3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Each free-body diagram shows all the forces exerted on an object, drawn to scale and labeled with their magnitudes; the caption gives the object's motion at the instant shown. In which diagram are the forces exerted on the object unbalanced?
Answer and reasoning
ADiagram 1 A student who adds the magnitudes picks this: 30 N + 40 N = 70 N on one side against 50 N. Added as vectors, the 30 N and 40 N forces at right angles have a resultant of 50 N, exactly opposite to the third force, so the net force is zero.
BDiagram 2 A student who thinks a moving object needs a net force in its direction of motion picks this. The two 25 N forces cancel; an object can move to the right with balanced forces on it, at constant velocity.
CDiagram 4 A student who thinks the largest force decides the motion picks this, because the 50 N force is the biggest. The 30 N and 20 N forces together add to 50 N in the opposite direction, so the net force is zero.
DDiagram 3Correct The upward 30 N force and downward 20 N force add to a net force of 10 N upward, so the forces are unbalanced, even though the object is at rest at this instant (its velocity is about to change). In each other diagram the forces add, as vectors, to zero.
Working Unbalanced means F⃗net ≠ 0. Diagram 1: 30 N right + 40 N up + 50 N at (−30 N, −40 N) = 0. Diagram 2: 25 N − 25 N = 0. Diagram 3: 30 N up − 20 N down = 10 N up ≠ 0. Diagram 4: 30 N + 20 N right − 50 N left = 0. Only diagram 3 is unbalanced; being at rest at an instant does not make the net force zero.
Two boxes, of mass 6.0 kg and 2.0 kg, sit side by side on a level floor. A worker pushes the 6.0 kg box horizontally with a force of 40 N, so that it pushes the 2.0 kg box ahead of it, and the boxes speed up together. The floor exerts frictional forces totaling 16 N on the two boxes, opposite to their motion. What is the magnitude of the boxes' acceleration?
Answer and reasoning
A4.0 m/s² A student who divides by the mass of the box the worker pushes picks this: 24 N/6.0 kg. Both boxes speed up together, so the 24 N net force accelerates the whole 8.0 kg system.
B3.0 m/s²Correct Take both boxes as the system, so the push between them is internal. The external horizontal forces are 40 N forward and 16 N backward, so Fnet = 24 N, and a = 24 N/8.0 kg = 3.0 m/s².
C5.0 m/s² A student who counts only the worker's push picks this: 40 N/8.0 kg. The floor's 16 N of friction is also a force on the system and reduces the net force to 24 N.
D7.0 m/s² A student who adds the magnitudes of the forces picks this: (40 N + 16 N)/8.0 kg. Friction acts opposite to the push, so the forces partly cancel: Fnet = 40 N − 16 N.
Working System: both boxes, msys = 8.0 kg. External horizontal forces: 40 N forward, 16 N backward; Fnet = 24 N. a = Fnet/msys = 24 N/8.0 kg = 3.0 m/s². (Only the pushed box's mass: 24/6.0 = 4.0 m/s². Friction left out: 40/8.0 = 5.0 m/s². Magnitudes added: 56/8.0 = 7.0 m/s².)
An astronaut floats at rest in deep space, far from any other objects, holding a wrench. She throws the wrench forward and moves backward herself. Consider the system of the astronaut and the wrench. What happens to the velocity of the system's center of mass during and after the throw?
Answer and reasoning
AIt becomes forward, since she exerts a forward force on the wrench. A student who thinks internal forces move the center of mass picks this. The wrench also pushes back on the astronaut with an equal and opposite force; both forces act within the system and cancel, so the center of mass does not start moving.
BIt stays zero, since no net external force is exerted on the system.Correct The forces that the astronaut and the wrench exert on each other during the throw are internal to the system and cancel in pairs. With no net external force, the velocity of the system's center of mass cannot change, so it stays zero: the wrench moves forward and the astronaut backward about a fixed center of mass.
CIt becomes backward, the way she moves, since she is most of the mass. A student who thinks the center of mass moves with the most massive part picks this. The wrench's forward motion also counts: the mass-weighted average of the two velocities is zero.
DIt changes, since both of the system's parts start to move during the throw. A student who thinks the system's velocity changes whenever its parts' velocities change picks this. The parts' changes are opposite and, weighted by mass, cancel, so the center of mass's velocity stays zero.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.5.A.1 Net force, F⃗netFix
Net force, F⃗net
The vector sum of all the forces exerted on a system by objects outside it: F⃗net = Σ F⃗. Forces are added as vectors, so their directions matter as well as their magnitudes. SI unit: N.
Unbalanced forces
A set of forces exerted on a system whose vector sum is not zero. Unbalanced forces can act on a system that is at rest at that instant, and balanced forces can act on a system that is moving.
Students often think A moving object must have a net force on it in its direction of motion; balanced forces mean the object is at rest. In fact No. A net force changes an object's velocity; it is not needed to keep the object moving. An object moving at constant velocity has balanced forces on it.
Students often think The net force on an object is found by adding the magnitudes of the forces, whatever their directions. In fact No. Forces are vectors and are added with their directions. Forces of 30 N and 40 N at right angles have a resultant of 50 N, and two opposite forces partly or wholly cancel.
2.5.A.2 Newton's second law Fix
Newton's second law
The acceleration of a system's center of mass is proportional to the net force exerted on the system, inversely proportional to the system's mass, and in the direction of the net force: a⃗sys = F⃗net/msys. 1 N = 1 kg·m/s².
Direction of acceleration
The acceleration of a system is in the direction of the net force, which need not be the direction of its velocity: a system moving upward with a downward net force slows down.
System mass, msys
The total mass of all the objects chosen as the system. For objects that move together, such as blocks joined by a string, Newton's second law can be applied to the whole system with msys equal to the sum of their masses, or to each object separately.
Velocity under a time-varying net force
When the net force varies with time, a(t) = Fnet(t)/m and v(t) = v0 + ∫₀ᵗ a dt; the constant-acceleration equations do not apply.
Students often think Only active agents, such as a person pushing or a motor pulling, exert forces that affect the acceleration; resisting surfaces such as a floor do not. In fact Yes. A floor that exerts a frictional force on a box is exerting a force just as a person pushing it is, and the friction must be included in the net force.
Students often think An object's acceleration is in the direction it is moving. In fact No. The acceleration is in the direction of the net force. An object moving left with a net force to the right has a rightward acceleration: it slows down and may reverse.
2.5.A.3 Internal and external forces Fix
Internal and external forces
Forces that objects within a system exert on each other are internal; they cancel in pairs in the sum for the whole system. Only a nonzero net external force changes the velocity of the system's center of mass.
Students often think Forces that the parts of a system exert on each other, such as a push between them, change the motion of the system's center of mass in the direction of the push. In fact No. Internal forces come in equal and opposite pairs exerted on two parts of the same system, so they cancel in the net external force. Only a nonzero net external force changes the velocity of the center of mass.
Students often think If the velocities of the parts of a system change, the velocity of the system's center of mass changes too. In fact No. The parts can change velocity in opposite ways while the mass-weighted average, the center of mass's velocity, stays the same. It changes only if the net external force is nonzero.
6 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 6
Block 1, of mass m, and block 2, of mass 2m, are joined by a light string and rest on a level, frictionless surface, with block 1 in front. A force of magnitude F, directed at an angle θ above the horizontal, pulls block 1 forward, and both blocks speed up together while staying on the surface. What is the tension in the string between the blocks?
Answer and reasoning
A0.67F cos θCorrect For both blocks together, the horizontal external force is F cos θ, so a = F cos θ/(3m). The string is the only horizontal force on block 2, so FT = (2m)a = (2/3)F cos θ ≈ 0.67F cos θ: less than the horizontal pull, because the pull also accelerates block 1.
B2.00F cos θ A student who uses only block 1's mass, the block the force is exerted on, gets a = F cos θ/m, and then the string must give block 2 a force (2m)a = 2F cos θ. The pull accelerates both blocks, so a = F cos θ/(3m) and FT = (2/3)F cos θ.
C0.50F cos θ A student who thinks the pull is shared equally between the two blocks picks this. Each block gets the force needed for the common acceleration, in proportion to its mass, so block 2, with two-thirds of the mass, needs two-thirds of F cos θ.
D1.00F cos θ A student who thinks the string passes the horizontal pull on at full strength picks this. If the tension equaled F cos θ, the net horizontal force on block 1 would be zero and it could not accelerate; the tension must be smaller.
Working System of both blocks: horizontal external force F cos θ, so a = F cos θ/(3m). Block 2 alone: the only horizontal force on it is the tension, so FT = (2m)a = (2/3)F cos θ ≈ 0.67F cos θ. (Block 1's mass alone, a = F cos θ/m, then (2m)a: 2F cos θ. Shared equally: (1/2)F cos θ. Passed on at full strength: F cos θ.) Checked with sympy.
The figure shows block A on a level, frictionless table, connected by a light string over an ideal pulley to block B, which hangs freely; each block's mass is labeled. When released, the blocks accelerate. Block A is then replaced by a block of mass 2m, and the blocks are released again. By what factor does the magnitude of the blocks' acceleration change?
Answer and reasoning
A×0.50 A student who uses only block A's mass picks this: Fg/m becoming Fg/(2m). Block B accelerates too, so the force must accelerate the total mass, 2m before and 3m after.
B×1.00 A student who thinks a block on a frictionless surface offers no resistance picks this. Without friction, block A still has inertia: its mass is part of the mass the force must accelerate.
C×0.67Correct Along the string, the only external force on the system of both blocks is the gravitational force on B, which does not change. That force must accelerate both blocks, so a = Fg/(mA + mB): from Fg/(2m) to Fg/(3m), a factor of 2/3.
D×1.50 A student who thinks a more massive system accelerates faster picks this, scaling a with the total mass from 2m to 3m. The extra mass on the table adds nothing to the driving force, so it reduces the acceleration.
Working Let Fg be the gravitational force on B (unchanged). Treat A, B and the string as one system moving along the string: the only external force along the motion is Fg, so a = Fg/msys. Before: a = Fg/(2m). After: a = Fg/(3m). Factor (1/3)/(1/2) = 2/3 ≈ ×0.67. (A's mass only: Fg/m to Fg/(2m), ×0.50. A's mass ignored on a frictionless table: ×1.00. Larger system faster, a ∝ msys: ×1.50.)
A cart of constant mass moves along a straight track. The graph shows its velocity v as a function of time t during three intervals, I, II and III. FI, FII and FIII are the magnitudes of the net force exerted on the cart during each interval. Which ranking is correct?
Answer and reasoning
AI > III > IICorrect The net force is proportional to the acceleration, the slope of the velocity–time graph. Magnitudes of the slopes: 3.0 m/s² in I, 0 in II and 1.5 m/s² in III. So FI > FIII > FII, with no net force at all while the cart moves at a constant 6 m/s.
BII > I = III A student who reads the net force from the velocity picks this: the cart is fastest in II. In II the velocity is constant, so the acceleration and the net force are zero.
CI > II > III A student who ranks the signed slopes picks this, putting the negative slope of III below zero. The question asks for magnitudes: 1.5 m/s² in III is larger than 0 in II.
DII > III > I A student who reads the net force from the area under the graph picks this: 18, 12 and 6 m. The area is the displacement in each interval; the net force is given by the slope.
Working |Fnet| = m|a| = m × |slope|. I: 6/2 = 3.0 m/s². II: 0. III: 6/4 = 1.5 m/s². FI > FIII > FII. (Heights: averages 3, 6, 3 m/s: II > I = III. Signed slopes +3, 0, −1.5: I > II > III. Areas 6, 18, 12 m: II > III > I.)
Two carts on a straight track collide during a short interval. A student claims that a nonzero net external force was exerted on the system of the two carts during that interval. Which observation, if made, would support the student's claim?
Answer and reasoning
AThe velocity of the first cart changed sharply while the carts pushed on each other. A student who thinks a change in one part's velocity shows a change in the system's motion picks this. The carts' push on each other changes each cart's velocity, but those forces are internal; the center of mass can keep the same velocity.
BThe carts exerted very large forces on each other during the time they were touching. A student who thinks forces between parts of a system change its overall motion picks this. However large, the carts' forces on each other are internal and cancel in pairs, so they are no evidence of a net external force.
CThe system's center-of-mass velocity was different after the collision than before.Correct By Newton's second law, the velocity of a system's center of mass changes only if a nonzero net external force is exerted on the system. A measured change in the center of mass's velocity is therefore direct evidence of a nonzero net external force, for example from friction with the track.
DThe center of mass of the two carts kept moving steadily throughout the whole interval. A student who thinks motion requires a net force picks this. A center of mass moving at constant velocity has zero net external force on its system; only a change in that velocity shows a nonzero net external force.
An object of mass m moves along the x-axis with velocity v₀ in the +x direction at time t = 0. From t = 0 on, the net force exerted on it is directed in the −x direction and has magnitude bt, where b is a positive constant. What is the object's x-velocity at time t, for times before it stops?
Answer and reasoning
A−bt²/(2m) A student who leaves out the initial velocity when integrating picks this. ∫ a dt gives the change in velocity; the object already had v₀ at t = 0, so v = v₀ − bt²/(2m).
Bv₀−bt/m A student who treats the acceleration at time t as the change in velocity picks this, adding a = Fx/m to v₀. The units show the error: bt/m is an acceleration, in m/s², not a velocity. The change in velocity is the integral of a over time.
Cv₀−bt²/m A student who uses v = v₀ + at with the acceleration at time t picks this. The acceleration grows from zero, so its average over the interval is half its final value; integrating gives v₀ − bt²/(2m).
Dv₀−bt²/(2m)Correct Newton's second law gives ax(t) = −bt/m, which changes with time, so the constant-acceleration equations do not apply. Integrating, v(t) = v₀ + ∫₀ᵗ (−bs/m) ds = v₀ − bt²/(2m), which correctly equals v₀ at t = 0.
Working a(t) = Fx/m = −bt/m. v(t) = v₀ + ∫₀ᵗ (−bs/m) ds = v₀ − bt²/(2m). Check: v(0) = v₀. (v₀ left out: −bt²/(2m). Acceleration at t taken as the change in velocity, v₀ + Fx/m: v₀ − bt/m, where bt/m has units m/s². Constant-acceleration equation with the acceleration at time t: v₀ − (bt/m)t = v₀ − bt²/m.) Checked with sympy.
A 2.0 kg cart on a level track with negligible friction is moving to the left at 3.0 m/s. At t = 0, a string starts to pull it to the right with a constant force of 6.0 N, while a hand pushes it to the left with a constant force of 2.0 N. Taking rightward as positive, what is the cart's velocity at t = 2.0 s?
Answer and reasoning
A−7.0 m/s A student who takes the acceleration to be in the direction of motion picks this: a = −2.0 m/s², so v = −3.0 − 4.0 m/s. The acceleration is in the direction of the net force, to the right, so the cart slows down.
B+4.0 m/s A student who leaves out the initial velocity picks this: v = at = (2.0 m/s²)(2.0 s). The cart was already moving at −3.0 m/s, so v = −3.0 m/s + 4.0 m/s.
C+1.0 m/sCorrect The net force is 6.0 N − 2.0 N = 4.0 N to the right, so a = +2.0 m/s², in the direction of the net force although the cart is moving left. Then v = −3.0 m/s + (2.0 m/s²)(2.0 s) = +1.0 m/s: the cart slows, stops and moves right.
D+3.0 m/s A student who uses only the largest force picks this: a = 6.0 N/2.0 kg = 3.0 m/s², giving −3.0 + 6.0 m/s. The hand's 2.0 N push also acts, so the net force is 4.0 N.
Working Fnet = +6.0 N − 2.0 N = +4.0 N; a = Fnet/m = +4.0 N/2.0 kg = +2.0 m/s² (rightward, opposite to the velocity). v = v₀ + at = −3.0 m/s + (2.0 m/s²)(2.0 s) = +1.0 m/s: the cart has stopped and reversed. (a taken in the direction of motion: −3.0 − 4.0 = −7.0 m/s. v₀ left out: +4.0 m/s. Largest force only, a = +3.0 m/s²: −3.0 + 6.0 = +3.0 m/s.)
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
