Newton’s laws: practice problems with step-by-step solutions

1What Newton’s laws say

Kinematics describes motion. Newton’s laws explain it: a body changes its velocity only when a net force acts on it.

• First law: if the forces on a body balance, it stays at rest or keeps moving at constant velocity.
• Second law: a net force gives an acceleration in the same direction. Bigger force, bigger acceleration. Bigger mass, smaller acceleration.
• Third law: if A pushes on B, then B pushes back on A with an equal force in the opposite direction.

The second law is the one you calculate with:

ΣF (“sum of F”) is the net force: all forces added with their signs. Force is in newtons (N), mass in kg and acceleration in m/s². One newton gives 1 kg an acceleration of 1 m/s².

2The forces you meet

Most questions use the same four kinds of force. Draw each one as an arrow on the body: this drawing is the free-body diagram.

• Weight W = m·g, straight down, with g = 9.81 m/s².
• Normal force N: the surface pushes back, at right angles to the surface. On a level floor with no other vertical forces, N = m·g.
• Friction F_f = μ·N, along the surface and against the motion. μ (mu) is the coefficient of friction, a number such as 0.3.
• Tension T in a string or rope, pulling along the string. A light string has the same tension at both ends.

Then add the forces along the direction of motion. In the diagram, ΣF = F − F_f, because friction points backwards.

3Bodies on a slope

On a slope, split the weight into two parts. Take the angle θ of the slope with the horizontal.

• m·g·sin θ pulls the body down along the slope.
• m·g·cos θ presses it into the slope. The normal force balances this part, so N = m·g·cos θ.

Friction is then μ·m·g·cos θ, pointing up the slope when the body slides down. A steeper slope gives a bigger sin θ and a smaller cos θ, so the body speeds up faster.

Two bodies joined by a string over a pulley move with the same acceleration. Treat them first as one system with the total mass, then look at one body alone to find the tension.

4The method step by step

1. Draw a free-body diagram with every force on the body. Forget no weight and no normal force.
2. Choose a positive direction along the motion.
3. Split forces into components along and at right angles to the motion. On a slope: m·g·sin θ along, m·g·cos θ into the slope.
4. Use the forces at right angles to find N. There is no acceleration in that direction.
5. Find friction with F_f = μ·N, against the motion.
6. Apply ΣF = m·a along the motion and solve for the unknown.

5A worked example

Follow every step. Try to predict the next one before you open it.

6Common mistakes

• Mixing up sin and cos on a slope. Along the slope is sin θ, into the slope is cos θ.
• Using N = m·g on a slope, or when a rope pulls partly upwards. Find N from the forces at right angles to the motion.
• Letting friction point the wrong way. It always opposes the motion.
• Forgetting a force, for example the weight of a hanging mass, or counting a force twice.
• Using ΣF = m·a with only the applied force instead of the net force.

7How exam questions look

• Easy: a push with friction on a level floor, two blocks pushed together, or a person on scales in a lift.
• Medium: an inclined plane: a block sliding down, a box pushed up at constant speed, or a sled pulled by a rope at an angle.
• Hard: a table and pulley: a block on a table or ramp tied over a pulley to a hanging mass. Find the common acceleration and the tension.

At constant speed the acceleration is zero, so the forces balance: ΣF = 0.