Physics

4.12 - Newton's 1st & 2nd Laws

8 min

Newton's first law

Resultant force and acceleration

F = ma

note 1 of 5

Newton's First Law - Resultant force changes motion

Newton's First Law - If the resultant force on an object is zero, its velocity doesn't change:
- A stationary object stays stationary.
- A moving object continues at the same velocity (same speed and direction).
When an object is stationary or moving at a constant velocity, the forces on it are balanced.
Velocity only changes if the forces are unbalanced, giving a non-zero resultant force that causes an acceleration in the direction of the force.

Two cubes either side of a dashed line: the blue one labelled balanced has equal arrows pointing in from both sides, the orange one labelled unbalanced has a long arrow meeting a short one.

A non-zero resultant force can cause an object to:

start moving
stop moving
speed up
slow down
change direction

note 2 of 5

Newton's Second Law - Acceleration and resultant force

The larger the resultant force on an object, the more it accelerates - force and acceleration are directly proportional ( $F \propto a$ ).
Acceleration is inversely proportional to mass: for a fixed resultant force, a larger mass accelerates less than a smaller one.

Formula for Newton's Second Law

$F = m a$

Where:

$F$ = Resultant force, in newtons (N)
$m$ = Mass, in kilograms (kg)
$a$ = Acceleration, in metres per second squared (m/s²)

$F$ is always the resultant force, found by adding or subtracting the forces acting along the same line.

note 3 of 5

Worked example - Acceleration of a car

A car of mass 2,000 kg has an engine providing a driving force of 8,000 N. The drag force on the car is 2,000 N. Find its acceleration.

A purple car labelled 2,000 kg with a magenta arrow pointing left labelled 2,000 N and a longer orange arrow pointing right labelled 8,000 N.

Step 1 - Work out the resultant force

The driving force and drag force act along the same line in opposite directions, so subtract them.

$F = 8,000 - 2,000 = 6,000 N$

Step 2 - Rearrange F = ma for acceleration

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

Step 3 - Substitution and calculation

$a = \frac{6,000}{2,000} = 3 {m/s}^{2}$

The car accelerates at 3 m/s² in the direction of the resultant force.

note 4 of 5

Large decelerations and vehicle safety features

A large deceleration requires a large force ( $F = m a$ ).
Large decelerations are dangerous because the large forces can cause serious injuries, as well as:
- causing brakes to overheat and lose effectiveness
- causing the vehicle to skid

Increasing the collision time

Safety features increase the collision time, which reduces deceleration and lowers the force on passengers:

Seat belts - Stretch slightly so the occupant slows down over a longer time
Air bags - Compress to slow the occupant down gradually
Crumple zones - Crumple on impact to increase the time taken to stop

When braking, friction does work on the wheels, transferring energy from the vehicle's kinetic energy store to the thermal energy store of the brakes.

note 5 of 5

Worked example - Force on a car stopping quickly

Estimate the resultant force on a car stopping quickly from 18 m/s.

Step 1 - Estimate the deceleration

The car comes to a stop in about 2 s, so:

$a = \frac{v - u}{t} = \frac{0 - 18}{2} = - 9 {m/s}^{2}$

The acceleration is negative because it acts against the motion.

Step 2 - Estimate the mass

The mass of a car is about 1,200 kg.

Step 3 - Apply Newton's Second Law

$F = m a = 1,200 \times (- 9) = - 10,800 N$

The force is about -10,800 N. The negative sign shows it acts opposite to the car's motion, and its large size is why a sharp stop is dangerous.

end of notes

Key points to remember

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4.13 - Reaction TimesNext up