6 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 6
A ball is thrown straight up. Air resistance is negligible. Which statement correctly describes the forces exerted on the ball at the instant it reaches its highest point?
Answer and reasoning
ANo forces at all, because the ball's velocity is zero then A student who thinks a momentarily stationary object has no forces on it picks this. Earth pulls on the ball whatever the ball's velocity, at the top of the flight as everywhere else.
BOnly the downward gravitational force exerted on it by EarthCorrect Every force on the ball is due to an interaction with another object. At the top of its flight the hand no longer touches the ball and air resistance is negligible, so only Earth interacts with it: the one force is Earth's downward gravitational force. The ball's velocity is zero for an instant, but the interaction with Earth does not depend on the velocity.
CEarth's pull and an equal upward force kept from the hand's throw A student who thinks a throw gives the ball a force that it carries along picks this. The hand exerts a force only while it touches the ball; after release no object pushes the ball upward, at the top or anywhere else.
DEarth's pull on the ball and the ball's equal pull on Earth A student who includes forces exerted by the ball picks this. The ball's pull on Earth is exerted on Earth, not on the ball, so it is not one of the forces on the ball.
A student presses a hand flat against a wall and holds it still. Which statement correctly describes the force that the wall exerts on the hand?
Answer and reasoning
AIt is a contact force due to electric forces between the atoms of the wall and hand.Correct The wall and the hand touch, so the wall exerts a contact force on the hand. Contact forces are the large-scale effect of electric forces between atoms: the hand compresses the wall's surface very slightly, and the electric forces between the atoms of the two surfaces push the hand away from the wall.
BThere is no such force, as a rigid wall just blocks the hand and does not push it. A student who thinks a support only blocks and cannot push picks this. No real wall is perfectly rigid: its surface is compressed very slightly, and the electric forces between atoms push back on the hand for as long as the two touch.
CIt is the hand's own push, returned unchanged to the hand by the wall's surface. A student who thinks a surface's push is another force passed back picks this. The hand's push is exerted on the wall; the wall's push is a different force, exerted on the hand by the wall.
DIt acted while the hand moved into the wall, and stopped when the hand was still. A student who thinks forces act only during events such as impacts picks this. The wall pushes on the hand for as long as the hand presses on it, whether or not anything is moving.
A block is on a ramp inclined at 37° above the horizontal, and friction is negligible. A horizontal force of magnitude F pushes on the block; its component along the ramp points up the ramp. The gravitational force on the block has magnitude Fg, and the ramp exerts a normal force on the block. A student uses an x-axis parallel to the ramp's surface, positive up the ramp. Use sin 37° = 0.60 and cos 37° = 0.80. Which equation correctly gives the sum of the x-components of the forces on the block?
Answer and reasoning
AΣFx = 0.80F − 0.80Fg A student who uses cos 37° for every x-component, whatever the angle is measured from, picks this. Fg makes 37° with the perpendicular to the ramp, not with the x-axis, so its component along the ramp is Fg sin 37° = 0.60Fg; on a level ramp the cosine version would wrongly give Fg a component along the surface.
BΣFx = 0.80F − 1.20Fg A student who adds the component of Fg along the ramp as an extra force, as well as Fg itself, counts it twice: 0.60Fg + 0.60Fg. The component is part of Fg, not another force.
CΣFx = 0.80F − 0.60FgCorrect The horizontal push makes 37° with the ramp, so its x-component is +F cos 37° = +0.80F. Fg makes 37° with the perpendicular to the ramp, so its component along the ramp is Fg sin 37° = 0.60Fg, pointing down the ramp. The normal force has no x-component. ΣFx = 0.80F − 0.60Fg.
DΣFx = 1.00F − 1.00Fg A student who combines forces by their magnitudes, as if each acted wholly along or against the axis, writes F − Fg. Neither force is parallel to the ramp, so each contributes only its component along the x-axis.
Working F is horizontal and the x-axis is tilted 37° from the horizontal, so the angle between F and the axis is 37°: Fx = +F cos 37° = +0.80F. Fg is vertical; the angle between Fg and the perpendicular to the ramp is 37°, so its component along the ramp is Fg sin 37° = 0.60Fg, directed down the ramp: −0.60Fg. The normal force is perpendicular to the ramp: x-component 0. ΣFx = 0.80F − 0.60Fg. Check: on a level surface (θ = 0) Fg would have no x-component.
A lamp hangs at rest from two cables: cable 1 pulls on it up and to the left, and cable 2 pulls on it up and to the right. In the diagrams shown, F₁ and F₂ are the forces exerted on the lamp by cables 1 and 2, Fg is the gravitational force exerted on the lamp by Earth, FC is a single arrow standing for the combined pull of both cables, and Flamp is the gravitational force exerted on Earth by the lamp. The arrows are not drawn to scale. Which diagram is a correct free-body diagram of the lamp?
Answer and reasoning
ADiagram 1 A student who thinks Earth does not pull on an object held up by supports picks the diagram with F₁ and F₂ only. Earth pulls on the lamp whether or not cables hold it, so Fg must be drawn.
BDiagram 2Correct The lamp interacts with three objects, cable 1, cable 2 and Earth, so its free-body diagram has three arrows from the dot: F₁ along cable 1, F₂ along cable 2 and Fg downward. Each force is drawn individually.
CDiagram 3 A student who combines the pulls of several objects into one arrow picks the diagram with FC and Fg. Each cable exerts its own force in its own direction, and an AP free-body diagram must show each force as an individual arrow.
DDiagram 4 A student who includes forces exerted by the lamp picks the diagram that adds Flamp. That force is exerted on Earth, so it belongs on Earth's diagram, not on the lamp's.
Working The lamp interacts with three objects: cable 1, cable 2 and Earth. Each gives one force on the lamp, drawn as its own arrow from the dot: F₁ up-left, F₂ up-right, Fg down. Flamp is exerted on Earth, and FC combines two forces into one arrow, so neither belongs.
A worker pushes a crate across a rough, level floor, pushing on its upper corner at an angle below the horizontal. A student represents the forces on the crate by drawing each arrow where the force is applied, as shown: FH by the hand, FN and Ff by the floor, and Fg by Earth. For analyzing the crate's translational motion, how should these forces be represented on a free-body diagram?
Answer and reasoning
AFrom one dot for the crate's center of mass, as if all its mass were located thereCorrect For translational motion the crate is treated as though all of its mass were at its center of mass. Its free-body diagram is a dot with all four forces drawn as individual arrows starting on it, FH still pointing down and to the right.
BAs shown, because a force acts only at the point of the crate where it is applied A student who thinks each force must start at its point of application picks this. For translational motion the crate is represented by a dot at its center of mass, and every force arrow starts on that dot.
CAs one arrow for the net force, starting on a dot at the crate's center of mass A student who treats the net force as a force of its own picks this. The net force is the sum of the four forces, not a force exerted by some object; the free-body diagram shows the four individual forces.
DFrom one dot, with FH replaced by its horizontal and vertical components A student who draws components as forces picks this. FH is one force exerted by one object, the hand; on AP free-body diagrams it is drawn as a single arrow in its own direction, and its components are used only in equations.
A block rests on a ramp inclined at 30° above the horizontal. The gravitational force on the block has magnitude Fg. Using axes parallel and perpendicular to the ramp's surface, F∥ and F⊥ are the magnitudes of the components of the gravitational force parallel and perpendicular to the ramp. Which ranking of Fg, F∥ and F⊥ is correct?
Answer and reasoning
AF⊥ < F∥ < Fg A student who uses the cosine for the component along the ramp, out of habit, gets F∥ = 0.87Fg and F⊥ = 0.50Fg. The angle between Fg and the perpendicular to the ramp is 30°, so the perpendicular component uses the cosine and is the larger one on a ramp below 45°.
BF∥ < F⊥ < FgCorrect Fg is the hypotenuse of the right triangle formed with its components, so both components are smaller than Fg. On a 30° ramp F∥ = Fg sin 30° = 0.50Fg and F⊥ = Fg cos 30° ≈ 0.87Fg, so F∥ < F⊥ < Fg.
CF∥ < F⊥ = Fg A student who thinks the whole gravitational force presses perpendicular into the ramp picks this. Only the component Fg cos 30° is perpendicular to the ramp; the rest of Fg's effect is along the ramp.
DFg < F⊥ < F∥ A student who divides by the trigonometric ratios gets F⊥ = Fg/cos 30° ≈ 1.15Fg and F∥ = Fg/sin 30° = 2Fg. A component is the force multiplied by a sine or cosine and can never exceed the force.
Working F∥ = Fg sin 30° = 0.50Fg; F⊥ = Fg cos 30° ≈ 0.87Fg. So F∥ < F⊥ < Fg.
In preparation: 0 of 6 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.2.A.1 Force Fix
Force
A vector quantity that describes an interaction between two objects or systems. A force has a magnitude and a direction and is always exerted on one object by another. SI unit: newton (N).
Agent of a force
The object that exerts a force. Every force on an object is due to its interaction with a specific other object (written, for example, as the force on the block by the ramp); if no such object can be named, the force does not exist.
Internal force
A force that one part of a system exerts on another part of the same system. A system cannot exert a net force on itself, so internal forces give no net force on the system and are left off the system's free-body diagram.
Gravitational force, Fg
The force that Earth exerts on an object near its surface, directed downward. It does not require contact, and it is exerted whether the object is at rest, moving up, moving down or supported.
Students often think A moving object has a force on it in its direction of motion, which keeps it moving. In fact No. A force on an object is exerted only by another object that it interacts with. An object can keep moving in a direction in which no force is exerted on it.
Students often think A throw or push gives an object a force that it carries along after contact ends and that gradually runs out. In fact No. The hand exerts a force only while it touches the object. Once contact ends, that interaction is over and its force no longer exists.
2.2.A.2 Contact force Fix
Contact force
A force between two objects that touch, such as a normal force, a friction force or the tension in a rope. At the atomic scale it is the large-scale effect of the electric forces between the atoms of the two objects.
Normal force, FN
The contact force that a surface exerts on an object pressing on it, directed perpendicular to the surface and away from it. On a ramp it is perpendicular to the ramp's surface, not vertical. SI unit: newton (N).
Tension force, FT
The pull that a string, rope or cable exerts on an object attached to it, directed along the string, away from the object. SI unit: newton (N).
Students often think Objects that only support, block or rest, such as walls, ramps, tables or an object lying on another, do not exert forces; only active agents such as people and engines push or pull. In fact Yes. Any object touching another exerts a contact force on it. A surface that is pressed on is deformed very slightly, and the electric forces between atoms push back; a resting object presses on whatever supports it.
Students often think A surface's push is not a force of its own but another force, such as the push it receives or the object's weight, sent back through it. In fact No. A surface's push on an object is a force of its own, exerted on the object by the surface. It arises from the electric forces between the atoms of the two objects where they touch.
2.2.B.1 Sum of force components, ΣFxFix
Sum of force components, ΣFx
The sum of the x-components of all the forces on a free-body diagram, each with the sign that gives its direction along the axis; together with ΣFy it translates the diagram into equations. SI unit: newton (N).
Students often think The normal force always points straight up, opposite to the gravitational force, even on a ramp. In fact No. The normal force is perpendicular to the surface that exerts it. On a ramp it is tilted from the vertical by the ramp's angle, so it has a horizontal component.
2.2.B.2 Free-body diagram Fix
Free-body diagram
A diagram of a single object or system that shows each force exerted on it by the environment as a separate straight arrow starting on a dot, pointing in the direction of the force, with a longer arrow for a larger force. Forces that the object exerts on other objects are not shown.
Students often think A free-body diagram of an object should also show the forces that the object exerts on other objects. In fact No. A free-body diagram of an object shows only the forces exerted on that object by its environment. A force that the object exerts belongs on the diagram of the object it is exerted on.
Students often think Forces from several objects may be combined and drawn as a single arrow, for example one upward arrow for the pull of two cables. In fact No. Each force is due to one interaction with one object and must be drawn as its own arrow. Two cables exert two forces, drawn as two arrows, even if they could be added into one.
2.2.B.3 Center-of-mass dot Fix
Center-of-mass dot
The dot from which every force arrow on a free-body diagram starts. It represents the center of mass, because for translational motion the object or system is treated as though all of its mass were located there.
Students often think The net force is a separate force, and a free-body diagram may show it as an arrow of its own, alongside or instead of the individual forces. In fact No. The net force is the vector sum of the forces already drawn, not an additional force exerted by some object. A free-body diagram shows the individual forces and does not include the net force.
Students often think Each force on an object must be drawn starting at the exact point where it is applied, since a force acts only at that point. In fact No. On a free-body diagram for translational motion every force starts on a single dot, because the object is treated as though all of its mass were located at its center of mass.
2.2.B.4 Component of a force Fix
Component of a force
The part of a force along a chosen axis: F cos α, where α is the angle between the force and that axis, with a sign for its direction along the axis. Components are used in equations; on AP free-body diagrams they are not drawn as extra arrows.
Tilted coordinate axes
Axes rotated so that one is parallel to the direction of the object's acceleration, for example parallel and perpendicular to the surface of a ramp. The forces are unchanged by this choice; only their components change, and the equations become simpler.
Students often think The components of a force are additional forces and should be drawn as their own arrows on a free-body diagram, alongside the force or in place of it. In fact No. A component is part of a force along an axis, not an additional force. On AP free-body diagrams only the forces themselves are drawn; components appear in the equations. Drawing both a force and its components counts the same force twice.
Students often think The x-component of a force is always found with the cosine, and the y-component with the sine, of whatever angle the problem gives. In fact No. A component is F cos α only when α is the angle between the force and that axis. If the given angle is measured from the other axis, or from a surface, the component along the x-axis uses the sine of that angle.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
Seen from above, two ropes pull horizontally on a crate. Using horizontal x- and y-axes, the force exerted on the crate by rope 1 is F⃗₁ = (30î + 40ĵ) N and the force exerted by rope 2 is F⃗₂ = (−20î) N. What is the magnitude of the sum of these two forces?
Answer and reasoning
A70 N A student who adds forces by their magnitudes picks this: |F⃗₁| = 50 N and |F⃗₂| = 20 N give 70 N. The two forces point in different directions, so their magnitudes would add only if they were parallel.
B50 N A student who adds all the components as ordinary numbers picks this: 30 N + 40 N − 20 N = 50 N. An x-component cannot be added to a y-component; the two sums are perpendicular and combine as the sides of a right triangle.
C41 NCorrect Forces add as vectors, component by component. The x-components give 30 N − 20 N = 10 N and the y-components give 40 N, so the magnitude of the sum is √((10 N)² + (40 N)²) ≈ 41 N.
D64 N A student who drops the minus sign on F⃗₂'s component gets (30 + 20)î + 40ĵ and a magnitude of about 64 N. The minus sign shows that rope 2 pulls in the −x direction, so it reduces the x-component of the sum to 10 N.
Working Add component by component: ΣFx = 30 N + (−20 N) = 10 N; ΣFy = 40 N + 0 = 40 N. |ΣF⃗| = √((10 N)² + (40 N)²) = √(1700) N = 41.2 N ≈ 41 N.
A student gives a block a quick push up a ramp. Friction and air resistance are negligible. After the hand leaves the block, the block slides up the ramp, slowing down. The diagram shows the block at an instant after the hand has left it and, on the right, the student's free-body diagram for the block at that instant. Fg is the gravitational force exerted by Earth, FN is the normal force exerted by the ramp, and Fpush is the student's label for the force of the push. Which change would make the free-body diagram correct?
Answer and reasoning
ARemove Fpush, since no object exerts that force once the hand has left the block.Correct A force is always due to an interaction with another object. The hand interacted with the block only while they touched; now the only objects interacting with the block are Earth (Fg) and the ramp (FN). The block keeps moving up the ramp because of the velocity it already has, not because something pushes it.
BKeep Fpush, since a block moving up the ramp needs a force pointing up the ramp. A student who thinks motion needs a force in its direction picks this. Once the hand has left, no object pushes the block up the ramp, so no such force can be drawn; the block moves up the ramp because of the velocity it already has.
CRemove FN, since the ramp merely blocks the block's path and does not push on the block. A student who thinks a surface only blocks motion picks this. The ramp touches the block and pushes on it, perpendicular to its surface and away from it; this contact force, FN, belongs on the diagram.
DAdd an arrow down the ramp for the part of Fg that is parallel to the ramp. A student who draws components as extra forces picks this. The component of Fg along the ramp is part of Fg, not a separate force; drawing it as well would count part of the gravitational force twice.
An astronaut floats inside a spacecraft far from any planet or star. Two thrusters exert forces of 10 N and 24 N on the spacecraft, perpendicular to each other. At the same time the astronaut pushes on an inside wall of the spacecraft with a 14 N force in the same direction as the 10 N thrust. During the push, what are the magnitudes of the net force on the spacecraft alone and of the net force on the system consisting of the astronaut and the spacecraft?
Answer and reasoning
ASpacecraft 34 N; system 34 N A student who counts a push between parts of a system as a force on the whole system picks this: the 14 N push added to the thrusts as a vector gives (24, 24) N, about 34 N. The astronaut is inside the system, so the push is internal; only the two thrusts act on the system from outside, giving 26 N.
BSpacecraft 26 N; system 26 N A student who thinks the astronaut's push and the wall's push back are exerted on the same object and cancel picks this, leaving only the thrusts, √(10² + 24²) = 26 N, on the spacecraft. The wall's push is exerted on the astronaut, not on the spacecraft, so the 14 N push still acts on the spacecraft alone.
CSpacecraft 48 N; system 34 N A student who adds forces by their magnitudes picks this: 10 N + 24 N + 14 N = 48 N on the spacecraft and 10 N + 24 N = 34 N on the system. The thrusts are perpendicular, so they combine as vectors: (24, 24) N, about 34 N, on the spacecraft and √(10² + 24²) = 26 N on the system.
DSpacecraft 34 N; system 26 NCorrect On the spacecraft alone act the two thrusts and the astronaut's 14 N push, which is along the 10 N thrust: (10 + 14, 24) N = (24, 24) N, magnitude about 34 N. For the astronaut-and-spacecraft system the push and the wall's push on the astronaut are internal, so only the two perpendicular thrusts are external: √(10² + 24²) = 26 N.
Working Spacecraft alone: the thrusts (10 N and 24 N, perpendicular) and the astronaut's 14 N push along the 10 N thrust: (10 + 14, 24) N = (24, 24) N, magnitude √(24² + 24²) = 33.9 N ≈ 34 N. System: the push and the wall's push on the astronaut are internal; the only external forces are the two thrusts: √(10² + 24²) = 26 N.
The diagram shows a block on a ramp inclined at 37° above the horizontal and, on the right, the free-body diagram of the block: the gravitational force Fg, the normal force FN exerted by the ramp, and the force FT exerted by a rope parallel to the ramp, with their magnitudes. Friction is negligible. A student analyzes the forces using the horizontal and vertical axes shown. Use sin 37° = 0.60 and cos 37° = 0.80. What is the magnitude of the sum of the horizontal components of the forces on the block?
Answer and reasoning
A48 N A student who uses cos 37° for every horizontal component picks this: 16 N − 80 N × 0.80 = −48 N. FN makes 37° with the vertical, not with the horizontal, so its horizontal component is 80 N × sin 37° = 48 N.
B16 N A student who takes the normal force to point straight up gives it no horizontal component and keeps only FT's 16 N. The normal force is perpendicular to the ramp's surface, so on a ramp it has a horizontal component, here 48 N toward −x.
C64 N A student who drops the minus sign on FN's horizontal component adds 16 N + 48 N = 64 N. FN's horizontal component points toward −x, opposite to FT's, so the two partly cancel.
D32 NCorrect FT points up the ramp, 37° above the horizontal, so its horizontal component is +20 N × 0.80 = +16 N. FN is perpendicular to the ramp, tilted 37° from the vertical toward −x, so its horizontal component is −80 N × 0.60 = −48 N. Fg is vertical. The sum is −32 N, of magnitude 32 N: with horizontal axes even the normal force must be resolved, which is why axes along the ramp are simpler here.
Working FT points up the ramp, 37° above +x: FTx = +20 N × cos 37° = +16 N. FN is perpendicular to the ramp, tilted 37° from the vertical toward −x: FNx = −80 N × sin 37° = −48 N. Fg is vertical: Fgx = 0. ΣFx = 16 N − 48 N = −32 N; magnitude 32 N.
A cup rests on a book, and the book rests on a level table. Which list gives all the forces exerted on the book?
Answer and reasoning
AEarth's pull on the book, the table's upward push and the cup's weight A student who thinks an object's weight acts on whatever supports it picks this. The cup's weight is the gravitational force exerted on the cup by Earth; the force on the book is a different force, the cup's contact push on the book.
BEarth's pull and the table's upward push; a resting cup exerts none A student who thinks a resting object exerts no force picks this. The cup touches the book and presses on it, so it exerts a downward contact force on the book, at rest or not.
CEarth's pull, the table's upward push and the cup's downward pushCorrect The book interacts with three objects: Earth, which pulls it down, and the two objects touching it, the table, which pushes it up, and the cup, which pushes it down. Each interaction gives one force exerted on the book.
DEarth's pull, the table's and cup's pushes, the book's push on the table A student who includes forces exerted by the book picks this. The book's push on the table is exerted on the table, so it belongs on the table's free-body diagram, not the book's.
A block is on a ramp with negligible friction, inclined at 30° above the horizontal. The component of the gravitational force on the block parallel to the ramp has magnitude F₁. The ramp is then raised until it is inclined at 60°. What is now the magnitude of the component of the gravitational force on the block parallel to the ramp?
Answer and reasoning
A2.00F₁ A student who takes the component to be proportional to the angle doubles it with the angle. The component is proportional to sin θ, and sin 60° is only √3 times sin 30°.
B0.58F₁ A student who uses the cosine for the component along the ramp gets cos 60°/cos 30° ≈ 0.58. The component along the ramp is Fg sin θ, which grows as the ramp is raised; Fg cos θ is the component perpendicular to it.
C1.00F₁ A student who thinks the components stay the same because Earth's pull is unchanged picks this. Fg is unchanged, but the axes have tilted with the ramp, so the share of Fg along the ramp has grown.
D1.73F₁Correct The component of Fg along a ramp of angle θ is Fg sin θ, and Fg does not change. The ratio of the new to the old component is sin 60°/sin 30° = √3 ≈ 1.73, so the new component is about 1.73F₁.
Working Parallel component = Fg sin θ. F₁ = Fg sin 30°; new = Fg sin 60°. Ratio = sin 60°/sin 30° = (√3/2)/(1/2) = √3 ≈ 1.73. New component ≈ 1.73F₁.
The diagram shows a top view of a crate pulled across the floor by two horizontal ropes. Each rope exerts a force of magnitude F₀ on the crate, at the angle shown from the dashed line along the crate's right face. Use sin 53° = 0.80 and cos 53° = 0.60. What is the magnitude of the sum of the two rope forces?
Answer and reasoning
A2.0F₀ A student who adds the two forces by their magnitudes picks this. The forces point in different directions, so only their components perpendicular to the dashed line add fully; their components along it cancel.
B1.2F₀ A student who uses the cosine of the given angle for the component along the direction of pull picks this: 2 × F₀ cos 53° = 1.2F₀. The 53° is measured from the face, not from the direction of pull, so that component is F₀ sin 53° = 0.80F₀.
C1.6F₀Correct Each rope makes 53° with the face, so its component perpendicular to the face, along the direction of pull, is F₀ sin 53° = 0.80F₀, and the two add. Along the face the components are +0.60F₀ and −0.60F₀, which cancel. The sum has magnitude 2 × 0.80F₀ = 1.6F₀.
D2.5F₀ A student who divides by sin 53° picks this: 2F₀/0.80. A component is the force multiplied by a sine or cosine, never larger than the force, so each rope contributes 0.80F₀ along the direction of pull.
Working Take x perpendicular to the crate's face, along the direction of pull. Each rope makes 53° with the face, so its x-component is F₀ sin 53° = 0.80F₀ and its components along the face, ±F₀ cos 53° = ±0.60F₀, cancel. Sum: 1.6F₀.
Compiled from the AP Physics C: Mechanics Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
