14.1 Force, Motion, and Newton's Laws

Physics on the USTET Science subtest expects Grade 11 fluency with motion and forces—not calculus, but clear definitions, unit sense, and the ability to apply Newton's second law to short numerical problems. This section builds from describing motion (kinematics) to explaining what causes changes in motion (dynamics).

Describing Motion: Scalars and Vectors

A scalar quantity has magnitude only. A vector quantity has both magnitude and direction. Distance is how far an object travels along its path (scalar). Displacement is the straight-line change from initial to final position (vector). If you walk 3 m east and then 4 m west, your distance is 7 m but your displacement is 1 m west.

Speed is distance divided by time (scalar). Velocity is displacement divided by time (vector). Acceleration is the rate of change of velocity. An object can accelerate by speeding up, slowing down, or changing direction—even if its speed stays constant, as in uniform circular motion.

On multiple-choice items, watch for traps that swap distance with displacement or speed with velocity. If a question mentions "back to the starting point," displacement is zero even when distance is not.

Average Velocity and Acceleration

Average velocity:

$v_{avg} = \frac{\Delta x}{\Delta t}$

Average acceleration:

$a_{avg} = \frac{\Delta v}{\Delta t}$

Worked example (kinematics). A jeepney goes from rest to 20 m/s in 5.0 s along a straight road. What is its average acceleration?

$a = \frac{20\ \mathrm{m/s} - 0}{5.0\ \mathrm{s}} = 4.0\ \mathrm{m/s^2}$

The positive sign means velocity increased in the forward direction. If the same jeepney later slows from 20 m/s to 8 m/s in 4.0 s, acceleration is $(8 - 20)/4.0 = -3.0\ \mathrm{m/s^2}$—speed decreases, so acceleration is opposite the velocity.

For constant acceleration from rest, a useful relation is $v = at$ and $d = \frac{1}{2}at^2$. USTET rarely demands the full kinematic equation set, but recognizing that doubling time at constant $a$ from rest quadruples distance (because of the $t^2$ dependence) can unlock estimate-style items.

Newton's First Law: Inertia

Newton's first law (law of inertia): An object at rest remains at rest, and an object in uniform straight-line motion continues that motion, unless acted on by a net (unbalanced) force.

Inertia is the tendency to resist changes in motion. Mass measures inertia: a fully loaded truck is harder to start or stop than a bicycle. On a frictionless horizontal surface (idealized), a puck sliding at constant velocity needs no continuing push—constant velocity already means zero net force. Real roads have friction, so engines must supply force to cancel drag and keep speed steady.

Passengers lurch forward when a vehicle brakes because their bodies tend to keep moving forward while the vehicle slows. Seat belts apply a force that changes the passenger's velocity safely with the cabin.

Newton's Second Law: F = ma

Newton's second law: The net force on an object equals its mass times its acceleration:

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

In magnitude for one-dimensional problems: $F_{net} = ma$. One newton is the force that accelerates 1 kg at 1 m/s²: $1\ \mathrm{N} = 1\ \mathrm{kg\cdot m/s^2}$.

Direction matters: acceleration is in the same direction as the net force. If several forces act, add them as vectors (or use signed components on a line) to find $F_{net}$ before dividing by mass.

Worked example 1 — Finding acceleration

A 2.0 kg cart on a smooth horizontal track is pulled by a horizontal force of 6.0 N. Friction is negligible. Find the acceleration.

$a = \frac{F_{net}}{m} = \frac{6.0\ \mathrm{N}}{2.0\ \mathrm{kg}} = 3.0\ \mathrm{m/s^2}$

Worked example 2 — Finding net force

A 5.0 kg box accelerates at 2.0 m/s² to the right. What net force acts on it?

$F_{net} = ma = (5.0\ \mathrm{kg})(2.0\ \mathrm{m/s^2}) = 10\ \mathrm{N}$ to the right.

Worked example 3 — Two opposing forces

A 4.0 kg crate is pushed right with 18 N while friction of 6.0 N acts left. Find acceleration.

$F_{net} = 18\ \mathrm{N} - 6.0\ \mathrm{N} = 12\ \mathrm{N}$ $a = \frac{12\ \mathrm{N}}{4.0\ \mathrm{kg}} = 3.0\ \mathrm{m/s^2}$ to the right.

If the push were only 6.0 N against 6.0 N of friction, $F_{net} = 0$ and $a = 0$—the crate could remain at rest or move at constant velocity, consistent with the first law.

Worked example 4 — Mass from F and a

A net force of 24 N gives an object an acceleration of 3.0 m/s². Find the mass.

$m = \frac{F}{a} = \frac{24\ \mathrm{N}}{3.0\ \mathrm{m/s^2}} = 8.0\ \mathrm{kg}$

Weight Versus Mass

Mass is the amount of matter (kg). Weight is the gravitational force on that mass:

$W = mg$

Near Earth's surface, $g \approx 9.8\ \mathrm{m/s^2}$ (many entrance-exam items use $10\ \mathrm{m/s^2}$ for speed). A 50 kg student has weight $W = 50 \times 9.8 = 490\ \mathrm{N}$ (or 500 N if $g = 10$). On the Moon, mass stays 50 kg but weight drops because $g$ is smaller. Bathroom "weight" in kilograms on consumer scales is actually reporting mass after converting from force—physically, weight is a force in newtons.

Newton's Third Law: Action–Reaction

Newton's third law: If object A exerts a force on object B, then B exerts an equal-magnitude, opposite-direction force on A. The two forces act on different objects, so they do not cancel on a free-body diagram of one object alone.

When you push a wall, the wall pushes you. A rocket pushes exhaust gases backward; gases push the rocket forward. Walking works because your foot pushes the ground backward and the ground pushes you forward (static friction).

Common mistake: thinking action and reaction cancel so nothing moves. They cannot cancel on the same object—they form a pair across two objects.

Free-Body Thinking for USTET

Sketch forces on one object: gravity (weight) downward, normal force from a surface perpendicular to the surface, friction opposing relative sliding (or the tendency to slide), tension along a string, and applied pushes or pulls. Then:

1. Choose a positive direction.
2. Write $F_{net} = \sum F$ in that direction.
3. Set $F_{net} = ma$ and solve for the unknown.

On a horizontal surface at rest, normal force balances weight: $N = mg$. If you push down on a box, $N$ increases; if you lift partly upward, $N$ decreases. Friction often satisfies $f \leq \mu N$ (static) or $f = \mu_k N$ (kinetic), but many USTET items simply give a friction force value rather than requiring $\mu$.

Connecting Kinematics to Forces

Once you have $a$ from $F = ma$, you can predict velocity changes. Example: a 1.5 kg toy car feels a constant net force of 3.0 N for 4.0 s starting from rest.

$a = \frac{3.0}{1.5} = 2.0\ \mathrm{m/s^2}$ $v = at = (2.0)(4.0) = 8.0\ \mathrm{m/s}$

That two-step pattern—force to acceleration, then acceleration to kinematics—is a frequent exam pathway.

Quick Check Habits

• Units: N with kg and m/s²; inconsistent units usually mean a wrong formula choice.
• Net force zero means $a = 0$, not necessarily "no forces."
• Heavier mass at the same net force means smaller acceleration ($a = F/m$).
• Third-law partners are equal in magnitude even when masses differ; the lighter object gets larger acceleration from the same-magnitude force.

Master these definitions and the four $F = ma$ patterns above, and most USTET force–motion items become routine algebra rather than memorization.