Newton's second law in a straight lineEdexcel A-Level Maths: Revision notes

Newton's second law: the resultant force on a body is proportional to its rate of change of momentum. For a body of constant mass $m$ this gives $\mathbf{F}=m\mathbf{a},$ where $\mathbf{F}$ is the resultant force (newtons), $m$ the mass (kilograms) and $\mathbf{a}$ the acceleration (m s$^{-2}$). The acceleration is in the same direction as the resultant force. One newton is the force that gives a mass of $1$ kg an acceleration of $1$ m s$^{-2}$. If the resultant force is zero then $\mathbf{a}=\mathbf{0}$, which is Newton's first law.

Choose a positive direction (the direction of motion or acceleration), add the forces acting in that direction (opposing forces negative), and write $F=ma$. Example: a sledge of mass $40$ kg is pulled by a rope with tension $100$ N against a resistance of $20$ N. $100-20=40a$, so $a=2$ m s$^{-2}$. Forces perpendicular to the motion cancel when there is no acceleration in that direction, so the normal reaction balances the weight on horizontal ground. Each problem is written as an equation of motion: $\text{driving force}-\text{resistance}=ma$.

Once $a$ is known from $F=ma$, use the constant-acceleration (suvat) equations: $v=u+at$, $s=ut+\frac12at^2$, $v^2=u^2+2as$, $s=\frac{(u+v)t}{2}$. Example: a lorry of mass $8000$ kg travelling at $13$ m s$^{-1}$ has its engine switched off, with resistance $1200$ N. Then $-1200=8000a$, so $a=-0.15$ m s$^{-2}$, and $0=13^2+2(-0.15)s$ gives $s=563$ m. A negative acceleration means the velocity is decreasing (a deceleration) in your chosen positive direction.

When forces are given as vectors, add them to get the resultant $\mathbf{F}$ and use $\mathbf{F}=m\mathbf{a}$ on the whole vector, or component by component. Example: $m=2$ kg with $\mathbf{F}_1=(6\mathbf{i}-4\mathbf{j})$ and $\mathbf{F}_2=(2\mathbf{i}+8\mathbf{j})$. Resultant $=(8\mathbf{i}+4\mathbf{j})$, so $\mathbf{a}=(4\mathbf{i}+2\mathbf{j})$ and $|\mathbf{a}|=\sqrt{20}=4.47$ m s$^{-2}$. Unknown forces: if a velocity changes from $\mathbf{u}$ to $\mathbf{v}$ in time $t$ at constant acceleration, then $\mathbf{a}=\frac{\mathbf{v}-\mathbf{u}}{t}$ and $\mathbf{F}=m\mathbf{a}$ gives the resultant, from which an unknown force can be found by subtraction.

1. Draw a force diagram, marking every force and the direction of the acceleration.
2. Write $F=ma$ in the direction of motion: resultant force on the left, $ma$ on the right.
3. Solve for $a$ (or an unknown force), then use suvat or vectors for the rest of the question.
4. Give units: N, kg, m s$^{-2}$.
5. For magnitudes of vectors, use Pythagoras, and give 3 significant figures unless exact.