Newton’s Second Law: Study Guide (Physics 1 2.5)

This guide is for the algebra-based Physics 1 course. Everything here uses algebra, diagrams and graphs. If you are taking the calculus-based Physics C: Mechanics course, it has its own separate guide.

From balanced to unbalanced forces

Topic 2.4 showed that when the forces on a system are balanced (net force zero), its velocity stays constant. Now take away that condition.

When the net force on a system is not zero, the forces are unbalanced. An unbalanced force changes the system’s velocity: it can speed the system up, slow it down, change its direction, or do several of these at once.

The net force ΣF is the vector sum of every external force on the system. “External” means exerted by something outside the system. Find it one axis at a time, with signs:

ΣF_x = (sum of all x-components of the forces)

The velocity changes only along an axis where the net force is not zero. A box sliding across a floor can have balanced vertical forces and an unbalanced horizontal force at the same time. Its vertical velocity stays zero while its horizontal velocity changes.

Newton’s second law

Newton’s second law: the acceleration of a system’s centre of mass is in the same direction as the net external force on the system, and its size is proportional to the size of that net force.

a = ΣF / m

Here m is the system’s mass in kg, ΣF is the net external force in newtons (N), and a is the acceleration of the centre of mass in m/s². The equation also defines the newton: 1 N = 1 kg·m/s², the net force that gives a 1 kg mass an acceleration of 1 m/s².

Use the law one axis at a time:

ΣF_x = m a_x and ΣF_y = m a_y

Functional dependence: what happens if you change F or m?

The law tells you how acceleration depends on each variable.

• a ∝ ΣF (fixed mass). Double the net force and the acceleration doubles.
• a ∝ 1/m (fixed net force). Double the mass and the acceleration halves.

Mass (kg)	Net force (N)	Acceleration (m/s²)
2.0	1.0	0.50
2.0	2.0	1.0
2.0	4.0	2.0
4.0	2.0	0.50

To predict a factor of change, write the new acceleration as a ratio. If the net force triples and the mass doubles:

a_new / a_old = (3ΣF / 2m) ÷ (ΣF / m) = 3/2

The new acceleration is 1.5 times the old one. No numbers are needed.

Figure 1 also shows how to measure a mass with the second law: vary the net force on a cart, measure each acceleration, plot a against ΣF, and take m = 1 / slope.

A routine for every second-law problem

1. Choose the system. Say in words what is inside it.
2. Draw a free-body diagram for that system. Each force is a separate straight arrow from a dot, labelled with what exerts it.
3. Choose axes, ideally with one axis along the acceleration.
4. Write ΣF = ma along each axis, with signs from your axes.
5. Solve, then check units, signs and a limiting case.

The quantity ma is the result of the forces, not another force. It never goes on the free-body diagram.

Worked example 1: dragging a crate

Question. Take +x to the right and +y up. A 25 kg crate starts from rest on a level floor. A worker pulls it with a horizontal rope force of 120 N to the right. The floor exerts a friction force of 45 N to the left (Topic 2.7 explains where this value comes from). Find (a) the normal force, (b) the crate’s acceleration and (c) its speed and displacement after 2.0 s.

1. System: the crate. Forces: rope, friction (floor), normal (floor), gravity (Earth).
2. Vertical axis. The crate stays on the floor, so a_y = 0: ΣF_y = F_N − F_g = 0, so F_N = F_g = mg = 25 kg × 9.8 m/s² = 245 N (about 250 N to 2 significant figures).
3. Horizontal axis. ΣF_x = +120 N − 45 N = +75 N. a_x = ΣF_x / m = 75 N ÷ 25 kg = +3.0 m/s² (to the right).
4. Kinematics (Topic 1.3), constant acceleration from rest: v_x = 0 + (3.0 m/s²)(2.0 s) = 6.0 m/s; Δx = ½ (3.0 m/s²)(2.0 s)² = 6.0 m to the right.

Check. Units: N ÷ kg = (kg·m/s²) ÷ kg = m/s². The acceleration points the same way as the net force (right). If the friction were 120 N, the net force would be zero and a_x = 0, as Topic 2.4 predicts.

Choosing the system: external and internal forces

The second law uses only external forces. Forces that one part of the system exerts on another part come in Newton’s third-law pairs (Topic 2.3). Both forces of a pair are inside the system, so they cancel in ΣF.

That gives a strong result: the velocity of a system’s centre of mass changes only if a non-zero net external force acts on the system. Internal pushes and pulls cannot change it.

Example. Two skaters stand at rest on smooth ice: A (60 kg) and B (40 kg). They push on each other with forces of 120 N for 0.50 s. Take +x towards B’s side.

• On A: a = −120 N ÷ 60 kg = −2.0 m/s², so A ends at −1.0 m/s.
• On B: a = +120 N ÷ 40 kg = +3.0 m/s², so B ends at +1.5 m/s.
• Centre-of-mass velocity: (60 × (−1.0) + 40 × 1.5) ÷ 100 = 0 m/s.

Each skater speeds up, but the two-skater system’s centre of mass does not move. The ice exerts no horizontal force, so there is no net external force on the system.

Worked example 2: two blocks, two systems

Question. Take +x to the right. Block A (4.0 kg) and block B (2.0 kg) sit on a frictionless horizontal surface, joined by a light string with B behind A. A student pulls A to the right with a force of 18 N. Find the acceleration and the string tension.

1. System 1: A and B together. The string forces are internal and cancel. The only horizontal external force is 18 N. a = 18 N ÷ (4.0 kg + 2.0 kg) = 3.0 m/s² to the right.
2. System 2: B alone. Its only horizontal force is the string tension T, pulling right. T = m_B a = 2.0 kg × 3.0 m/s² = 6.0 N.
3. Check with System 3: A alone. Forces: 18 N right, 6.0 N left (string). ΣF_x = 18 − 6.0 = 12 N, and m_A a = 4.0 × 3.0 = 12 N. ✓

Interpretation. The whole-system view gives the acceleration fastest, because the unknown tension does not appear. Splitting the system is how you find an internal force. The tension (6.0 N) is less than the pull (18 N) because the string only has to accelerate B.

Worked example 3: deriving a symbolic expression

Question. A cart of mass m₁ sits on a level, frictionless track. A light string runs from the cart over a light, frictionless pulley to a hanging block of mass m₂. The system is released from rest. Derive expressions for the acceleration a and the string tension T, then evaluate them for m₁ = 0.80 kg and m₂ = 0.20 kg.

1. Cart (m₁), horizontal: the only horizontal force is the tension. T = m₁a.
2. Hanging block (m₂), +down: gravity m₂g down, tension T up. m₂g − T = m₂a.
3. Add the two equations to remove T: m₂g = (m₁ + m₂)a, so a = m₂g / (m₁ + m₂).
4. Substitute back: T = m₁a = m₁m₂g / (m₁ + m₂).
5. Numbers: a = (0.20 × 9.8) ÷ 1.00 = 1.96 m/s² ≈ 2.0 m/s²; T = 0.80 × 1.96 = 1.568 N ≈ 1.6 N.

Check with limiting cases.

• If m₁ → 0, then a → g: with no cart to drag, the block falls freely. ✓
• If m₂ is very small compared with m₁, a ≈ m₂g / m₁: a tiny weight barely moves a heavy cart. ✓
• T (1.6 N) is less than m₂g (1.96 N). It must be, or the block could not accelerate downward. ✓

Shortcut. Treat cart, string and block as one system. The only unbalanced external force along the string’s path is m₂g, and the mass being accelerated is m₁ + m₂. That gives step 3 in one line.

Common misconceptions

• “An object moves in the direction of the net force.” The acceleration is in the direction of the net force. A car moving east while braking has a net force west.
• “Something must push an object forward to keep it moving.” A non-zero net force changes velocity. Constant velocity needs zero net force (Topic 2.4).
• “ma is a force, so draw it on the diagram.” ma is what the forces produce. Drawing it counts it twice.
• “The tension in a string always equals the hanging weight.” Only when the hanging object is not accelerating. In Worked example 3 the tension is 1.6 N, not 1.96 N.
• “Internal forces can move a system.” The skaters speed up, but their combined centre of mass stays put.
• Using a force instead of the net force. In Worked example 1, 120 N ÷ 25 kg = 4.8 m/s² is wrong: friction also acts.
• Mixing up mass and weight. Mass (kg) goes in the denominator; weight is a force (N) and goes in ΣF.

Where this leads

The second law turns every force model into motion. Topic 2.6 (Gravitational Force) explains the weight force mg and shows why an object’s apparent weight changes when it accelerates; Topics 2.7 and 2.8 add friction and spring forces; Topic 2.9 applies a = ΣF / m to circular motion. Try the practice questions now, then use the revision notes and the checklist to consolidate. Next topic: Gravitational Force. You can also return to the course roadmap.