Page:Calculus Made Easy.pdf/77

or

$f=m{\frac {dv}{dt}}$;

or

$f=m{\frac {d^{2}y}{dt^{2}}}$.

The product of a mass by the speed at which it is going is called its momentum, and is in symbols $mv$. If we differentiate momentum with respect to time we shall get ${\dfrac {d(mv)}{dt}}$ for the rate of change of momentum. But, since $m$ is a constant quantity, this may be written $m{\dfrac {dv}{dt}}$ which we see above is the same as $f$. That is to say, force may be expressed either as mass times acceleration, or as rate of change of momentum.

Again, if a force is employed to move something (against an equal and opposite counter-force), it does work; and the amount of work done is measured by the product of the force into the distance (in its own direction) through which its point of application moves forward. So if a force $f$ moves forward through a length $y$, the work done (which we may call $w$) will be

$w=f\times y$;

where we take $f$ as a constant force. If the force varies at different parts of the range $y$, then we must find an expression for its value from point to point. If $f$ be the force along the small element of length $dy$, the amount of work done will be $f\times dy$. But as $dy$ is only an element of length, only an element of