Newton's second law, weight and gravityAQA A-Level Maths: Revision notes

Newton's second law: the resultant force on a particle equals its mass multiplied by its acceleration, in the direction of the resultant force. $F=ma$ $F$ is the resultant (net) force in newtons, $m$ is mass in kg and $a$ is acceleration in m s⁻². A force of $20$ N on an $8$ kg box on smooth ground gives $a=\frac{20}{8}=2.5$ m s⁻². Use the second law in one direction at a time, writing "resultant force in the direction of motion $=ma$". If the resultant is zero then $a=0$, which is Newton's first law.

For motion in a straight line, take the direction of motion as positive and add the forces along it, with forces opposing motion negative. Combine with the constant acceleration equations ($v=u+at$, $s=ut+\frac12at^2$, $v^2=u^2+2as$, $s=\frac{(u+v)}{2}t$) to find speeds, times and distances. Example: a box from rest with $a=2.5$ m s⁻². After $4$ s, $v=10$ m s⁻¹ and $s=\frac12(2.5)(16)=20$ m. A lift of mass $500$ kg accelerating upwards at $1.2$ m s⁻² has $T-500g=500(1.2)$, so $T=5500$ N.

When forces are given as 2D vectors, find the resultant by adding the $\mathbf{i}$ and $\mathbf{j}$ parts, then use $\mathbf{F}=m\mathbf{a}$. For $m=2$, $\mathbf{F}_1=(6\mathbf{i}-2\mathbf{j})$ N and $\mathbf{F}_2=(\mathbf{i}+8\mathbf{j})$ N: $\mathbf{F}=(7\mathbf{i}+6\mathbf{j})$ N and $\mathbf{a}=(3.5\mathbf{i}+3\mathbf{j})$ m s⁻². From rest, $\mathbf{v}=\mathbf{a}t$, and speed is the magnitude $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}$. At AS, forces are restricted to two perpendicular directions, or simple cases of 2D vectors.

The weight of a body is the gravitational force on it: $W=mg$, acting vertically downwards. Mass (kg) is the amount of matter and does not change with location. Weight (N) depends on the gravitational acceleration $g$. Near Earth's surface $g=9.8$ m s⁻² (often given as $9.81$ m s⁻² when more accuracy is needed). A body falling freely under gravity alone has acceleration $g$ downwards, whatever its mass. $g$ is not a universal constant: it depends on location, for example about $1.6$ m s⁻² on the Moon, so the same body has a smaller weight there but the same mass. In AS Mathematics $g$ is assumed constant, and the inverse square law is not needed.

For vertical motion under gravity alone, use $a=g$ downwards (or $-g$ if upwards is positive). A lift whose cable snaps at $2$ m s⁻¹ upwards has $0=2^2-2(9.8)s$, so it rises $0.204$ m further. When a cable, string or rope also acts, include it in the resultant: for upward acceleration $a$, $T-mg=ma$. For a probe of mass $12$ kg raised at $0.5$ m s⁻²: on a planet with $g_p=1.6$, $T=12(1.6+0.5)=25.2$ N; on Earth, $T=12(9.8+0.5)=123.6$ N.