Free-Body Diagrams & Net Force — Practice · Grade 12 Science

Chapter: Forces, Rotation & Equilibrium

What this practice covers

By the end of Free-Body Diagrams & Net Force you should be able to:

10 questions, numeric, multiple choice and written answer, each with a worked explanation.

Vocabulary used: free-body diagram, net force (ΣF), equilibrium, normal force, vector.

Problem 1

A 10 kg box on level ground is pushed right with 50 N while friction resists with 20 N. Find the net horizontal force.
Answer and explanation

30 N

Net force = 50 − 20 = 30 N to the right (the vertical weight and normal force cancel).

Problem 2

Two forces act on an object at a right angle: 6 N east and 8 N north. Find the magnitude of the net force.
Answer and explanation

10 N

Perpendicular forces combine by the Pythagorean theorem: √(6² + 8²) = √(36 + 64) = √100 = 10 N.

Problem 3

In a free-body diagram of a book resting on a flat table, which forces should be drawn?
  1. Only the normal force, upward
  2. An applied force and friction
  3. Weight downward and the normal force upward
  4. Only the weight, downward
Answer and explanation

Weight downward and the normal force upward

A resting book feels gravity pulling it down and the table's normal force pushing it up. The two are equal, so it stays in equilibrium.

Problem 4

An object moves in a straight line at constant velocity. Its free-body diagram must show forces that…
  1. include no forces at all
  2. grow larger over time
  3. point in the direction of motion
  4. add up to zero
Answer and explanation

add up to zero

Constant velocity means zero acceleration, so the net force is zero — the forces present must cancel out. (Forces can exist; they simply balance.)

Problem 5

List the forces you would draw on a free-body diagram of a sled being pulled across snow by a horizontal rope, and give the direction of each.
Answer and explanation

Weight pulls straight down; the normal force from the snow pushes straight up; the rope's tension pulls forward (horizontal); and kinetic friction from the snow pushes backward, opposing the sled's motion. If the sled moves at steady speed, the up/down pair cancels and tension equals friction.

Problem 6

A 6.5 kg object on frictionless ice feels two perpendicular forces: 12 N east and 5 N north. Find its acceleration.
Answer and explanation

2 m/s²

Net force = √(12² + 5²) = √(144 + 25) = √169 = 13 N. Then a = ΣF ÷ m = 13 ÷ 6.5 = 2 m/s².

Problem 7

A hanging mass is at rest on one vertical rope, with weight 30 N. Find tension. Show the reasoning, not just the conclusion.
Hints

Normal equals weight only for the specified force and acceleration conditions.

Answer and explanation

30 N upward, balancing weight in the given model.

Problem 8

Why draw only forces acting on the selected object?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Newton's second law concerns that object's force sum; partner forces on other objects belong in other diagrams.

Problem 9

Write independent equations along the selected axesWrite independent equations along the selected axesAxisForce equationMotion constraintHorizontalTcosθ−f=maxAcceleration can benonzeroVerticalN+Tsinθ−mg=mayay=0 on a level surfaceTwo component equations share one physical diagram.
Original force equations for a level-table contact model with a single angled tension.
Interpret the supplied visual. Why must N be found before μN in this pulled-block example?
Answer and explanation

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

A vector equation becomes one scalar equation per perpendicular axis. Independent does not mean unrelated: the same tension angle enters both, and contact constraints can set a component of acceleration. For a block that remains on a horizontal table, ay=0 while ax may be nonzero. Assign signs to every component before substituting values; then solve the vertical support account before using a friction model that depends on N.

The upward tension component changes N, so using mg directly in μN would use the wrong contact force.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the physical or environmental relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 10

A 3 kg block is pulled by 15 N at 30° above horizontal. No vertical acceleration; resistance is 4 N. Find N and ax.
Answer and explanation

N=29.4−7.5=21.9 N. Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s². The tension is counted once through its components.

N=29.4−7.5=21.9 N.

Horizontal net=15cos 30°−4≈8.99 N; ax≈3.00 m/s².

The tension is counted once through its components.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the physical or environmental mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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