Forces, Rotation & Equilibrium — Problem set, Grade 12

Use force diagrams, friction, circular motion, springs, and torque to explain motion and balance.

Review the key ideas

Newton's Three Laws of Motion

First law: objects resist changes in motion (inertia). Second law: ΣF = ma. Third law: every force comes with an equal, opposite force on a different object.

Free-Body Diagrams & Net Force

A free-body diagram includes external forces on one selected body. Their vector sum determines acceleration; static rigid-body equilibrium also requires torque balance.

Friction & Common Forces

Static friction adjusts up to a modeled limit μsN; kinetic friction is approximately μkN for specified sliding conditions. Find N from actual force components before using either model.

Circular Motion & Gravitation

Circular motion needs a center-seeking net force: a = v²/r and F = mv²/r, supplied by a real force. Gravity is universal, F = Gm₁m₂/r² (inverse-square), and it supplies the centripetal force that holds satellites in orbit.

Springs, Torque & Static Equilibrium

Equilibrium requires balanced forces and torques; spring energy grows with displacement squared.

Problem 1

A net force of 60 N acts on a 12 kg cart. Find its acceleration.
Answer and explanation

5 m/s²

a = ΣF ÷ m = 60 ÷ 12 = 5 m/s².

Problem 2

What net force gives a 0.145 kg baseball an acceleration of 250 m/s²?
Answer and explanation

36.25 N

ΣF = ma = 0.145 × 250 = 36.25 N.

Problem 3

Two skaters at rest push off each other. The 50 kg skater accelerates at 3 m/s². Using Newton's third law (equal force on each), find the force involved and the acceleration of the 75 kg skater.
Answer and explanation

150 N on each; the 75 kg skater accelerates at 2 m/s²

Force on the 50 kg skater: F = ma = 50 × 3 = 150 N. By Newton's third law the 75 kg skater feels the same 150 N: a = F ÷ m = 150 ÷ 75 = 2 m/s². The heavier skater accelerates less from the same force.

Problem 4

Two horizontal forces act on a box: 80 N right and 30 N left. Find the net force and its direction.
Answer and explanation

50 N to the right

Net force = 80 − 30 = 50 N. The larger force points right, so the net force is 50 N to the right.

Problem 5

A 10 kg box rests on a frictionless 30° incline. Find (a) the component of weight pulling it down the slope and (b) its acceleration down the slope.
Answer and explanation

(a) 49 N (b) 4.9 m/s²

(a) Down-slope weight component = mg sinθ = 10 × 9.8 × sin 30° = 98 × 0.5 = 49 N. (b) a = g sinθ = 9.8 × 0.5 = 4.9 m/s² (mass cancels).

Problem 6

A block presses on a surface with a normal force of 100 N and slides with μₖ = 0.25. Find the kinetic friction force.
Answer and explanation

25 N

f = μₖN = 0.25 × 100 = 25 N.

Problem 7

A 20 kg crate slides across level ground with μₖ = 0.30. Find the kinetic friction force.
Answer and explanation

58.8 N

On level ground N = mg = 20 × 9.8 = 196 N. Then f = μₖN = 0.30 × 196 = 58.8 N.

Problem 8

An 8 kg box slides down a 30° incline where μₖ = 0.20. Find its acceleration down the slope.
Answer and explanation

About 3.2 m/s²

Along the slope: a = g(sinθ − μₖcosθ) = 9.8(sin 30° − 0.20 cos 30°) = 9.8(0.5 − 0.20 × 0.866) = 9.8(0.5 − 0.173) = 9.8 × 0.327 ≈ 3.2 m/s². (Mass cancels.)

Problem 9

A ball moves in a circle of radius 2 m at a speed of 4 m/s. Find its centripetal acceleration.
Answer and explanation

8 m/s²

a = v²/r = 4² ÷ 2 = 16 ÷ 2 = 8 m/s², directed toward the center.

Problem 10

A 3 kg ball on a string is whirled in a circle of radius 1.5 m at 6 m/s. Find the centripetal force in the string.
Answer and explanation

72 N

F = mv²/r = 3 × 6² ÷ 1.5 = 3 × 36 ÷ 1.5 = 108 ÷ 1.5 = 72 N (supplied by the string's tension).

Problem 11

Find the gravitational force between two 800 kg cars whose centers are 4 m apart.
Answer and explanation

About 2.7 × 10⁻⁶ N

F = Gm₁m₂/r² = (6.67 × 10⁻¹¹)(800)(800) ÷ 4² = (6.67 × 10⁻¹¹)(640,000) ÷ 16 = (6.67 × 10⁻¹¹)(40,000) ≈ 2.67 × 10⁻⁶ N — far too small to notice, which is why we never feel the pull between ordinary objects.

Problem 12

A 1,500 kg car rounds a flat curve of radius 50 m at 20 m/s. What centripetal force must friction supply to keep it on the road?
Answer and explanation

12,000 N

F = mv²/r = 1,500 × 20² ÷ 50 = 1,500 × 400 ÷ 50 = 1,500 × 8 = 12,000 N, directed toward the center of the curve.

Problem 13

The Moon has a mass of 7.35 × 10²² kg and a radius of 1.74 × 10⁶ m. Find the acceleration due to gravity at its surface using g = GM/r².
Answer and explanation

About 1.6 m/s²

g = GM/r² = (6.67 × 10⁻¹¹)(7.35 × 10²²) ÷ (1.74 × 10⁶)² = (4.90 × 10¹²) ÷ (3.03 × 10¹²) ≈ 1.6 m/s² — about one-sixth of Earth's 9.8 m/s², which is why astronauts could bound across the Moon.

Problem 14

A3 kg cart has 6N net force. Find acceleration.
Answer and explanation

a=Fnet/m=2 m/s² along the net force.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 15

Chapter review: A 4 kg cart has forces 18 N east and 6 N west. Find acceleration and the partner of the 18 N force.
Answer and explanation

Net=12 N east; a=3 m/s² east. Its third-law partner is the cart’s 18 N force on the pulling agent, not the 6 N resistance on the cart.

Net=12 N east; a=3 m/s² east.

Its third-law partner is the cart’s 18 N force on the pulling agent, not the 6 N resistance on the cart.

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.

Problem 16

A4 kg block has 20N right and 8N left. Find acceleration.
Answer and explanation

Net 12N right; a=12/4=3 m/s² right.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 17

Chapter review: 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=15cos30°−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=15cos30°−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.

Problem 18

An object rests on a horizontal floor with a 2N push and max static friction 5N. Find friction.
Answer and explanation

Static friction is 2N opposite the push, so net force is zero. It need not reach 5N.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 19

Chapter review: A 5 kg block has μs=0.40 and μk=0.30 on a level surface. For a 15 N push, find static friction; for a 25 N push during sliding, find acceleration.
Answer and explanation

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction. During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

N=49 N; static limit=19.6 N, so 15 N is balanced by 15 N friction.

During sliding fk=14.7 N; a=(25−14.7)/5=2.06 m/s² along the push.

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.

Problem 20

v=4 m/s,r=2 m. Find inward acceleration.
Answer and explanation

ac=v²/r=8 m/s². Speed can be constant while velocity direction changes.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 21

Chapter review: A 0.50 kg mass moves at 4.0 m/s in a 2.0 m horizontal circle. Find inward acceleration and required net force; name what else is needed to identify tension.
Answer and explanation

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N. Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

ac=16/2=8.0 m/s² and inward net force=0.50×8=4.0 N.

Tension equals that net only if the specified geometry and other forces establish it as the sole inward contribution.

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.

Problem 22

A perpendicular 8N force acts .25 m from pivot. Find torque.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

2 N·m

.25×8=2.

Problem 23

Why is spring work not final force times extension?
Answer and explanation

Force varies with extension; integrate or use the triangular graph area.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 24

Chapter review: A weightless 2 m beam is pivoted at its left end. A 60 N load acts 0.50 m from the pivot. An upward force acts at the right end. Find that force and vertical pivot reaction for equilibrium.
Answer and explanation

Torque balance: F×2=60×0.50, so F=15 N upward. Vertical balance gives pivot reaction=60−15=45 N upward. Both balances are necessary.

Torque balance: F×2=60×0.50, so F=15 N upward.

Vertical balance gives pivot reaction=60−15=45 N upward.

Both balances are necessary.

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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