Free-body diagrams and Newton's three laws — the foundation for every mechanics problem after this one. Nearly every force question comes down to correctly adding up every force acting on an object before applying F = ma. Every question comes with a written walkthrough of the forces involved.
F = ma requires the net force — the vector sum of every force acting on the object — not just one of the forces present, like an applied push ignoring friction and gravity.
An action-reaction pair always acts on two different objects — a table pushing up on a book (normal force) pairs with the book pushing down on the table, not with gravity pulling down on the book, which is a completely separate force.
Mass is a fixed amount of matter, measured in kilograms; weight is the force of gravity on that mass, measured in newtons, and changes depending on the local gravitational field — the two aren't interchangeable, even though they're proportional on Earth.
Straight from the bank — one per difficulty tier. Reveal the answer to see the explanation you'd get in a real session.
A 10 kg object experiences a net force of 25 N. What is its acceleration?
A — 2.5 m/s². Newton's Second Law: F = ma, so a = F/m = 25/10 = 2.5 m/s². 0.4 m/s² (C) comes from inverting the formula — computing m/F (10/25) instead of F/m.
A 60 kg skater pushes off a wall, and the wall exerts a force of 120 N back on the skater. According to Newton's Third Law, what force does the skater exert on the wall?
A — 120 N, in the opposite direction. Newton's Third Law states that for every action force, there's an equal and opposite reaction force acting on a different object. If the wall exerts 120 N on the skater, the skater exerts exactly 120 N back on the wall, in the opposite direction. 60 N, in the opposite direction (C) incorrectly scales the reaction force down — Newton's Third Law pairs are always equal in magnitude, regardless of the two objects' masses.
A 5 kg block sits on a horizontal surface with a coefficient of kinetic friction of 0.3. A horizontal force of 20 N is applied to the block, causing it to slide. Using g = 10 m/s², find the block's acceleration.
A — 1 m/s². First find the friction force: f = μN = 0.3 × (5)(10) = 15 N, opposing the applied force. The net force is 20 − 15 = 5 N, so the acceleration is a = F_net/m = 5/5 = 1 m/s². 4 m/s² (B) comes from ignoring friction entirely and computing 20/5 = 4 — treating the applied force as if it were the net force.
Force problems reward drawing the free-body diagram first — nearly every error comes from missing a force or forgetting that F=ma needs the net force, not just one contributor.
Sketch a free-body diagram listing every force acting on the object — gravity, normal, friction, applied — before summing them. Untimed practice is where that habit prevents missed forces.
Move to timed sessions once free-body diagrams and net force calculations are fast.
Pair forces and Newton's laws with kinematics in a mock — most multi-step mechanics problems combine both.
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