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Free Body Diagram Net Force & Acceleration Quiz - Test Your Skills

Ready to master free body diagram acceleration and net force? Start the quiz now!

Difficulty: Moderate
2-5mins
Learning OutcomesCheat Sheet
Paper art illustration of quiz on net force acceleration and free body diagrams on teal background

Get ready to master free body diagram net force problems with our free quiz! Designed for physics learners, this net force acceleration quiz puts your skills to the test in calculating forces and determining motion through accurate diagrams. You'll explore acceleration free body diagram scenarios, tackle free body diagram acceleration challenges, and gain confidence in computing net force. Whether you review core concepts in a quick force, mass, and acceleration review or dive deeper with our Newton's Second Law quiz , you'll sharpen your understanding. Challenge yourself now and see your score - let's start!

A 5 kg block sits on a frictionless horizontal surface. A single horizontal force of 20 N is applied to the right. What is the magnitude of the net force acting on the block?
20 N to the right
5 N to the right
0 N
100 N to the right
On a frictionless surface, the normal and gravity cancel vertically, so the only horizontal force is the applied 20 N. Thus the net force is 20 N. More on net force.
Two horizontal forces of 10 N to the right and 4 N to the left act on an object. What is the net force on the object?
6 N to the right
14 N to the right
6 N to the left
14 N to the left
Net force is the vector sum: 10 N right minus 4 N left equals 6 N to the right. See force addition.
A 2 kg mass is subjected to a constant net force of 10 N to the right. What is its acceleration?
5 m/sē to the right
2 m/sē to the right
12 m/sē to the right
0.2 m/sē to the right
Newton’s second law gives a = F/m = 10 N / 2 kg = 5 m/sē. Learn more.
Which free body diagram correctly represents a book at rest on a horizontal table?
One downward arrow labeled weight
One upward arrow labeled normal
One downward arrow labeled weight and one upward arrow labeled normal
One rightward arrow labeled friction
At rest, the only forces on the book are its weight downward and the normal force upward, which balance. Free body diagrams explained.
A 5 kg block is pulled to the right on a horizontal surface with 50 N. The coefficient of kinetic friction is 0.3. What is the net force on the block?
35.3 N to the right
64.7 N to the right
50 N to the right
14.7 N to the left
Friction = ? m g = 0.3·5·9.8 = 14.7 N. Net force = 50 N ? 14.7 N = 35.3 N to the right. Friction concepts.
A box is pulled by two perpendicular forces: 8 N north and 6 N east. What is the magnitude of the resultant net force?
10 N
2 N
14 N
36 N
Use the Pythagorean theorem: ?(8ē + 6ē) = ?(64 + 36) = 10 N. Vector addition.
A 10 kg block is pulled by a 50 N force at 30° above the horizontal on a frictionless surface. What is its acceleration?
4.33 m/sē
2.50 m/sē
3.00 m/sē
5.00 m/sē
Horizontal component = 50 N·cos30° ? 43.3 N. a = 43.3 N / 10 kg = 4.33 m/sē. Component forces.
A 3 kg object is pulled to the right by a 30 N force at 30° above horizontal on a frictionless surface. What is its horizontal acceleration?
8.66 m/sē
5.00 m/sē
4.33 m/sē
2.50 m/sē
Horizontal component = 30 N·cos30° ? 25.98 N. a = 25.98 N / 3 kg = 8.66 m/sē. See details.
Two masses, 3 kg and 2 kg, are connected by a light string over a frictionless pulley. What is the magnitude of their acceleration?
1.96 m/sē
2.94 m/sē
3.27 m/sē
0.98 m/sē
For an Atwood machine, a = (m? ? m?)g/(m? + m?) = (3?2)·9.8/5 = 1.96 m/sē. Atwood machine.
A 5 kg object is suspended in equilibrium by two ropes attached to the ceiling at 60° from the horizontal on both sides. What is the tension in each rope?
28.3 N
14.1 N
25.0 N
35.0 N
Vertical equilibrium: 2T sin60° = mg, so T = mg/(2 sin60°) = (5·9.8)/(1.732) ? 28.3 N. More on tension.
A 2 kg block on a frictionless table has two forces: 6 N to the right and 8 N upward. What is the magnitude of its acceleration?
5 m/sē
7 m/sē
10 m/sē
2 m/sē
Resultant force = ?(6ē + 8ē) = 10 N. Then a = 10 N/2 kg = 5 m/sē. Vector resolution.
A 10 kg block slides down a 30° incline with a kinetic friction coefficient of 0.2. What is its acceleration?
3.20 m/sē
1.64 m/sē
4.90 m/sē
2.45 m/sē
Parallel component = mg sin30° = 49 N; friction = ? mg cos30° ? 16.97 N; net = 32.03 N; a = 32.03/10 = 3.20 m/sē. Incline with friction.
A 2 kg block rests on a frictionless 30° wedge of mass 10 kg on a frictionless floor. A horizontal force F is applied to the wedge so that the block remains at rest relative to the wedge. What is the required magnitude of F?
68 N
49 N
59 N
78 N
To keep the block stationary, the system accelerates at a = g tan30° ? 5.66 m/sē. The total mass is 12 kg, so F = (12 kg)(5.66 m/sē) ? 68 N. Accelerating wedge analysis.
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Study Outcomes

  1. Analyze Free Body Diagrams -

    Recognize and label individual forces in a free body diagram net force context to set up problems correctly.

  2. Calculate Net Force -

    Use vector addition and subtraction to compute the resultant force acting on an object in physics scenarios.

  3. Predict Acceleration -

    Apply Newton's second law to determine acceleration free body diagram outcomes from calculated net force.

  4. Interpret Acceleration Diagrams -

    Deduce the direction and magnitude of motion in free body diagram acceleration problems.

  5. Apply Concepts to Real-World Scenarios -

    Employ net force acceleration quiz techniques to solve practical physics questions accurately.

  6. Refine Problem-Solving Skills -

    Use instant feedback to evaluate your answers and strengthen free body diagram net force strategies.

Cheat Sheet

  1. Vector Addition of Forces -

    In any free body diagram net force (ÎĢF) is found by tip-to-tail vector addition of all forces acting on an object. For example, if F₝ = 5 N to the right and F₂ = 3 N to the left, the net force is 2 N to the right (5 N − 3 N). A quick mnemonic is "All Forces, All Together" to remember you must include every arrow in your ÎĢF calculation.

  2. Newton's Second Law and Acceleration -

    Newton's second law states that ÎĢF = ma, linking net force and acceleration free body diagram analysis directly. For instance, a 10 N net force applied to a 2 kg block produces an acceleration of 5 m/sÂē (a = F/m). Remember the phrase "Force Equals Mass Accelerates" to recall the formula instantly.

  3. Identifying and Labeling Forces -

    Accurately drawing each force - gravity (mg), normal force, friction, tension - ensures your free body diagram acceleration predictions are reliable. A classic example is a book on a table, where mg acts downward while the normal force balances it upward. Tip: label each arrow with magnitude and direction to avoid missing hidden forces like static friction.

  4. Force Decomposition on Inclines -

    On an incline, decompose the weight vector into mg·sinÎļ parallel and mg·cosÎļ perpendicular to the surface for correct net force acceleration calculations. For a 30° slope, the parallel component is mg·sin30° = 0.5 mg guiding how the block slides. A handy mnemonic is "Sine Slides Down, Cosine Clings Up" to split weight components quickly.

  5. Consistent Axes and Sign Conventions -

    Selecting consistent axes (e.g., right/positive, up/positive) is crucial for solving net force acceleration quiz problems without sign errors. For example, if you define right as positive, a leftward friction force of 4 N becomes F = −4 N in ÎĢF = ma. Establish your convention before summing forces to avoid confusion and ensure correct acceleration direction.

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