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The second law, F = ma

Forces and motion · level 5, lesson 3 of 6

A question to start

An astronaut on Mars pushes a 200 kg cart of equipment with a steady 100 N. Friction holds it back with 40 N. How fast does it speed up? Would the answer change on Earth?

The idea

Newton's second law: the overall force on an object equals its mass times its acceleration,

\[F_\text{net} = m a.\]

With \(F\) in newtons, \(m\) in kilograms and \(a\) in m/s², it fits exactly; in fact this is what defines the newton: \(1 \text{ N} = 1 \text{ kg m/s}^2\).

Three things to remember:

  • Use the net force: add the forces with their directions, then use what is left.
  • The acceleration points the same way as the net force.
  • Rearranged, \(a = F_\text{net}/m\): more force, more acceleration; more mass, less.
200 kgpush 100 Nfriction 40 Nnet 60 N forward, so a = 60 / 200 = 0.3 m/s²

The law holds the same everywhere, on Mars, on Earth and in space. What changes between planets is some of the forces, like weight and friction, not the law.

Worked examples

1. The cart. \(F_\text{net} = 100 - 40 = 60\) N. \(a = 60 \div 200 = 0.3\) m/s². On Earth, the push and the mass are the same, but the friction would be bigger (the cart presses harder on the floor), so it would accelerate less.

2. Braking a rover. A 1,000 kg rover moving at 0.05 m/s stops in 0.5 s. \(a = -0.05 \div 0.5 = -0.1\) m/s². \(F = 1000 \times 0.1 = 100\) N of braking force.

3. Finding a mass. A net force of 3,000 N gives a cargo pod an acceleration of 1.5 m/s². \(m = 3000 \div 1.5 = 2{,}000\) kg.

Try it

  1. A net force of 50 N acts on a 10 kg box. What is its acceleration?
  2. What net force gives a 1,200 kg car an acceleration of 2.5 m/s²?
  3. A 60 kg skater is pushed with 90 N, and friction is 30 N. What is her acceleration?
  4. A 900 kg rover needs to reach 0.04 m/s in 2 s from rest. What net force is needed?
Answers
  1. \(50 \div 10 = 5\) m/s².
  2. \(1200 \times 2.5 = 3{,}000\) N.
  3. \(F_\text{net} = 60\) N, so \(a = 60 \div 60 = 1\) m/s².
  4. \(a = 0.04 \div 2 = 0.02\) m/s², so \(F = 900 \times 0.02 = 18\) N.

Common mistake

Using one force instead of the net force. Put only the push into \(F = ma\) and you forget friction. First add up all the forces, with their directions; then use the total.

Where it leads