Newton's Second Law

What this law says

The net external force acting on a body equals the rate of change of its momentum:

$\vec{F} = \frac{d\vec{p}}{dt}$

For a body of constant mass, this simplifies to the familiar form:

$\vec{F} = m\vec{a}$

Force, acceleration, and momentum are all vectors — the equation holds in every direction simultaneously.

The more fundamental form: $\vec{F} = \frac{d\vec{p}}{dt}$

Newton originally stated the law in terms of momentum, not acceleration. Momentum is $\vec{p} = m\vec{v}$, so:

$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$

If mass is constant:

$\vec{F} = m\frac{d\vec{v}}{dt} = m\vec{a}$

This gives the familiar $\vec{F} = m\vec{a}$.

But the momentum form is more general. When mass changes — a rocket burning fuel, a raindrop collecting mass as it falls — the momentum form is correct and $\vec{F} = m\vec{a}$ breaks down.

What the law actually means

The Second Law quantifies what the First Law described qualitatively. The First Law said force causes change in motion. The Second Law says how much change:

$\vec{a} = \frac{\vec{F}}{m}$

• Larger force → larger acceleration (proportional)
• Larger mass → smaller acceleration (inversely proportional)

Push a cricket ball and a cannonball with the same force. The cricket ball accelerates far more — it has far less mass.

The law applies to net force

$\vec{F}$ in $\vec{F} = m\vec{a}$ is always the net (resultant) force — the vector sum of all forces acting on the body.

If three forces act on a body:

$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = m\vec{a}$

Each component can be written separately:

$F_x = ma_x, \quad F_y = ma_y, \quad F_z = ma_z$

This is extremely useful — it means we can analyse horizontal and vertical motions independently.

Units

From $F = ma$:

$[F] = [m][a] = \text{kg} \cdot \text{m/s}^2 = \text{N (Newton)}$

One Newton is the force that gives a 1 kg mass an acceleration of 1 m/s².

Applying the Second Law — the method

1. Identify the system (which body are we applying the law to?)
2. Draw a free body diagram — show all forces on that body
3. Choose a coordinate system
4. Write $F_{net} = ma$ for each direction
5. Solve for the unknown

Example: A 5 kg block is pushed by a 20 N force on a frictionless surface. Find acceleration.

$F_{net} = 20 \text{ N}, \quad m = 5 \text{ kg}$

$a = \frac{F}{m} = \frac{20}{5} = 4 \text{ m/s}^2$

Example with friction: Same block, but now $\mu_k = 0.3$, $g = 10$ m/s².

Normal force: $N = mg = 50$ N

Friction: $f_k = \mu_k N = 0.3 \times 50 = 15$ N (opposing motion)

Net force: $F_{net} = 20 - 15 = 5$ N

$a = \frac{5}{5} = 1 \text{ m/s}^2$

The Second Law and the First Law

When $\vec{F}_{net} = 0$:

$\frac{d\vec{p}}{dt} = 0 \implies \vec{p} = \text{constant} \implies \vec{v} = \text{constant}$

The First Law is recovered as a special case. A body with no net force has constant momentum — it stays at rest or moves in a straight line at constant speed.