masses, or incorrectly setting up the force equations.

Question 7

A $2.0\,\text{kg}$ block rests on a $40^\circ$ incline with $\mu_s=0.50$ and no applied force. Consider the scenario described above, with the block initially at rest and $g=9.8\,\text{m/s}^2$ . Static friction acts along the plane as needed, up to $\mu_s N$ . The normal force is perpendicular to the plane and weight is vertical. Is the object in equilibrium? Justify your answer.

Yes; $mg\sin\theta < \mu_s mg\cos\theta$
No; $mg\sin\theta > \mu_s mg\cos\theta$ (correct answer)
Yes; $N=mg\sin\theta$ guarantees rest
No; static friction is always $\mu_s N$
Yes; friction points down the plane here

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams on inclined planes require careful analysis of static friction, which can vary from zero up to μₛN depending on what's needed to prevent motion. In this scenario, a 2.0 kg block rests on a 40° incline with static friction coefficient 0.50, and we must determine if it remains in equilibrium. Choice B is correct because the component of weight down the plane is mg sin(40°) = 12.6 N, while the maximum static friction is μₛmg cos(40°) = 0.50(2.0)(9.8)cos(40°) = 7.5 N; since 12.6 N > 7.5 N, static friction cannot balance the weight component, so the block will slide. Choice A is incorrect because it reverses the inequality - students often confuse when to use sine versus cosine or compare the forces incorrectly. To help students: Teach the critical angle concept where tan(θ) = μₛ determines the steepest angle for equilibrium, and practice comparing the weight component down the plane with maximum static friction. Emphasize that static friction adjusts to match the applied force up to its maximum value. Watch for: students always using μₛN for static friction instead of recognizing it as a maximum, confusing the conditions for static equilibrium, or mixing up force components.

Question 8

An object is being pulled up a rough inclined plane at a constant velocity by a rope parallel to the plane. Which of the following sets of forces should be shown on a correct free-body diagram for the object?

Tension, gravity, normal force, and kinetic friction, with the net force directed up the incline.
Tension, gravity, and normal force, with the tension equal to the parallel component of gravity.
Tension, gravity, normal force, and static friction, with the net force equal to zero.
Tension, gravity, normal force, and kinetic friction, with the net force equal to zero. (correct answer)

Explanation: The forces acting on the object are: tension from the rope (up the incline), gravity (straight down), the normal force from the plane (perpendicular to the surface), and kinetic friction (down the incline, opposing motion). Since the velocity is constant, the acceleration is zero, and thus the net force must be zero. The friction is kinetic because the object is moving.

Question 9

A projectile is launched at an angle and is at the highest point of its trajectory. Neglecting air resistance, which forces should be included on the free-body diagram of the projectile at this instant?

Only the downward force of gravity. (correct answer)
The downward force of gravity and a horizontal force in the direction of motion.
The downward force of gravity and an upward force that becomes zero at the peak.
No forces, as the projectile is momentarily at rest in the vertical direction.

Explanation: Once a projectile is launched, and neglecting air resistance, the only force acting on it throughout its flight is the gravitational force exerted by the Earth, which always acts straight down. There is no force propelling it forward after it has left the launcher. Its vertical velocity is momentarily zero at the peak, but this does not mean the net force is zero; it is still accelerating downwards due to gravity.

Question 10

A mass is suspended from the ceiling by a string, forming a simple pendulum. It is pulled to one side and released. When the mass is at the lowest point of its swing, which of the following best describes the forces on its free-body diagram?

The upward tension force is equal in magnitude to the downward gravitational force.
The upward tension force is greater in magnitude than the downward gravitational force. (correct answer)
The upward tension force is less in magnitude than the downward gravitational force.
Only the tension force and the gravitational force, plus a centripetal force directed upwards.

Explanation: At the lowest point of the swing, the mass is moving along a circular arc and has a net upward acceleration (centripetal acceleration). The forces acting on it are tension (up) and gravity (down). For the net force to be upward, the magnitude of the tension force must be greater than the magnitude of the gravitational force. Centripetal force is a net force, not a separate force to be drawn on the diagram.

Question 11

The purpose of a free-body diagram is to isolate an object and consider only the external forces acting upon it. Which of the following is an example of an internal force that would NOT be shown on the free-body diagram of a car?

The gravitational force exerted by the Earth on the car.
The friction force exerted by the road on the car's tires.
The force exerted by the car's engine on its transmission. (correct answer)
The force of air resistance exerted by the air on the car's body.

Explanation: Internal forces are forces that parts of a system exert on each other. The force of the engine on the transmission is internal to the car as a system. External forces are exerted on the system by objects in its environment. Gravity (from Earth), friction (from the road), and air resistance (from the air) are all external forces and would be included on the free-body diagram of the car.

Question 12

A block is pulled by a rope at an angle $\theta$ above the horizontal across a rough surface, causing the block to accelerate. How does the magnitude of the normal force $F_N$ compare to the magnitude of the gravitational force $mg$ ?

$F_N = mg$
$F_N > mg$
$F_N < mg$ (correct answer)
The relationship cannot be determined without knowing the acceleration.

Explanation: The free-body diagram includes gravity (down), tension (at angle

\theta

), normal force (up), and friction (horizontal). Analyzing the vertical forces, the upward normal force plus the upward component of tension must balance the downward gravitational force. So,

F_N + T \sin{\theta} = mg

. Since

T \sin{\theta}

is a positive value,

F_N

must be less than

mg

Question 13

A block hangs at rest from two strings of unequal length, which are attached to the ceiling at different points. The block is not centered between the attachment points. What must be true about the forces on the free-body diagram for the block?

The magnitudes of the tension in the two strings must be equal.
The sum of the magnitudes of the two tensions must equal the magnitude of the gravitational force.
The horizontal components of the two tension forces must be equal in magnitude and opposite in direction. (correct answer)
The vertical components of the two tension forces must be equal to each other.

Explanation: Since the block is in static equilibrium, the net force in any direction must be zero. Analyzing the horizontal forces, the horizontal component of tension from one string must exactly balance the horizontal component of tension from the other string. Because the angles are likely different, the magnitudes of the tensions will not be equal, and their vertical components will sum to balance the gravitational force but are not necessarily equal to each other.

Question 14

For a block sliding down a rough incline, a student considers setting up a coordinate system to analyze the forces. Which choice of coordinate system best simplifies the application of Newton's second law, and why?

Horizontal and vertical axes, because the gravitational force aligns with the vertical axis.
Axes parallel and perpendicular to the incline, because the acceleration and the normal force align with the axes. (correct answer)
Horizontal and vertical axes, because it avoids having to resolve the normal force into components.
Axes parallel and perpendicular to the incline, because this makes the gravitational force components equal.

Explanation: Aligning the coordinate system with the incline (one axis parallel, one perpendicular) is most convenient. In this system, the acceleration is entirely along one axis. The normal force and the friction force are also aligned with the axes. Only the gravitational force needs to be resolved into components. This simplifies the equations for Newton's second law, as

a_y=0

Question 15

A car is traveling at a constant speed as it rounds a flat, unbanked circular curve. What is the primary purpose of drawing a free-body diagram for the car in this situation?

To identify the outward centrifugal force that keeps the car on the curve.
To determine the net force, which must be zero since the speed is constant.
To identify the real forces whose vector sum provides the net centripetal force. (correct answer)
To show that the normal force is greater than the gravitational force.

Explanation: A free-body diagram is used to identify all the real, physical forces acting on an object. For an object in uniform circular motion, there must be a net force directed towards the center of the circle (the centripetal force). The FBD helps identify which force or combination of forces (in this case, static friction) provides this required net force. Centrifugal force is a fictitious force, and the net force is not zero because the direction of velocity is changing.

Question 16

A 0.50 kg ball hangs from a string while inside an elevator. When the elevator accelerates upward at $2.0 \text{ m/s}^2$ , the tension in the string is $T_1$ . When the elevator accelerates downward at $2.0 \text{ m/s}^2$ , the tension is $T_2$ . Using free-body diagram analysis and taking $g = 10 \text{ m/s}^2$ , what is the ratio $\frac{T_1}{T_2}$ ?

$\frac{T_1}{T_2} = 1.0$ because the acceleration magnitudes are equal in both cases
$\frac{T_1}{T_2} = 2.0$ because the tension changes proportionally with acceleration direction
$\frac{T_1}{T_2} = 1.5$ because upward acceleration increases tension while downward decreases it (correct answer)
$\frac{T_1}{T_2} = 1.25$ based on the specific values of mass and acceleration given

Explanation: When you encounter problems involving objects in accelerating reference frames like elevators, you need to carefully apply Newton's second law while considering the direction of all forces and accelerations. Let's analyze each case using free-body diagrams. For the ball, two forces act: weight (

mg = 0.50 \times 10 = 5.0 \text{ N}

downward) and tension (upward). Case 1 (upward acceleration): The net force must be upward at

ma = 0.50 \times 2.0 = 1.0 \text{ N}

. Using Newton's second law:

T_1 - mg = ma

, so

T_1 = mg + ma = 5.0 + 1.0 = 6.0 \text{ N}

. Case 2 (downward acceleration): The net force must be downward at

1.0 \text{ N}

. Since we define upward as positive:

T_2 - mg = -ma

, so

T_2 = mg - ma = 5.0 - 1.0 = 4.0 \text{ N}

. Therefore:

\frac{T_1}{T_2} = \frac{6.0}{4.0} = 1.5

Why wrong answers fail:

A incorrectly assumes equal acceleration magnitudes mean equal tensions, ignoring that direction matters for the force balance
B suggests a simple doubling relationship that has no physical basis in the force equations
D appears to result from calculation errors or misapplying the given values

Key strategy: In elevator problems, always remember that upward acceleration requires extra tension (beyond supporting weight), while downward acceleration reduces the needed tension. Set up your coordinate system clearly and apply

\Sigma F = ma

methodically in the chosen direction.

Question 17

A car travels around a horizontal circular curve of radius $R$ at constant speed $v$ . The road surface provides the centripetal force through friction. In the free-body diagram for the car, which statement correctly describes the direction and magnitude of the friction force?

The friction force is tangent to the circular path with magnitude $\mu mg$ in the direction opposing motion
The friction force points radially inward toward the center with magnitude $\frac{mv^2}{R}$ to provide centripetal acceleration (correct answer)
The friction force points radially outward from the center with magnitude $\frac{mv^2}{R}$ to balance centrifugal force
The friction force is zero because the car moves at constant speed around the curve

Explanation: For circular motion at constant speed, the car needs centripetal force

F_c = \frac{mv^2}{R}

directed toward the center of the circle. Friction provides this force, so it points radially inward with magnitude

\frac{mv^2}{R}

. Choice A incorrectly describes kinetic friction opposing tangential motion. Choice C wrongly suggests friction balances a centrifugal force (which doesn't exist in an inertial frame). Choice D misunderstands that constant speed still requires centripetal acceleration and force.

Question 18

A box sits on the bed of a pickup truck that is accelerating forward. The box does not slide relative to the truck bed. In the free-body diagram for the box, which force provides the acceleration, and what can be concluded about its magnitude?

The normal force from the truck bed provides acceleration, and its horizontal component equals $ma$
Air resistance provides the acceleration by pushing the box forward as the truck moves through air
The truck's engine force is transmitted directly to the box, creating the acceleration $a$
Static friction from the truck bed provides acceleration, and its magnitude equals $ma$ in the forward direction (correct answer)

Explanation: When analyzing forces on objects that move together without slipping, you need to identify which force actually causes the acceleration by applying Newton's second law to each object separately. Since the box doesn't slide relative to the truck bed, both have the same acceleration

a

forward. For the box to accelerate forward, there must be a net horizontal force acting on it. Looking at the box's free-body diagram, the only horizontal force available is static friction from the truck bed surface. This friction force points forward (in the direction of acceleration) and must equal

ma

to satisfy Newton's second law:

F_{net} = ma

. Choice A incorrectly identifies the normal force as providing acceleration. The normal force acts vertically upward, balancing the box's weight. Even if the truck bed were tilted, any horizontal component of the normal force alone couldn't account for the full acceleration. Choice B misunderstands the physics entirely. Air resistance opposes motion and would act backward on the box, not forward. Air doesn't "push" objects forward as vehicles move through it. Choice C confuses the system analysis. The engine accelerates the truck, and the truck then accelerates the box through friction. There's no direct mechanical connection transmitting engine force to the box. Remember this key principle: when objects accelerate together without slipping, look for the contact force between them that prevents relative motion. That contact force (usually friction) is what accelerates the "passenger" object. This applies to boxes on trucks, people in elevators, or any situation where objects move together.

Question 19

A block of mass $m$ is placed on a wedge of mass $M$ that can slide freely on a horizontal frictionless surface. The wedge has a frictionless inclined surface at angle $\theta$ . When the system is released from rest, both the block and wedge accelerate. In the free-body diagram analysis for the block, which statement about the normal force from the wedge is correct?

The normal force equals $mg\cos\theta$ perpendicular to the wedge surface, just as if the wedge were fixed
The normal force equals $mg$ because the wedge's motion doesn't affect the contact force between surfaces
The normal force is greater than $mg\cos\theta$ because the block must accelerate relative to the ground frame
The normal force is less than $mg\cos\theta$ because the wedge accelerates horizontally, reducing the apparent weight (correct answer)

Explanation: When you encounter problems involving accelerating reference frames like this wedge system, remember that forces between objects change when the reference frame itself is accelerating. This isn't the same as analyzing motion on a fixed incline. Since the wedge can slide freely on the frictionless surface, both the block and wedge will accelerate when released. The wedge accelerates horizontally to the right, while the block accelerates both down the incline relative to the wedge and horizontally with respect to the ground. This creates a situation where the block experiences a reduced "apparent weight" against the wedge surface. Think of it like being in an accelerating elevator - when the elevator accelerates upward, you feel heavier; when it accelerates downward, you feel lighter. Here, the wedge's horizontal acceleration effectively reduces the component of the block's weight that presses against the inclined surface, so the normal force becomes less than

mg\cos\theta

. Option A incorrectly assumes the wedge is fixed, ignoring its acceleration. Option B makes the fundamental error of thinking the normal force always equals the full weight

mg

, which would only be true for a horizontal surface with no vertical acceleration. Option C suggests the opposite effect - that the normal force increases - which would occur if the wedge were somehow accelerating in the opposite direction. The key insight: whenever your reference frame (the wedge) is accelerating, you must account for how that acceleration affects the apparent forces within that frame. Always consider whether surfaces are truly fixed or can move freely.

Question 20

A block is pulled up a rough inclined plane at constant velocity by a force $\vec{F}$ applied parallel to the incline. The incline makes angle $\theta$ with the horizontal, and the coefficient of kinetic friction is $\mu_k$ . In the free-body diagram analysis, which expression correctly represents the relationship between the applied force and other forces?

$F = mg\sin\theta + \mu_k mg\cos\theta$ because friction opposes motion up the incline (correct answer)
$F = mg\sin\theta - \mu_k mg\cos\theta$ because friction aids motion up the incline
$F = mg\cos\theta + \mu_k mg\sin\theta$ due to the perpendicular force components
$F = \mu_k mg\cos\theta - mg\sin\theta$ when friction exceeds the gravitational component

Explanation: At constant velocity, the net force is zero, so the applied force must balance both the component of weight down the incline (

mg\sin\theta

) and the friction force opposing motion (

\mu_k N = \mu_k mg\cos\theta

). Both oppose the applied force, so

F = mg\sin\theta + \mu_k mg\cos\theta

. Choice B incorrectly subtracts friction, suggesting it aids upward motion. Choice C confuses sine and cosine components. Choice D has the wrong sign relationship and would only apply if friction somehow overcame gravity.