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AP Physics 1 Unit 2 practice questions

Try 10 multiple-choice questions from Unit 2 here. Pick an answer to see why it's right or wrong, then take the timed test above for a full AP-style set.

  1. Question 1

    A student pushes a heavy crate across a rough horizontal floor at a constant velocity of 1.5 m/s. The student claims that the net force on the crate is zero. Which statement best justifies this claim using Newton's laws?

    Valid because constant velocity implies zero acceleration, so the applied force balances the kinetic friction.

    Valid because constant velocity implies zero acceleration, so the applied force balances the static friction.

    Invalid because motion requires a net force, so the applied force must be greater than the kinetic friction.

    Invalid because motion requires a net force, so the applied force must be greater than the static friction.

    Show answer

    Answer: (A)

    Connecting constant velocity to zero acceleration applies Newton's first law, confirming that the vector sum of forces (applied force and kinetic friction) must be zero.

  2. Question 2

    A crane lifts a steel beam vertically upward at a constant speed. A student claims that the tension in the cable is equal in magnitude to the weight of the beam. Which of the following best supports this claim?

    The claim is correct because the constant velocity implies zero acceleration, so the net vertical force is zero.

    The claim is correct because the constant velocity implies zero acceleration, so tension and weight are an action-reaction pair.

    The claim is incorrect because upward motion implies upward acceleration, so the tension must be greater than the weight.

    The claim is incorrect because upward motion implies upward acceleration, so the tension must be greater than the normal force.

    Show answer

    Answer: (A)

    Linking constant velocity to zero acceleration justifies why the upward tension must exactly balance the downward weight to maintain equilibrium.

  3. Question 3

    A car travels around a circular track at a constant speed. A student claims that the car is in a state of translational equilibrium because its speed is not changing. Which statement best evaluates this claim?

    The claim is incorrect because the changing direction implies a changing velocity, requiring a nonzero net force toward the center.

    The claim is incorrect because the changing direction implies a changing velocity, requiring a nonzero net force away from the center.

    The claim is correct because the constant speed implies a constant velocity, requiring a zero net force to maintain the motion.

    The claim is correct because the constant speed implies a constant velocity, requiring a zero net force to balance the friction.

    Show answer

    Answer: (A)

    Recognizing that velocity is a vector quantity justifies why a change in direction constitutes acceleration, requiring a net centripetal force.

  4. Question 4

    A block slides down a rough incline at a constant speed. A student claims that the frictional force acting up the incline is equal in magnitude to the weight of the block. Which analysis best evaluates this claim?

    Incorrect; friction balances the component of weight parallel to the surface, which is less than the block's total weight.

    Incorrect; friction balances the component of weight perpendicular to the surface, which is less than the block's total weight.

    Correct; friction balances the total weight of the block because the block is in equilibrium and moving at a constant speed.

    Correct; friction balances the normal force on the block because the block is in equilibrium and moving at a constant speed.

    Show answer

    Answer: (A)

    Decomposing forces on an incline justifies why friction only balances the parallel component of gravity (mg sin θ), not the full weight (mg).

  5. Question 5

    A rectangular block with unequal side lengths slides across a horizontal surface. In Trial 1, the block slides on its largest face. In Trial 2, the block slides on its smallest face. The block has the same initial speed in both trials. How do the magnitude of the kinetic friction force and the magnitude of the acceleration of the block compare between the two trials?

    Force is equal in both trials; acceleration is equal in both trials

    Force is greater in Trial 1; acceleration is greater in Trial 1

    Force is greater in Trial 1; acceleration is equal in both trials

    Force is equal in both trials; acceleration is greater in Trial 2

    Show answer

    Answer: (A)

    Applying the model of kinetic friction shows that magnitude depends only on the coefficient of friction and normal force (Fk=μkFnF_k = \mu_k F_n), not contact area. Since the block's weight and the surface material remain constant, both the friction force and resulting acceleration remain unchanged.

  6. Question 6

    Block X of mass mm and Block Y of mass 3m3m slide across the same rough horizontal surface with the same initial speed. The coefficient of kinetic friction is the same for both blocks. Which of the following correctly compares the magnitude of the kinetic friction force exerted on the blocks and the stopping distance of the blocks?

    Force on Y is three times force on X; stopping distance is equal

    Force on Y is three times force on X; stopping distance is tripled

    Force on Y is equal to force on X; stopping distance is equal

    Force on Y is equal to force on X; stopping distance is tripled

    Show answer

    Answer: (A)

    Scaling the normal force by mass indicates the friction force triples (Fk=μk(3mg)F_k = \mu_k (3mg)), while Newton's second law shows acceleration remains constant (a=Fk/m=μkga = F_k/m = \mu_k g). Since acceleration and initial speed are identical, kinematic relationships dictate the stopping distance must be the same.

  7. Question 7

    A block slides down a rough incline. In Scenario 1, the incline angle is 30∘30^\circ relative to the horizontal. In Scenario 2, the incline angle is 60∘60^\circ relative to the horizontal. How do the magnitude of the normal force and the magnitude of the kinetic friction force exerted on the block compare between the two scenarios?

    Normal force is greater in Scenario 1, and friction is greater in Scenario 1

    Normal force is greater in Scenario 2, and friction is greater in Scenario 2

    Normal force is greater in Scenario 1, but friction is greater in Scenario 2

    Normal force is greater in Scenario 2, but friction is greater in Scenario 1

    Show answer

    Answer: (A)

    Analyzing the components of gravitational force shows the normal force equals mgcos⁡θmg \cos\theta, which decreases as the angle increases (cos⁡60∘<cos⁡30∘\cos 60^\circ < \cos 30^\circ). Since kinetic friction is directly proportional to normal force (Fk=μkFnF_k = \mu_k F_n), it also decreases in the steeper scenario.

  8. Question 8

    A box is pushed so that it slides across a rough horizontal floor. In Trial 1, the box slides with an initial speed vv. In Trial 2, the box slides with an initial speed 2v2v. How do the magnitude of the kinetic friction force exerted on the box and the magnitude of the box's acceleration compare between the two trials?

    Force and acceleration are both equal in Trials 1 and 2

    Force is greater in Trial 2, but acceleration is equal in both trials

    Force and acceleration are both greater in Trial 2 than in Trial 1

    Force is equal in both trials, but acceleration is greater in Trial 2

    Show answer

    Answer: (A)

    Recognizing that kinetic friction is a constant force determined by surface materials and normal force reveals it is independent of sliding speed. Consequently, the net force and acceleration derived from Newton's second law remain identical regardless of the initial velocity.

  9. Question 9

    A satellite orbits Earth in a circular path with radius RR and period TT. If the satellite is moved to a new circular orbit with radius 4R4R, which of the following expressions represents the new period?

    8T8T

    2T2T

    4T4T

    16T16T

    Show answer

    Answer: (A)

    Kepler's third law states that T2∝R3T^2 \propto R^3, so T∝R3/2T \propto R^{3/2}. Increasing the radius by a factor of 4 changes the period by a factor of 43/2=(22)3/2=23=84^{3/2} = (2^2)^{3/2} = 2^3 = 8.

  10. Question 10

    A planet orbits a star in a circular path with orbital speed vv at a distance RR. If the planet's orbital radius were increased to 4R4R, what would be its new orbital speed?

    v/2v/2

    v/4v/4

    v/16v/16

    2v2v

    Show answer

    Answer: (A)

    Orbital speed is given by v=GM/Rv = \sqrt{GM/R}, meaning v∝1/Rv \propto 1/\sqrt{R}. Increasing RR by a factor of 4 decreases vv by a factor of 4=2\sqrt{4} = 2.

What's on the AP Physics 1 Unit 2 practice test

The standard version takes about 88 minutes and follows the timing and directions of the AP exam, sized to one unit:

  • 20 Multiple Choice Questions (40 minutes)
  • 1 FRQ – FRQ 1 – Mathematical Routines (24 minutes)
  • 1 FRQ – FRQ 2 – Translation Between Representations (24 minutes)

Questions are drawn from 2045 multiple-choice questions and 35 FRQs written for Unit 2, Force and Translational Dynamics.

Practice AP Physics 1 Unit 2 FRQs

Answer the unit's free-response questions one at a time and get a score on every part.

Review Unit 2 before you start

Other AP Physics 1 unit tests