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AP Physics 1

AP Physics 1 Unit 2 Forces: Conceptual Practice 2 (Elevators & Data)

Shared by ashtont · Oct 29, 2026 · 0 plays

21 conceptual questions for the Unit 2 1D and 2D Forces test: elevators and apparent weight, Newton's third law, modified Atwood, friction, inclines, springs and center of mass. Most come with a data table, graph or diagram to read, and every choice has an explanation.

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21 questions across 21 sections. After every answer you’ll see why each choice is right or wrong, not just the one you picked.

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Sections (none selected means all)

Sample questions

Open a question to see the answer choices and the correct answer.

Which description of the elevator's motion matches the data?
  1. A. It moves upward, speeding up from 0 to 2 s. From 2 to 8 s it is stopped between floors. From 8 to 10 s it moves downward.
  2. B. It moves upward the whole time: it speeds up, then moves at constant speed, then slows down. ✓
  3. C. It moves downward the whole time: it speeds up, then moves at constant speed, then slows down.
  4. D. It moves upward from 0 to 2 s and then downward for the rest of the trip, because the reading only falls after the first interval.

Correct answer: B. Reading > mg means the net force (and acceleration) is upward, so starting from rest it speeds up going up. Reading = mg means a = 0, so constant velocity. Reading < mg means the acceleration is downward, so the upward-moving elevator slows to a stop at the top floor.

Which ranks the scale readings from greatest to least?
  1. A. W > Z > Y > X
  2. B. Y > W > Z > X
  3. C. W = Y > Z > X
  4. D. Y > W = Z > X ✓

Correct answer: D. N − mg = ma with up positive. Y: moving down and slowing means a is upward, so N > mg. W and Z: a = 0, so N = mg. X: moving down and speeding up means a is downward, so N < mg.

At t = 4 s, how does the scale reading of a passenger compare to the passenger's weight, and what is the elevator doing?
  1. A. The reading equals mg, and the elevator is moving upward at a constant speed. ✓
  2. B. The reading equals mg, and the elevator is at rest between floors.
  3. C. The reading is greater than mg, because the elevator is still moving upward.
  4. D. The reading is less than mg, because the elevator has stopped speeding up.

Correct answer: A. With a = 0 the net force is zero, so N = mg. But the elevator gained upward velocity from 0 to 2 s (area under the a–t graph = +3 m/s), so it is still moving up at 3 m/s.

Which motions could the elevator have during the part of the trip when the stretch is 12.0 cm?
  1. A. Only moving upward and speeding up
  2. B. Moving upward at a constant speed, or moving upward and speeding up
  3. C. Moving upward and speeding up, or moving downward and slowing down ✓
  4. D. Moving downward and speeding up, or moving upward and slowing down

Correct answer: C. A larger stretch means the spring force kx is greater than mg, so the net force and acceleration are upward. An upward acceleration happens when the elevator speeds up going up or slows down going down.

Which relationship among the magnitudes is correct?
  1. A. F₁ = F₂ > F₃ ✓
  2. B. F₁ > F₂ = F₃
  3. C. F₁ = F₂ = F₃
  4. D. F₁ > F₃ > F₂

Correct answer: A. F₁ and F₂ are a Newton's third-law pair, so they are always equal. The passenger accelerates upward, so the upward force on the passenger (F₁) must be bigger than the downward force (F₃).