5 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 5
A puck rests on level ice with negligible friction. The diagram shows, viewed from above, the only two horizontal forces exerted on it, F⃗1 and F⃗2. The vertical forces on the puck balance. What is the magnitude of the net force on the puck?
Answer and reasoning
A7.0 N A student who adds magnitudes regardless of direction picks this: 3.0 N + 4.0 N = 7.0 N. That would be right only if both forces pointed the same way; perpendicular forces combine as the sides of a right triangle.
B5.0 NCorrect The net force is the vector sum of the forces. Placed tip to tail, the 3.0 N and 4.0 N forces form the two perpendicular sides of a right triangle, and the net force is its hypotenuse: √(3.0² + 4.0²) N = 5.0 N, directed between east and north.
C1.0 N A student who thinks forces at an angle partly cancel picks this: 4.0 N − 3.0 N = 1.0 N. Only opposite forces subtract, and perpendicular forces have no parts that oppose each other.
D4.0 N A student who takes the net force to be the largest single force picks this. The 3.0 N force also contributes: the net force is the vector sum of both forces, 5.0 N.
Working The two horizontal forces are perpendicular, so their vector sum is the hypotenuse of a right triangle with sides 3.0 N and 4.0 N: |F⃗net| = √((3.0 N)² + (4.0 N)²) = √(25 N²) = 5.0 N. The vertical forces add to zero, so this is the magnitude of the net force.
Which of the following objects is in translational equilibrium?
Answer and reasoning
AA ball at the top of its flight, momentarily at rest A student who thinks any object at rest is in equilibrium picks this. The ball is at rest for only an instant: Earth's pull is still the only force on it, so the net force is not zero, and its velocity keeps changing, from upward to downward.
BA car rounding a curve on a level road at a constant speed A student who treats constant speed as constant velocity picks this. The car's direction keeps changing as it rounds the curve, so its velocity is not constant and the net force on it is not zero.
CA puck slowing down on rough ice after a stick stops pushing it A student who thinks moving objects slow down on their own, without a force, picks this. The puck slows because the rough ice exerts a backward force on it; its velocity is changing, so the net force is not zero.
DA skydiver falling straight down at a constant velocityCorrect In translational equilibrium the net force is zero, and then the velocity stays constant. A skydiver whose velocity, both speed and direction, is constant has zero net force: the air's upward force balances Earth's downward pull.
A space probe is so far from any star or planet that the forces exerted on it are negligible. Its engines fire briefly and then shut off. Which statement describes the probe's motion after the engines have shut off?
Answer and reasoning
AIt slows down and stops, as all moving objects do in time. A student who thinks rest is the natural state of moving objects picks this. Objects on Earth stop because friction and air resistance act on them; with no forces on the probe, nothing changes its velocity.
BIt slows down as the force from the engine firing is used up. A student who thinks the engines gave the probe a store of force picks this. The engines exert a force only while they fire. Afterward no force acts, so there is nothing to be used up and the probe's velocity stays constant.
CIt moves in a straight line at a constant speed.Correct With zero net force, Newton's first law says that the probe's velocity stays constant: it keeps the speed and direction it had when the engines shut off. No force is needed to keep it moving.
DIt stops almost at once, as no force now pushes it forward. A student who thinks motion needs a continuing force picks this. A force is needed to change velocity, not to keep it: with zero net force, the probe continues at the velocity it had when the engines shut off.
A puck slides east at constant velocity across level ice with negligible friction. Starting at time t, a constant force directed north is exerted on the puck, and no other horizontal force acts. Which statement describes the puck's motion after time t?
Answer and reasoning
AIt turns and moves due north, in the direction of the force on it. A student who thinks an object moves in the direction of the force on it picks this. The force changes only the north–south part of the velocity; the puck keeps its eastward velocity, so it never moves due north.
BIts eastward velocity decreases while its northward velocity increases. A student who thinks a sideways force also slows the original motion picks this. No east–west force acts on the puck, so nothing changes the eastward component of its velocity.
CIt moves at a constant velocity in a direction between east and north. A student who thinks the original motion and the force combine into one new constant velocity picks this. A constant unbalanced force keeps changing the velocity: the northward component keeps increasing while the eastward component stays the same, so the puck's direction keeps turning toward the north.
DIts eastward velocity stays the same while its northward velocity increases.Correct The horizontal forces are balanced in the east–west direction and unbalanced in the north–south direction. The velocity changes only in the direction of the unbalanced force: the eastward component stays constant while the northward component grows, so the path curves toward the north.
A train is at rest on a straight, level track. A ball rests on the smooth floor of one of its carriages; friction and air resistance are negligible. The train then starts to speed up forward, and a passenger sees the ball slide toward the back of the carriage, although no horizontal force is exerted on the ball. Which statement is correct?
Answer and reasoning
AThe carriage is not an inertial frame while the train speeds up.Correct An inertial frame is one in which Newton's first law holds. In the speeding-up carriage the ball starts to move although no horizontal force acts on it, so the first law fails there and the carriage is not an inertial frame. Seen from the ground, an inertial frame, the ball simply stays at rest while the carriage speeds up beneath it.
BA backward force caused by the train's motion is exerted on the ball. A student who takes the apparent push as a real force picks this. No object exerts a backward force on the ball; it only seems to be pushed because the passenger observes from a frame that is speeding up.
CThe carriage is an inertial frame, as the train moves in a straight line. A student who thinks straight-line motion makes a frame inertial picks this. What matters is whether the frame's velocity changes. The carriage is speeding up, so the first law fails in it, as the sliding ball shows.
DNo moving vehicle is an inertial frame, even one at constant velocity. A student who thinks only frames at rest on the ground are inertial picks this. A carriage moving at constant velocity is an inertial frame: a ball on its floor would stay at rest, just as on the ground. This carriage is not inertial because it is speeding up.
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 Fix
Net force
F⃗net = ΣF⃗: the vector sum of all the forces exerted on a system; unit newton (N). It is found by adding the forces' components along each axis, not by adding their magnitudes.
Net force in one dimension
Along a single axis each force is a signed component, with opposite directions given opposite signs, so the net force is the algebraic sum of the components.
Students often think The net force of several forces is found by adding their magnitudes, subtracting only forces that point in exactly opposite directions, whatever the angles between them. In fact Only when they all point the same way. The net force is a vector sum: opposite components subtract, and perpendicular forces combine as the sides of a right triangle.
Students often think Two forces that do not point in the same direction partly oppose each other, so the magnitude of their net force is the difference between their magnitudes. In fact Only if they have parts in opposite directions. Two perpendicular forces have no opposing parts: their net force, √(F1² + F2²), is larger than either force.
2.4.A.2 Translational equilibrium Fix
Translational equilibrium
The configuration of forces in which the net force exerted on a system is zero, Σi F⃗i = 0. It holds for a system at rest and for one moving at constant velocity.
Balanced forces
Forces exerted on the same system whose vector sum is zero. Balanced forces are present, not absent; they leave the system's velocity unchanged.
Students often think An object is in translational equilibrium exactly when it is at rest, even if it is at rest for only an instant. In fact No to both. An object moving at constant velocity is in equilibrium, and an object at rest for only an instant, such as a ball at the top of its flight, is not.
Students often think An object in translational equilibrium, or at rest, has no forces exerted on it. In fact No. It means that the forces exerted on the object add to zero. A book at rest on a table has two forces exerted on it, and they balance.
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: a system at rest stays at rest, and a moving system keeps the same speed and direction.
Constant velocity
Motion with unchanging speed and unchanging direction, that is, in a straight line at constant speed. An object moving at constant speed along a curve does not have constant velocity.
Students often think An object keeps moving only while a force pushes it in its direction of motion; once the push stops, the object stops. In fact No. A force is needed to change an object's velocity, not to keep it. With zero net force, a moving object keeps moving at constant velocity.
Students often think A constant net force gives an object a constant velocity, in the direction of the force, with a speed set by the size of the force. In fact No. A constant unbalanced force changes the velocity continually. Constant velocity requires zero net force.
2.4.A.4 Balance along separate axes Fix
Balance along separate axes
Forces may be balanced along one axis and unbalanced along another. The velocity changes only in the direction of the unbalanced force; the velocity component perpendicular to it does not change.
Students often think An object moves in the direction of the force exerted on it, so a sideways force turns a moving object to move along the force. In fact No. A force changes the velocity in the direction of the force; the object keeps any velocity it already has in other directions, so it need not move along the force.
Students often think A force exerted at right angles to an object's motion also reduces its velocity in the original direction. In fact No. A force has no effect on the velocity component perpendicular to it; only the component along the force changes.
2.4.A.5 Reference frame Fix
Reference frame
The viewpoint, with its own coordinate system, from which positions and velocities are measured, such as the ground or the inside of a moving train.
Inertial reference frame
A reference frame from which an observer would verify Newton's first law: an object with zero net force stays at rest or keeps a constant velocity. A frame at rest on the ground, or moving at constant velocity relative to it, is inertial to a good approximation.
Noninertial reference frame
A frame that is speeding up, slowing down or changing direction. Observed from it, objects with zero net force appear to change velocity, so Newton's first law is not verified.
Students often think In a vehicle that speeds up, slows down or turns, a force caused by the vehicle's motion pushes loose objects, such as a ball on the floor, backward or outward. In fact No. No object exerts that backward force. Seen from the ground, the loose object keeps its velocity while the vehicle speeds up; it only appears to be pushed back because the vehicle is not an inertial frame.
Students often think A frame that moves in a straight line is an inertial frame, even while it speeds up or slows down. In fact No. A frame is inertial only if Newton's first law holds in it, which requires its velocity to be constant. A vehicle speeding up in a straight line is not inertial.
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
A crate rests on level ice with negligible friction. Viewed from above, two horizontal forces, each of magnitude 10 N, are exerted on it in three different setups. In setup 1 both forces point east. In setup 2 one points east and the other points north. In setup 3 one points east and the other points west. The vertical forces on the crate balance in every setup. If F1, F2 and F3 are the magnitudes of the net force on the crate in setups 1, 2 and 3, which ranking is correct?
Answer and reasoning
AF1 = F2 > F3 A student who adds magnitudes unless the forces are exactly opposite picks this, finding 20 N for both setup 1 and setup 2. Perpendicular forces combine to about 14 N, less than 20 N, because they do not point the same way.
BF1 > F2 = F3 A student who thinks forces at an angle partly cancel picks this, treating the perpendicular forces of setup 2 like the opposite forces of setup 3. Perpendicular forces do not oppose each other, so their net force is about 14 N, not zero.
CF1 > F2 > F3Correct The net force is a vector sum. Forces in the same direction add fully (20 N), perpendicular forces combine as the sides of a right triangle (√(10² + 10²) N ≈ 14 N), and opposite forces cancel (0). So F1 > F2 > F3.
DF1 = F2 = F3 A student who takes the net force to be the largest single force picks this, finding 10 N in every setup. The net force depends on the directions of all the forces: 20 N, about 14 N and 0.
Working Setup 1: 10 N + 10 N = 20 N. Setup 2: √((10 N)² + (10 N)²) ≈ 14 N. Setup 3: 10 N − 10 N = 0. So F1 > F2 > F3.
A block is held at rest on a ramp by a light string that is parallel to the ramp and tied to a post at the top, as shown in the diagram. The ramp is inclined at 37° above the horizontal, friction between the block and the ramp is negligible, and the gravitational force on the block is 50 N. Use sin 37° = 0.60 and cos 37° = 0.80. What is the tension in the string?
Answer and reasoning
A30 NCorrect The block is at rest, so the net force on it is zero. Along the ramp only two forces have components: the string's pull up the ramp and the component of the gravitational force down the ramp, Fg sin 37° = 50 N × 0.60 = 30 N. These must balance, so the tension is 30 N.
B40 N A student who uses cos 37° for the component along the ramp picks this: 50 N × 0.80 = 40 N. That is the component perpendicular to the ramp, which the normal force balances; the component along the ramp is Fg sin 37° = 30 N.
C50 N A student who thinks a string always pulls with the full weight of what it holds picks this. Here the ramp's normal force supports part of the block, and the string balances only the component of the gravitational force along the ramp, 30 N.
D83 N A student who divides by the trigonometric ratio picks this: 50 N ÷ 0.60 ≈ 83 N. A component is found by multiplying by sin 37° or cos 37°, and no component of the 50 N gravitational force can be larger than 50 N.
Working The block is in translational equilibrium, so the force components along each axis sum to zero. Take axes parallel and perpendicular to the ramp. Along the ramp: the string pulls up the ramp with FT; the gravitational force has a component Fg sin 37° = 50 N × 0.60 = 30 N down the ramp; the normal force has no component along the ramp. FT − 30 N = 0, so FT = 30 N. (Perpendicular to the ramp: FN = Fg cos 37° = 40 N.)
A cart moves along a straight, level track. The graph shows its velocity v as a function of time t. Which claim about the net force on the cart is supported by the graph?
Answer and reasoning
AIt is zero from 2 s to 5 s and from 7 s to 9 s, where the velocity is constant.Correct By Newton's first law, zero net force goes with constant velocity, and a changing velocity needs a nonzero net force. The graph is level from 2 s to 5 s (constant 4 m/s) and from 7 s to 9 s (at rest), so the net force is zero in both intervals.
BIt is zero from 7 s to 9 s alone, the interval in which the cart is at rest. A student who thinks only an object at rest can be in equilibrium picks this. From 2 s to 5 s the cart moves at a constant 4 m/s; constant velocity, not only zero velocity, goes with zero net force.
CIt is zero from 5 s to 9 s, as no force is needed to slow or to hold the cart. A student who thinks moving objects slow down without a force picks this. From 5 s to 7 s the velocity decreases, so it is changing, and a net force opposite to the motion must act.
DIt is not zero at any time, because forces act on the cart throughout. A student who thinks equilibrium means that no forces act picks this. Forces act on the cart the whole time, but where they balance the net force is zero, as the level parts of the graph show.
Working Zero net force means constant velocity (Newton's first law), and a changing velocity means a nonzero net force. From the graph: 0–2 s, v increases (net force not zero); 2–5 s, v constant at 4 m/s (net force zero); 5–7 s, v decreases (net force not zero); 7–9 s, v constant at 0 (net force zero).
A crate slides to the right across a level floor. The free-body diagram shows all the forces exerted on it: the normal force FN, the gravitational force Fg, the force FH exerted by a hand pushing to the right, and the friction force Ff exerted by the floor. Which claim about the crate's motion is supported by the diagram?
Answer and reasoning
AIt moves right at a constant velocity, as the push is larger than friction. A student who thinks a constant unbalanced force gives a constant velocity picks this. With 8 N more to the right than to the left, the horizontal forces are unbalanced, so the crate's velocity cannot stay constant; it speeds up.
BIt speeds up to the right, while its vertical velocity remains zero.Correct Vertically, FN = Fg = 50 N, so these forces balance and the vertical velocity stays zero. Horizontally, 20 N to the right and 12 N to the left leave an unbalanced 8 N to the right. The velocity changes only in that direction, and since the crate already moves right, it speeds up.
CIt is in equilibrium, since the vertical forces on it are balanced. A student who thinks balance in one direction is enough picks this. Translational equilibrium requires zero net force in every direction, and the horizontal forces here are unbalanced.
DIts velocity changes both vertically and horizontally, as the forces are unbalanced. A student who thinks unbalanced forces change the velocity in every direction picks this. The velocity changes only in the direction of the unbalanced force. Vertically the forces balance, so the crate stays on the floor with zero vertical velocity.
Working Vertical: FN − Fg = 50 N − 50 N = 0, so the vertical velocity (zero) does not change. Horizontal: FH − Ff = 20 N − 12 N = 8 N to the right, so the forces are unbalanced and the velocity changes toward the right; the crate is already moving right, so it speeds up.
A lamp hangs at rest from a single vertical cord. A student claims: “The lamp is in translational equilibrium because the cord pulls up on the lamp and the lamp pulls down on the cord with forces of equal magnitude.” Which statement correctly evaluates the student's reasoning?
Answer and reasoning
AIt is right: the two forces are equal and opposite, so the net force is zero. A student who treats any equal and opposite pair as balancing forces picks this. The lamp's pull on the cord is exerted on the cord; forces on different objects cannot be added to give the net force on the lamp.
BIt is wrong: the lamp is at rest, so no forces of any kind act on the lamp. A student who thinks equilibrium means that no forces act picks this. Two forces act on the lamp, the cord's pull and Earth's pull; they are balanced, not absent.
CIt is wrong: the cord pulls up harder than Earth pulls, so the lamp does not fall. A student who thinks a support must out-pull gravity picks this. If the cord pulled harder than Earth, the net force would be upward and the lamp's velocity would change. For the lamp to stay at rest, the two forces on it must be equal in magnitude.
DIt is wrong: the cord's pull on the lamp is balanced by Earth's pull on the lamp.Correct Equilibrium depends only on the forces exerted on the lamp. These are the cord's upward pull and Earth's downward gravitational pull, and they balance. The lamp's pull on the cord is exerted on the cord, so it cannot balance a force on the lamp.
A block rests on a ramp inclined at angle θ above the horizontal; the ramp rises to the right. Friction between the block and the ramp is negligible. The block is held at rest by a horizontal force of magnitude F directed to the right. The only other forces on the block are the gravitational force, of magnitude Fg, and the normal force exerted by the ramp. Which expression gives F?
Answer and reasoning
AFg sin θ A student who assumes the normal force equals Fg and then balances the horizontal forces, F = FN sin θ, picks this. The normal force is whatever equilibrium requires: here its vertical component alone must equal Fg, so FN = Fg/cos θ, which is larger than Fg.
BFg/tan θ A student who finds components by dividing by the trigonometric ratio, taking FN/cos θ as the vertical component and FN/sin θ as the horizontal one, gets FN = Fg cos θ and then F = Fg/tan θ. The components of FN are FN cos θ and FN sin θ; each is smaller than FN itself.
CFg tan θCorrect The normal force is perpendicular to the ramp, at θ to the vertical. Vertically, FN cos θ = Fg, so FN = Fg/cos θ. Horizontally, the push balances the normal force's horizontal component: F = FN sin θ = Fg tan θ.
DFg/cos θ A student who balances the push against the whole normal force, F = FN, with FN = Fg/cos θ, picks this. The normal force is not horizontal: only its horizontal component, FN sin θ, balances the horizontal push.
Working The block is in translational equilibrium, so the forces sum to zero along each axis. The normal force is perpendicular to the ramp's surface, so it makes angle θ with the vertical and has a component to the left. Vertical: FN cos θ − Fg = 0, so FN = Fg/cos θ. Horizontal: F − FN sin θ = 0, so F = FN sin θ = Fg sin θ/cos θ = Fg tan θ. Check along the ramp: the push's component up the ramp, F cos θ, balances Fg sin θ, giving the same result. (FN taken as Fg: F = Fg sin θ. Components found by dividing, FN/cos θ and FN/sin θ: FN = Fg cos θ and F = Fg/tan θ. Push balanced against the whole normal force: F = FN = Fg/cos θ.)
Compiled from the AP Physics 1 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
