Unit 2 · 18–23% of the exam Beta

Unit 2 practice test: Force and Translational Dynamics

12 exam-style questions from the unit, free, with an explanation for every option. Most have four answer choices, as on the real exam, and a few ask you to type a value; data sets share one table, graph or scene diagram.

Covers: Systems and Center of Mass; Forces and Free-Body Diagrams; Newton's Third Law; Newton's First Law; Newton's Second Law; Gravitational Force; Kinetic and Static Friction; Spring Forces; Circular Motion.

Diagram

A sign hung from two cords

A 5.0 kg sign hangs at rest from two identical light cords. Each cord makes a 30° angle with the horizontal. Use g = 9.8 m/s².

OPEN30°30°sign, 5.0 kg

1. Why do the horizontal pulls of the two cords on the sign cancel?

  1. Equal tensions at equal angles pull equally hard to opposite sides
  2. The cords pull up, and horizontal pulls do not affect a hanging sign
  3. The weight of the sign cancels the horizontal pulls
  4. The ceiling pushes sideways on the sign to cancel them
Show the answer

The left cord pulls up and to the left, the right cord up and to the right. With equal tensions and equal angles, the sideways parts F_T cos 30° are equal and opposite, so the horizontal net force is zero, as it must be for a sign at rest.

  • Correct: Equal tensions at equal angles pull equally hard to opposite sides: Right: equal tensions, equal angles, opposite sideways parts.
  • The cords pull up, and horizontal pulls do not affect a hanging sign: Each cord does pull partly sideways; those parts cancel each other rather than having no effect.
  • The weight of the sign cancels the horizontal pulls: The weight points straight down, so it cannot cancel a horizontal force.
  • The ceiling pushes sideways on the sign to cancel them: The ceiling touches the cords, not the sign; the only forces on the sign are the two cords and Earth.

2. The cords are replaced by longer ones attached farther apart, so each makes a smaller angle (15°) with the horizontal. The sign still hangs at rest. How does the tension in each cord change?

  1. It increases, since less of each pull is upward
  2. It decreases, since the cords are longer
  3. It stays the same, since the weight is the same
  4. It increases, since the sign's weight increases
Show the answer

The cords still hold up the same 49 N. Each cord's upward part is F_T sin θ, and sin 15° is smaller than sin 30°, so each cord must pull harder: F_T = mg ÷ (2 sin 15°) ≈ 95 N.

  • Correct: It increases, since less of each pull is upward: Right: a flatter cord needs a larger tension for the same upward pull.
  • It decreases, since the cords are longer: The length of the cord does not set the tension; the angle does.
  • It stays the same, since the weight is the same: The total upward pull is the same, but each cord gives less of its tension upward, so the tension rises.
  • It increases, since the sign's weight increases: The weight, mg, does not change when the cords change.

3. Which free-body diagram is correct for the sign?

  1. Three arrows: two cord tensions up and outward, and F_g down
  2. Seven arrows: two tensions, their four parts, and F_g down
  3. Two arrows: one total tension up and F_g down
  4. Three arrows: two tensions and a net force arrow down
Show the answer

Each cord is a separate agent, so each gets one whole arrow along its cord, plus the weight from Earth. Components belong in your equations beside the diagram, not on it, and the net force (zero here) is never an arrow.

  • Correct: Three arrows: two cord tensions up and outward, and F_g down: Right: one whole arrow per agent.
  • Seven arrows: two tensions, their four parts, and F_g down: Drawing components as extra arrows double-counts each pull; the exam does not accept components on a free-body diagram.
  • Two arrows: one total tension up and F_g down: Two cords are two agents and two forces; merging them hides their directions.
  • Three arrows: two tensions and a net force arrow down: The net force is the sum of the arrows, not an arrow of its own, and here it is zero.

Diagram

A bucket raised on a rope

A 4.0 kg bucket of sand hangs from a single rope that runs over a pulley on the ceiling to a person's hand. The person pulls the rope, raising the bucket straight up at a constant speed. Air resistance is negligible.

hand pulls downv (constant)

4. Which free-body diagram matches the bucket while it rises at a constant speed?

  1. F_T up and F_g down, the same length
  2. F_T up, longer than F_g down
  3. F_T up, F_g down and a lifting force up
  4. F_T up alone, since the bucket rises
Show the answer

Two objects act on the bucket: the rope pulls up (F_T) and Earth pulls down (F_g). When the velocity stays constant the forces in each direction are equal in size (2.4 shows why), so the arrows are the same length.

  • Correct: F_T up and F_g down, the same length: Right: two forces, equal lengths, because the velocity does not change.
  • F_T up, longer than F_g down: A longer tension arrow would mean a net force upward, which would make the bucket speed up. Rising does not need a larger upward force.
  • F_T up, F_g down and a lifting force up: A "lifting force" has no separate agent: the rope is what lifts, and its pull is already F_T.
  • F_T up alone, since the bucket rises: Earth pulls on the bucket whether it rises or falls, so F_g is always on its diagram near Earth.

5. Which object exerts the upward force on the bucket?

  1. The rope
  2. The person's hands
  3. Earth
  4. The sand inside it
Show the answer

Only objects that touch the bucket can exert contact forces on it, and the rope is the only one attached to it. The hands pull on the rope, not on the bucket.

  • Correct: The rope: Right: the rope is attached to the bucket and pulls up on it.
  • The person's hands: The hands touch the rope, not the bucket; their pull reaches the bucket through the rope as tension.
  • Earth: Earth pulls the bucket down, not up.
  • The sand inside it: The sand is part of the bucket-and-sand object; it does not push the object as a whole up.

6. The person stops pulling and holds the rope so the bucket hangs at rest. How does the free-body diagram change?

  1. It does not change: F_T and F_g, equal in size
  2. F_T shrinks, since the bucket is no longer rising
  3. F_T disappears, leaving F_g by itself
  4. F_g grows, since the bucket is now at rest
Show the answer

At rest and at a constant velocity the velocity is not changing, so in both cases the rope's pull equals the bucket's weight. The diagram is the same.

  • Correct: It does not change: F_T and F_g, equal in size: Right: rest is just a constant velocity of zero.
  • F_T shrinks, since the bucket is no longer rising: Moving up at a steady speed needs no extra pull, so stopping does not reduce the tension.
  • F_T disappears, leaving F_g by itself: The rope still holds the bucket up; if it pulled with no force the bucket would fall.
  • F_g grows, since the bucket is now at rest: Weight is mg; it does not depend on whether the bucket moves.

7. Three small objects lie on a level floor: 1.0 kg at (0, 0), 1.0 kg at (2.0 m, 0) and 2.0 kg at (1.0 m, 2.0 m). What is the y-coordinate of their center of mass, in meters?

Type a number in m.

Show the answer

Treat each direction on its own: y_cm = (1.0 kg × 0 + 1.0 kg × 0 + 2.0 kg × 2.0 m) ÷ 4.0 kg = 1.0 m.

  • Answer: 1.0 m

8. A 1000 kg car hits a 0.10 g bug. How does the force of the car on the bug compare with the force of the bug on the car?

  1. They are equal in size
  2. The car exerts the larger force
  3. The bug exerts the larger force
  4. The car exerts a force but the bug does not
Show the answer

The car and the bug interact, so they push on each other with forces of equal size, whatever their masses. The bug is squashed because the same force has a far larger effect on its tiny mass.

  • Correct: They are equal in size: Right: third-law partners are always equal in size.
  • The car exerts the larger force: A larger mass does not let one side of an interaction push harder.
  • The bug exerts the larger force: The bug cannot push harder than it is pushed.
  • The car exerts a force but the bug does not: Every interaction is two-sided: the bug pushes back on the car.

9. A 1200 kg car speeds up from rest to 15 m/s in 5.0 s along a straight, level road. What is the size of the net force on the car, in newtons?

Type a number in N.

Show the answer

First the acceleration: a = Δv ÷ Δt = (15 m/s − 0) ÷ 5.0 s = 3.0 m/s². Then F_net = ma = 1200 kg × 3.0 m/s² = 3600 N (3.6 × 10³ N).

  • Answer: 3.6 × 103 N

10. A student says: "Astronauts in the space station are weightless because there is no gravity up there." Which evidence best contradicts this?

  1. Earth's field there is about 90% of its surface value
  2. The astronauts feel no push from the floor
  3. The station is above most of the atmosphere
  4. The astronauts float freely inside the station
Show the answer

Gravity at the station's orbit is only about 10% weaker than at the ground; it is what keeps the station in orbit. The astronauts feel weightless because they and the station fall together, so the floor pushes on them with no force: zero apparent weight, like the falling elevator.

  • Correct: Earth's field there is about 90% of its surface value: Right: gravity is nearly as strong; the normal force is what vanishes.
  • The astronauts feel no push from the floor: That is the effect, zero normal force, not evidence about whether gravity is present.
  • The station is above most of the atmosphere: The air has nothing to do with gravity's strength.
  • The astronauts float freely inside the station: Floating is what the claim is trying to explain, so it cannot test it.

11. A 4.0 kg box slides across a level floor. The coefficient of kinetic friction is 0.30. What is the size of its acceleration as it slows down, in m/s²? Use g = 9.8 m/s².

Type a number in m/s^2.

Show the answer

The only horizontal force is kinetic friction, f_k = μ_k mg = 0.30 × 4.0 kg × 9.8 m/s² = 11.8 N. a = f_k ÷ m = μ_k g = 0.30 × 9.8 m/s² = 2.9 m/s². The mass cancels.

  • Answer: 2.9 m/s^2

12. A force of 6.0 N stretches a spring by 0.040 m. What is the spring constant? Give the value and its unit.

Type a number and its unit.

Show the answer

k = F ÷ |Δx| = 6.0 N ÷ 0.040 m = 150 N/m (1.5 × 10² N/m).

  • Answer: 150 N/m

Keep going

Practice has every question in the unit, with feedback after each one; this unit's lessons are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.