# Question #c57c0

Mar 5, 2016

Yes.

#### Explanation:

Force $F = m \cdot a$
Where $m$ is the mass and $a$ its acceleration of the object.
Also Kinetic Energy $K E = \frac{1}{2} m {v}^{2}$
Given that $F \propto \frac{1}{v}$
We know that acceleration $a$ is $\equiv \frac{\mathrm{dv}}{\mathrm{dt}}$
$\therefore$ one of the $v$ in the expression for Kinetic energy can be written as
$K E = \frac{1}{2} m {v}^{2} \propto m \left(a . t\right) . v$
Since the factor $\frac{1}{2}$ is a constant, it has been absorbed in the propotionality. Rearranging

$K E \propto \left(m . a\right) . \left(t . v\right)$

Substituting for force and its given inverse proportionality we get
$K E \propto \frac{1}{\cancel{v}} . \left(t . \cancel{v}\right)$
$K E \propto t$

$t$ is time

Does this help?