5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
Which statement correctly describes the net force on an object?
Answer and reasoning
AIt is one more force on the object, drawn along with the others. A student who treats the net force as a force of its own picks this. No object exerts the net force; it is the sum of the forces already acting, and counting it as well would count every force twice.
BIt is what all the forces on the object add up to, directions included.Correct The net force on an object is the vector sum of all the forces that other objects exert on it, F⃗net = ΣF⃗i, found by adding the forces' components along each axis.
CIt is equal to the largest of the forces exerted on the object. A student who thinks the strongest force 'wins' picks this. A book resting on a table has a gravitational force on it, but the table's push balances it, so the net force is zero, smaller than either force.
DIt points in the direction in which the object is moving. A student who thinks motion needs a force in its direction picks this. A ball thrown upward moves up while the only force on it, Earth's pull, points down, so the net force points opposite to its motion.
The diagram shows a block held at rest on a ramp with negligible friction by a horizontal force of magnitude F. The gravitational force on the block has magnitude Fg. Use sin 37° = 0.60 and cos 37° = 0.80. Which expression gives F?
Answer and reasoning
A0.60Fg A student who takes the push to be the component of the weight along the ramp, F = Fg sin 37°, picks this. The push is horizontal, so only its component F cos 37° acts along the ramp, and F must be larger: F = Fg tan 37°.
B1.33Fg A student who swaps sine and cosine writes F sin 37° = Fg cos 37° and gets F = Fg/tan 37°. The push makes 37° with the ramp's surface, so its component along the ramp is F cos 37°, and Fg's is Fg sin 37°.
C1.00Fg A student who thinks a force holding an object at rest must equal its weight picks this. Here the ramp's normal force also supports the block, and only the components along the ramp must balance: F = Fg tan 37°.
D0.75FgCorrect The block is in equilibrium, so the components of the forces along the ramp add to zero. The horizontal push makes 37° with the ramp, giving F cos 37° up the ramp; Fg gives Fg sin 37° down the ramp; the normal force has no component along the ramp. F cos 37° = Fg sin 37°, so F = Fg tan 37° = 0.75Fg.
Working Equilibrium along the ramp (x up the ramp): F is horizontal, 37° from the ramp: +F cos 37°; Fg has component −Fg sin 37° along the ramp; FN has none. F cos 37° − Fg sin 37° = 0 ⇒ F = Fg tan 37° = (0.60/0.80)Fg = 0.75Fg. (Perpendicular: FN = Fg cos 37° + F sin 37° > 0, so the block stays on the ramp.)
A puck slides on a horizontal frictionless table, guided around a curve by a barrier fixed to the table, and then leaves the end of the barrier. Air resistance is negligible, so once the puck has left the barrier the net force on it is zero. Seen from above, how does the puck move after it leaves the barrier?
Answer and reasoning
AIn a straight line, moving at constant speedCorrect By Newton's first law, with zero net force the puck's velocity stays constant in magnitude and direction. It therefore moves in a straight line, along the direction it had at the end of the barrier, at constant speed.
BAlong a curve that continues the barrier's curve A student who thinks curved motion persists after the force that caused it stops picks this. The barrier's push bent the puck's path only while they were in contact; with zero net force the velocity, and so the direction of motion, stays constant.
CIn a straight line, slowing until it stops A student who thinks motion wears out picks this. The net force is zero, so the speed cannot change; on the frictionless table nothing slows the puck.
DStraight outward, away from the curve's center A student who thinks an object leaving a curve is flung outward picks this. With zero net force the puck keeps the velocity it had as it left the barrier, which is tangent to the curve, not directed away from its center.
A space probe far from any star or planet drifts in the +x direction at constant velocity with its engines off. A side thruster then fires, exerting a constant force on the probe in the +y direction only. While the thruster fires, which statement describes the probe's velocity?
Answer and reasoning
AIt turns at once to point in +y, the direction of the force. A student who thinks an object moves in the direction of the latest force picks this. No force acts in the x-direction, so vx does not change; the probe keeps drifting in +x while it gains velocity in +y.
BIts x-component stays constant while its y-component increases.Correct The forces on the probe are balanced in the x-direction (there are none) and unbalanced in the y-direction. The velocity changes only in the direction of the unbalanced force: vx keeps its original value while vy grows for as long as the thruster fires, so the probe's path curves gradually toward +y.
CIt becomes a new constant velocity between +x and +y. A student who thinks a constant force produces a constant velocity picks this. The force in +y is unbalanced, so vy keeps changing for as long as it acts; the velocity does not settle to a constant value.
DIts x-component decreases while its y-component increases. A student who thinks motion is shared between directions picks this. The forces in the x-direction are balanced, so vx stays constant; the force in +y adds a y-component without taking anything from vx.
A ball rests on the smooth, level floor of a train car; friction between the ball and the floor is negligible. While the train speeds up along a straight, level track, a passenger sees the ball start to slide toward the back of the car, although no object exerts a horizontal force on the ball. A person standing on the platform sees the ball's horizontal position stay the same. Which conclusion do these observations support?
Answer and reasoning
AThe train's frame is inertial; a backward force acts on the ball. A student who takes the backward 'push' felt in a speeding-up vehicle to be a real force picks this. No object exerts a backward force on the ball, and the platform observer sees no change in its horizontal motion; the ball's apparent acceleration comes from observing in the accelerating train's frame.
BThe train's frame is noninertial at all times, since it moves. A student who thinks any moving frame is noninertial picks this. A train moving at constant velocity is an inertial frame, in which the ball would stay at rest; the train's frame is noninertial only while it accelerates.
CBoth frames are inertial; the observers just describe the ball differently. A student who thinks every frame is inertial picks this. In the train's frame an object with no horizontal force on it changes velocity, which the first law forbids in an inertial frame, so the two frames are not equivalent.
DThe train's frame is noninertial during the time the train speeds up.Correct In the train's frame, the ball's velocity changes although no horizontal force is exerted on it, so Newton's first law fails there: while it speeds up, the train's frame is not an inertial frame. From the platform, the ball keeps its velocity, as the first law predicts, consistent with the platform being an inertial frame.
In preparation: 0 of 5 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
2.4.A.1 Net force, F⃗netFix
Net force, F⃗net
The vector sum of all the forces exerted on a system, F⃗net = ΣF⃗i, found by adding the forces' components along each axis. It is not an additional force exerted by any object. SI unit: newton (N).
Students often think The net force is one more force exerted on an object, to be drawn and counted along with the other forces. In fact No. The net force is the vector sum of the forces that other objects exert on the object. It is not exerted by any object of its own and is not drawn as an extra arrow alongside the individual forces.
Students often think The net force on an object is the largest of the forces exerted on it, the one that 'wins'. In fact No. The net force is the vector sum of all the forces. A book at rest on a table has a large gravitational force on it, but the table's push balances it, and the net force is zero.
2.4.A.2 Translational equilibrium Fix
Translational equilibrium
The configuration of forces for which the net force on a system is zero: ΣF⃗i = 0, so the forces' components add to zero along every axis (ΣFx = 0 and ΣFy = 0). A system in translational equilibrium can be at rest or moving with constant velocity.
Students often think The horizontal component of a force is always found with the cosine, and the vertical component with the sine, of whatever angle is given. 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 uses the sine.
Students often think To hold an object at rest, a force equal in magnitude to the object's weight must be exerted on it, whatever its direction and whatever other forces act. In fact No. In equilibrium the forces on the object add to zero as vectors. If it is the only force besides the weight, a supporting force must point straight up and equal the weight; but a force in another direction, or one of several supporting forces, can be larger or smaller than the weight.
2.4.A.3 Newton's first law Fix
Newton's first law
If the net force exerted on a system is zero, the velocity of the system remains constant, in magnitude and in direction: a system at rest stays at rest, and a moving system keeps moving in a straight line at constant speed. Equivalently, if a system's velocity changes, the net force on it is not zero.
Students often think An object moves only while a net force acts on it in its direction of motion, so a moving object must have a net force on it. In fact No. By Newton's first law an object with zero net force keeps moving with constant velocity. A net force is needed to change the velocity, not to maintain it.
Students often think A moving object slows down and stops by itself as its motion is 'used up', even when the net force on it is zero. In fact No. With zero net force, the velocity stays constant: the object keeps its speed and its direction. Everyday objects slow down because friction or air resistance gives them a nonzero net force.
2.4.A.4 Forces balanced along one axis Fix
Forces balanced along one axis
Forces whose components add to zero along one axis but not along another. The component of the system's velocity along the balanced axis stays constant; only the component along the direction of the unbalanced force changes.
Students often think When a new force is applied to a moving object, the object immediately moves in the direction of that force. In fact No. A force changes only the component of the velocity along the direction in which the forces are unbalanced. The velocity component perpendicular to the force is unchanged, so the object curves gradually rather than turning at once.
Students often think A constant sideways force gives a moving object a new constant velocity, in a direction between its original motion and the force. In fact No. A constant unbalanced force changes the velocity continuously: the component along the force keeps changing for as long as the force acts, so the velocity does not settle to a new constant value.
2.4.A.5 Inertial reference frame Fix
Inertial reference frame
A reference frame from which an observer would verify Newton's first law: an object on which the net force is zero is seen to move with constant velocity. A frame moving with constant velocity relative to an inertial frame is also inertial.
Noninertial reference frame
A reference frame that accelerates (speeds up, slows down or changes direction) relative to an inertial frame. From it, objects on which the net force is zero can be seen to change velocity, so Newton's first law does not hold in it.
Students often think When a vehicle speeds up, a real backward force is exerted on unattached objects inside it, pushing them toward the back. In fact No. No object exerts such a force. Seen from the ground, the unattached object keeps its velocity while the vehicle speeds up beneath it. In the vehicle's frame, which is noninertial, the object appears to accelerate with no net force on it.
Students often think Any reference frame that moves is noninertial; only a frame at rest is inertial. In fact No. A frame moving with constant velocity relative to an inertial frame is itself inertial; Newton's first law holds in it. Only a frame that accelerates relative to an inertial frame is noninertial.
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
The diagram shows a top view of a ring on a horizontal table and the three horizontal forces exerted on it by three ropes, with their magnitudes and directions. Friction is negligible. Use sin 53° = 0.80 and cos 53° = 0.60. What is the magnitude of the net horizontal force on the ring?
Answer and reasoning
A60 N A student who adds the forces' magnitudes picks this: 10 N + 30 N + 20 N. The forces point in different directions and partly cancel, so the net force is much smaller than the sum of the magnitudes.
B24 N A student who adds the sizes of the net force's components picks this: 6 N + 18 N = 24 N. The components are perpendicular, so they combine as the sides of a right triangle: √(6² + 18²) N ≈ 19 N.
C19 NCorrect Add the forces component by component. The 20 N force is 53° from the −y direction, so its components are −20 N × sin 53° = −16 N and −20 N × cos 53° = −12 N. ΣFx = 10 N − 16 N = −6 N and ΣFy = 30 N − 12 N = 18 N, so |F⃗net| = √((6 N)² + (18 N)²) ≈ 19 N.
D14 N A student who always uses the cosine for the x-component and the sine for the y-component, whatever the angle is measured from, gets (−12 N, −16 N) for the 20 N force and a net force of about 14 N. The 53° angle is measured from the y-direction, so the x-component uses the sine.
Working F⃗₁ = (10, 0) N; F⃗₂ = (0, 30) N; F⃗₃ makes 53° with the −y direction, toward −x: F⃗₃ = (−20 sin 53°, −20 cos 53°) N = (−16, −12) N. Sum: ΣFx = 10 − 16 = −6 N; ΣFy = 30 − 12 = 18 N. |F⃗net| = √(6² + 18²) N = √360 N = 18.97 N ≈ 19 N.
A picture hangs at rest from two identical cords attached to its upper corners. When both cords are vertical, the tension in each cord is T₀. The picture is rehung so that the two cords meet at a single nail above the picture, each cord making an angle of 60° with the vertical. What is now the tension in each cord?
Answer and reasoning
A2.00T₀Correct With vertical cords, 2T₀ = Fg. With each cord at 60° to the vertical, the horizontal components cancel and the vertical components must add to Fg: 2T cos 60° = Fg, so T = Fg/(2 × 0.50) = Fg = 2T₀.
B1.00T₀ A student who thinks each cord carries half the weight whatever its angle picks this. Only the vertical component of each tension supports the picture, so slanted cords need a larger tension: T cos 60° = T₀.
C0.50T₀ A student who treats the tension as a component of the half weight it supports writes T = T₀ cos 60°. It is the tension's vertical component that equals T₀, so T = T₀/cos 60° = 2T₀.
D1.15T₀ A student who uses the sine for the vertical component writes 2T sin 60° = Fg and gets T = T₀/sin 60° ≈ 1.15T₀. The 60° angle is measured from the vertical, so the vertical component is T cos 60°.
Working Vertical cords: 2T₀ = Fg. At 60° from the vertical, by symmetry the horizontal components cancel, and ΣFy = 0 gives 2T cos 60° = Fg, so T = Fg/(2 cos 60°) = Fg = 2T₀.
A sign of weight Fg hangs at rest below a knot that is held by two cords attached to the ceiling, at the angles shown. T₁ and T₂ are the tensions in cords 1 and 2. Which ranking of T₁, T₂ and Fg is correct?
Answer and reasoning
AT₂ < T₁ < Fg A student who thinks the flatter cord always has the larger tension picks this. The horizontal components must be equal, and the flatter cord 1 has the larger fraction of its tension horizontal, so it is cord 1 that needs the smaller tension.
BT₁ = T₂ < Fg A student who thinks each cord carries half the weight whatever its angle picks this. With different angles, equal horizontal components require different tensions.
CT₁ < T₂ < FgCorrect The knot is in equilibrium. Horizontally, T₁ cos 37° = T₂ cos 53°, so the steeper cord 2 has the larger tension, T₂ = (4/3)T₁. Vertically, T₁ sin 37° + T₂ sin 53° = Fg, which gives T₁ = 0.60Fg and T₂ = 0.80Fg: both less than Fg.
DFg < T₂ < T₁ A student who thinks a slanted cord always needs more than the weight, and more the farther it slants from vertical, picks this. Here the cords are steep enough, and share the load, so that T₁ = 0.60Fg and T₂ = 0.80Fg, both less than Fg, and the steeper cord 2 carries more.
Working Knot in equilibrium. x: T₁ cos 37° = T₂ cos 53° ⇒ 0.80T₁ = 0.60T₂ ⇒ T₂ = (4/3)T₁. y: T₁ sin 37° + T₂ sin 53° = Fg ⇒ 0.60T₁ + 0.80(4/3)T₁ = Fg ⇒ T₁ = 0.60Fg, T₂ = 0.80Fg. So T₁ < T₂ < Fg.
The graph shows the position x of a cart moving along a straight, level track as a function of time t. A student claims that the net force on the cart was zero for the whole time from 0 to 6 s. During which interval do the data show that the student's claim is wrong?
Answer and reasoning
AFrom 0 to 2 s, where the cart moves at its highest speed A student who links net force with speed picks this. From 0 to 2 s the graph is a straight line, so the velocity is constant, as Newton's first law predicts for zero net force, however fast the cart moves.
BFrom 0 to 4 s, the whole time during which the cart moves A student who thinks a moving object needs a net force picks this. From 0 to 2 s the cart moves at constant velocity, which is consistent with zero net force; only the change in velocity from 2 s to 4 s shows a nonzero net force.
CFrom 4 s to 6 s, where the cart's position is at its largest A student who reads the height of the graph as a sign of force picks this. From 4 s to 6 s the graph is horizontal: the cart is at rest and its velocity does not change, which is consistent with zero net force.
DFrom 2 s to 4 s, where the slope of the graph changesCorrect The slope of a position–time graph is the velocity. From 2 s to 4 s the slope falls from 3 m/s to zero, so the cart's velocity changes. By Newton's first law, a zero net force would keep the velocity constant, so the net force cannot have been zero in this interval. The straight segments are consistent with zero net force.
Working 0–2 s: straight line, constant slope 3 m/s: constant velocity, consistent with zero net force. 2–4 s: slope decreases from 3 m/s to 0: the velocity changes, so by Newton's first law the net force cannot have been zero. 4–6 s: horizontal line, cart at rest: consistent with zero net force.
A puck slides on a horizontal frictionless table with velocity (4.0 m/s)î. At t = 0, air jets begin to exert a constant horizontal force on the puck in the −y direction only. From then on, the puck's y-coordinate is y(t) = −(0.75 m/s²)t². What is the speed of the puck at t = 2.0 s?
Answer and reasoning
A5.0 m/sCorrect The forces are balanced in the x-direction, so vx stays 4.0 m/s. In the y-direction, vy = dy/dt = −(1.5 m/s²)t = −3.0 m/s at 2.0 s. The speed is √((4.0 m/s)² + (3.0 m/s)²) = 5.0 m/s.
B3.0 m/s A student who thinks the puck now moves only in the direction of the force keeps only vy = 3.0 m/s. Nothing acts in the x-direction, so the puck keeps its 4.0 m/s in that direction as well.
C7.0 m/s A student who adds the velocity components as numbers gets 4.0 m/s + 3.0 m/s. The components are perpendicular, so the speed is √(4.0² + 3.0²) m/s = 5.0 m/s.
D1.0 m/s A student who thinks the perpendicular force takes motion away from the x-direction subtracts: 4.0 m/s − 3.0 m/s. The x-direction forces are balanced, so vx stays 4.0 m/s, and the speed grows to 5.0 m/s.
Working x-direction: no force, so the forces are balanced and vx stays 4.0 m/s. y-direction: vy = dy/dt = −(1.5 m/s²)t = −3.0 m/s at t = 2.0 s. Speed = √((4.0 m/s)² + (3.0 m/s)²) = 5.0 m/s.
A lamp of weight Fg hangs at rest from a knot. Cord 1 runs horizontally from the knot to a wall, and cord 2 runs from the knot to the ceiling at an angle of 53° from the vertical. Use sin 53° = 4/5 and cos 53° = 3/5. Which expression gives the tension in cord 1?
Answer and reasoning
A0.75Fg A student who always uses the sine for a vertical component and the cosine for a horizontal one writes T₂ sin 53° = Fg and T₁ = T₂ cos 53°, getting (3/4)Fg. The 53° angle is measured from the vertical, so cord 2's vertical component is T₂ cos 53°.
B1.33FgCorrect The knot is in equilibrium. Only cord 2 has a vertical component, so T₂ cos 53° = Fg and T₂ = (5/3)Fg. Horizontally, cord 1 balances cord 2's horizontal component: T₁ = T₂ sin 53° = (5/3)(4/5)Fg = (4/3)Fg ≈ 1.33Fg.
C0.00Fg A student who thinks a horizontal cord takes no part in holding up a weight picks this. Cord 2 pulls the knot sideways as well as up, and only cord 1 can balance that horizontal pull, so cord 1 carries a tension of (4/3)Fg.
D1.00Fg A student who thinks a cord holding an object must carry its weight picks this. Cord 1 is horizontal and supplies no upward force; its tension is set by the horizontal balance, T₁ = T₂ sin 53° = (4/3)Fg.
Working Knot in equilibrium. y: T₂ cos 53° = Fg ⇒ T₂ = (5/3)Fg. x: T₁ = T₂ sin 53° = (5/3)(4/5)Fg = (4/3)Fg ≈ 1.33Fg.
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
