SPH4U Dynamics Practice: Free-Body Diagrams and Newton’s Laws

To solve a dynamics problem, choose the object or system, draw only the external forces acting on it, choose axes, and apply the net-force equation in each direction. Newton’s second law uses the vector sum of forces: ΣF = ma, not just whichever force is easiest to see.

This original SPH4U practice set develops free-body diagrams, components and force equations. Use g = 9.8 m/s². Treat strings and pulleys as ideal where stated, and use the friction model specified in each question. The set is not a complete Grade 12 physics assessment.

What belongs on a free-body diagram?

Choose one object and represent it separately from its surroundings. Draw forces exerted on that object: for example, gravity from Earth, a normal force from a surface, tension from a string or friction from a contact. Do not add velocity or acceleration arrows as if they were forces.

A practical connection to earlier motion work is useful here. An Ontario physics discussion highlights students’ difficulty with motion interpretation. Dynamics adds the explanation of what net force produces the acceleration; review the motion model before adding forces. Student discussion

Eight original problems with worked answers

1. Find the net force on a horizontal surface

A 4.0 kg object is pulled horizontally with 18 N while friction acts oppositely with magnitude 6.0 N. Find acceleration.

Take the pull direction as positive. Fnet = 18 − 6.0 = 12 N, so a = 12/4.0 = 3.0 m/s² in the pull direction. On a level surface with no vertical acceleration or other vertical forces, the normal force balances weight; those vertical forces do not add to the horizontal net force.

2. Interpret constant velocity

A 10 kg crate moves across a level surface at constant velocity while a horizontal pull of 25 N acts on it. Find the friction force magnitude under this model.

Constant velocity means a = 0, so the horizontal net force is zero. Friction must be 25 N opposite the motion. Zero net force does not mean no forces act.

3. Find apparent weight in an accelerating lift

A 60.0 kg person stands on a scale in a lift accelerating upward at 1.50 m/s². Find the normal force on the person.

Choose upward positive: N − mg = ma. Thus N = m(g + a) = 60.0(9.8 + 1.50) = 678 N using the given values, approximately 6.8 × 10² N to two significant figures. The gravitational force remains mg; the scale responds to the contact force.

4. Resolve gravity on a frictionless incline

A 5.0 kg block rests on a frictionless plane inclined at 30° to the horizontal and is released. Find its acceleration down the plane and the normal force.

The parallel weight component is mg sin 30°; dividing by m gives a = 4.9 m/s² down the plane. Perpendicular to the plane, N = mg cos 30° ≈ 42 N. The normal force is not equal to mg in this geometry.

5. Calculate kinetic friction before acceleration

A 10.0 kg crate on a level surface is pulled horizontally with 50.0 N. It is already sliding, with coefficient of kinetic friction 0.20. Find its acceleration.

With no vertical acceleration, N = mg = 98 N. Kinetic friction is μₖN = 0.20(98) = 19.6 N. The horizontal net force is 50.0 − 19.6 = 30.4 N, giving a = 3.04 m/s², approximately 3.0 m/s² to two significant figures. The friction model applies because the crate is stated to be sliding.

6. Separate a connected system from one object

Two blocks of masses 2.0 kg and 3.0 kg are connected by a taut, light string on a frictionless horizontal surface. A 20 N horizontal force pulls the 3.0 kg block away from the 2.0 kg block. Find their acceleration and the string tension.

For the combined system, total mass is 5.0 kg and a = 20/5.0 = 4.0 m/s². For the 2.0 kg block alone, tension is its horizontal net force: T = 2.0(4.0) = 8.0 N. For the 3.0 kg block, 20 − 8.0 = 3.0(4.0), confirming the result.

7. Explain an action–reaction pair

A book rests on a table. Are the book’s weight and the table’s normal force on the book a Newton’s third-law pair?

No. They both act on the book. Third-law pairs act on different interacting objects. The partner to the table’s force on the book is the book’s force on the table. Forces on one object may balance without being a third-law pair.

8. Identify the required net force for circular motion

A 0.50 kg object moves at constant speed 4.0 m/s in a circle of radius 2.0 m. Find the net radial force magnitude and direction.

Fnet,radial = mv²/r = 0.50(4.0²)/2.0 = 4.0 N toward the centre. Constant speed does not imply constant velocity because direction changes. “Centripetal force” names the inward net-force requirement; do not automatically add it as an extra force alongside the actual interactions.

A check before submitting your solution

Confirm that every force belongs to the chosen object. State positive directions, keep perpendicular and parallel components separate, and write the force equation before numbers. Check whether the acceleration’s direction agrees with the net force.

For connected objects, identify which forces are internal to the combined system. They cancel in the system equation but can be essential when isolating one object. Question 6 uses that distinction to find tension.

Get support with force diagrams

Read the SPH4U guide, explore Grade 12 science tutoring, or book a free one-hour assessment.

Curriculum reference: Ontario Grades 11–12 Science. These are original practice problems, not an official exam.