AP Physics 1
AP Physics 1 Unit 2 Forces: FRQ Practice
Shared by Ashton · Oct 25, 2026 · 0 plays
5 AP-style free-response questions on forces (modified Atwood with friction, the elevator ride, the ramp, finding k, two boxes with an angled rope), split into 23 parts. Each part has a point-by-point rubric and a model answer. Use g = 9.8 m/s².
23 questions across 5 sections. Write your answer, then check it against the rubric and score yourself.
Sample questions
Open an item to see its answer.
Block A (M = 4.0 kg) is on a horizontal table with μₖ = 0.25. It's connected by a light string over an ideal pulley to block B (m = 2.0 kg), which hangs 0.80 m above the floor. The blocks are released from rest. Use g = 9.8 m/s². — (a) Describe the free-body diagram for block A and for block B. For each block, list every force, what exerts it, and its direction. (On the test you'd draw these, with each force as a separate arrow starting on the dot.)
Sample answer: Block A: F_N (table on A) points up, F_g (Earth on A) points down, F_T (string on A) points right toward the pulley, and f_k (table on A) points left, opposite the sliding. Block B: F_T (string on B) points up and F_g (Earth on B) points down. Nothing else acts on either block.
Block A (M = 4.0 kg) is on a horizontal table with μₖ = 0.25. It's connected by a light string over an ideal pulley to block B (m = 2.0 kg), which hangs 0.80 m above the floor. The blocks are released from rest. Use g = 9.8 m/s². — (b) Derive an expression for the magnitude of the acceleration of the blocks in terms of M, m, μₖ and physical constants. Begin by writing Newton's second law for each block.
Sample answer: Block A, right as +: F_T − μₖMg = Ma. Block B, down as +: mg − F_T = ma. Adding the two equations cancels F_T: mg − μₖMg = (m + M)a, so a = g(m − μₖM)/(m + M).
Block A (M = 4.0 kg) is on a horizontal table with μₖ = 0.25. It's connected by a light string over an ideal pulley to block B (m = 2.0 kg), which hangs 0.80 m above the floor. The blocks are released from rest. Use g = 9.8 m/s². — (c) Calculate the acceleration and the string tension.
Sample answer: a = 9.8(2.0 − 0.25 × 4.0)/6.0 = 9.8(1.0)/6.0 = 1.63 m/s². F_T = m(g − a) = 2.0(9.8 − 1.63) = 16.3 N. Check with block A: 16.3 − 0.25(4.0)(9.8) = 16.3 − 9.8 = 6.5 N ≈ 4.0(1.63) ✓.
Block A (M = 4.0 kg) is on a horizontal table with μₖ = 0.25. It's connected by a light string over an ideal pulley to block B (m = 2.0 kg), which hangs 0.80 m above the floor. The blocks are released from rest. Use g = 9.8 m/s². — (d) Calculate how long block B takes to hit the floor.
Sample answer: From rest, 0.80 = ½(1.63)t², so t² = 1.60/1.63 = 0.98 and t = 0.99 s.
Block A (M = 4.0 kg) is on a horizontal table with μₖ = 0.25. It's connected by a light string over an ideal pulley to block B (m = 2.0 kg), which hangs 0.80 m above the floor. The blocks are released from rest. Use g = 9.8 m/s². — (e) The mass of block A is increased slightly, and the blocks still slide. Does the acceleration increase, decrease or stay the same? Justify your answer using your expression from (b).
Sample answer: The acceleration decreases. In a = g(m − μₖM)/(m + M), increasing M makes the numerator smaller, because there's more friction opposing the motion, and makes the denominator larger, because there's more mass to accelerate. Both changes reduce a.