Newton's Second Law: Force, Mass & Acceleration
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Lesson overview
Applying F = ma with resultant forces, linking force to rate of change of momentum, and apparent weight in a lift.
Go to lesson quiz →Exam guidance · JAMB UTME Speed Lens
Core idea (Kinematics,SUVAT & Newton's Laws of Motion): Pick a positive direction first, list knowns/unknowns (u, v, a, s, t), draw the free-body diagram before writing ΣF = ma, and check vector signs.
Pacing
~40 seconds per MCQ (180 questions in 120 minutes, no negative marking).
How JAMB marks this
Fast dimensional/sign elimination and standard formula shortcuts — skip multi-line algebra when substitution tests options faster.
Watch out for
UTME distractors routinely include the intermediate value (e.g., forgetting to convert cm³ to dm³ or °C to Kelvin).
Lesson content
Overview
Push an empty wheelbarrow and it speeds up easily; push the same barrow loaded with cement bags and it barely moves. Push harder and it speeds up faster. Newton's second law turns these everyday observations into one precise equation, F = ma. It is the most important equation in mechanics. Every later topic, from momentum and projectiles to circular motion and the forces in a lift, depends on it. In this lesson you will meet all three of Newton's laws, learn to find the resultant force before applying F = ma, use the law in its momentum form, and solve the lift problem that UTME sets so often. Expect at least one question built on this law in every Physics paper.
Key concept: Newton's laws of motion
First law: a body continues in its state of rest or of uniform motion in a straight line unless it is acted on by an external resultant force. The tendency of a body to resist a change in its motion is its inertia, and mass is the measure of inertia. Second law: the rate of change of momentum of a body is directly proportional to the resultant force acting on it and takes place in the direction of that force. With suitable units this gives F = ma. Third law: to every action there is an equal and opposite reaction. The two forces act on different bodies. The newton (N) is the force that gives a mass of 1 kg an acceleration of 1 m/s². So 1 N = 1 kg m/s².
Using F = ma properly
The F in F = ma is the resultant (net) force: the single force left after all the forces on the body are combined. If you push a box with 50 N and friction opposes with 20 N, the resultant force is 50 − 20 = 30 N, and only that 30 N produces acceleration. If the resultant force is zero, the acceleration is zero. The body is then either at rest or moving at constant velocity (Newton's first law). A car cruising at a steady 80 km/h on a straight road has a resultant force of zero: the engine's driving force exactly balances friction and air resistance. The acceleration always points in the direction of the resultant force, not necessarily in the direction of motion. A car braking while moving forward has a backward resultant force and a backward acceleration (a deceleration). The second law comes from momentum. Momentum is p = mv. If a force F acts for a time t and changes the velocity from u to v, then F = (mv − mu) ÷ t = m(v − u) ÷ t = ma, since (v − u) ÷ t is the acceleration. The product F × t is the impulse, and it equals the change in momentum. Weight is a force: W = mg, with g = 10 m/s² near the Earth's surface. A 50 kg student has a weight of 500 N. Mass stays the same everywhere; weight depends on the gravitational field.
Formulae to know
Newton's second law: F = ma (F = resultant force in N, m in kg, a in m/s²) Momentum form: F = (mv − mu) ÷ t = change in momentum ÷ time Impulse: F × t = mv − mu (unit N s, equal to kg m/s) Weight: W = mg, with g = 10 m/s² Reaction (apparent weight) in a lift accelerating upwards, or decelerating while moving down: R = m(g + a) Reaction in a lift accelerating downwards, or decelerating while moving up: R = m(g − a) Free fall (a = g): R = 0, so the person feels weightless.
Worked example 1: resultant force first
A 2.0 kg box on a floor is pushed with a horizontal force of 10 N. Friction between the box and the floor is 4.0 N. Find the acceleration of the box.
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Worked example 2: force from a change in momentum
A ball of mass 0.50 kg moving at 20 m/s hits a wall at right angles and rebounds along the same line at 15 m/s. The ball is in contact with the wall for 0.050 s. Find the average force exerted on the ball.
Active Worked Example · Predict Step 1
4 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Worked example 3 (UTME standard): the lift problem
A man of mass 60 kg stands on a weighing scale in a lift. Find the reading of the scale when the lift (a) accelerates upwards at 2.0 m/s², (b) moves at constant velocity, and (c) accelerates downwards at 2.0 m/s². Take g = 10 m/s².
Active Worked Example · Predict Step 1
5 stepsWrite down the formula, rule, or first move you would make before revealing Step 1.
Common traps
• Always find the resultant force before using F = ma. Subtract opposing forces such as friction. • Zero resultant force does not mean the body is at rest; it may be moving at constant velocity. • In momentum questions, a rebound reverses the sign of the velocity. Equal speeds before and after still give a change in momentum. • In lift problems, look at the direction of the acceleration, not the direction of motion. 'Moving up and slowing down' means accelerating downwards. • Action and reaction (third law) act on different bodies, so they never cancel each other out on one body.
Summary
• First law: no resultant force means no change in velocity. Mass measures inertia. • Second law: resultant force = rate of change of momentum, giving F = ma. 1 N = 1 kg m/s². • Use the resultant force in F = ma; the acceleration is in the direction of that resultant force. • Impulse Ft = mv − mu; take care with signs when a body rebounds. • In a lift, the scale reads R = m(g + a) for upward acceleration and R = m(g − a) for downward acceleration; in free fall R = 0.
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