Newton's second law is the one law of motion you actually calculate with. The first law tells you what happens with no net force; the third law tells you forces come in pairs; the second law tells you how much something accelerates, and in which direction. Get this one right and most of mechanics follows.
The statement, and what it means
The rate of change of momentum of a body is proportional to the net external force on it, and happens in the direction of that force:
F = dp/dt, where p = mv is momentum.
When the mass does not change, dp/dt = m·dv/dt = ma, which gives the version everyone remembers:
F = ma
Three words in that statement carry all the marks:
- Net. Add every force as a vector first. Only the leftover force accelerates the body.
- External. Forces between parts of the same system (you pushing on your own car from inside) do not count.
- Direction. Acceleration is along the net force — not necessarily along the velocity. A ball thrown upward accelerates downward the whole time.
Units and a sanity check
The SI unit of force is the newton: 1 N = 1 kg·m/s². The CGS unit is the dyne; 1 N = 10⁵ dyne. A quick feel for size: holding a 100 g phone takes about 1 N.
Before writing an answer, check the direction of the acceleration against the direction of the net force. If they disagree, the free-body diagram is wrong.
Worked example 1 — a straight push
A 5 kg block on a frictionless floor is pushed with 20 N. Find the acceleration.
a = F/m = 20/5 = 4 m/s² along the push.
Worked example 2 — two forces, one body
The same block is pushed with 20 N to the right and pulled with 8 N to the left.
Net force = 20 − 8 = 12 N to the right. a = 12/5 = 2.4 m/s² to the right. (Students who write 28/5 have added magnitudes instead of vectors.)
Worked example 3 — friction included
A 10 kg crate is pulled with 50 N across a floor with kinetic friction coefficient 0.3. Take g = 10 m/s².
Friction = μN = 0.3 × (10 × 10) = 30 N, opposing motion. Net = 50 − 30 = 20 N. a = 20/10 = 2 m/s².
Worked example 4 — from the momentum form
A cricket ball of mass 0.15 kg moving at 20 m/s is stopped by a fielder in 0.1 s. Find the average force.
Change in momentum = 0.15 × (0 − 20) = −3 kg·m/s. F = Δp/Δt = −3/0.1 = −30 N, that is 30 N opposite to the ball's motion. Pulling the hands back increases Δt and so reduces the force — which is why fielders do it.
Worked example 5 — the lift problem
A 60 kg person stands on a weighing scale in a lift accelerating upward at 2 m/s². What does the scale read? (g = 10 m/s²)
Forces on the person: normal reaction N up, weight 600 N down. Net upward = N − 600 = ma = 60 × 2 = 120, so N = 720 N. The scale reads 72 kg. If the lift accelerated downward at 2 m/s², N = 600 − 120 = 480 N.
Where students go wrong
- Using mass in grams. 500 g is 0.5 kg. The unit of F = ma only works in kg, m and s.
- Forgetting friction is a force. It goes in the net force like any other.
- Taking g as 9.8 in one line and 10 in the next. Pick the value the question gives and keep it.
- Confusing F = ma with "force causes velocity". Force causes acceleration. A body can move at constant velocity with zero net force.
How it is tested
Class 9 asks for the statement, the unit, and one-step numericals. Class 11, JEE and NEET build the same law into pulleys, inclines, lifts and connected bodies — always the same procedure: draw the free-body diagram, resolve along the direction of motion, write ΣF = ma for each body, and solve the simultaneous equations.