# How do i prove this?

## a particle of mass $m$ moves under the influence of the force F= a(sinωt i+cosωtj) if the particle is initially at rest at the origin, prove that the work done on the particle up to time $t$ is given by (a^2 /(mω^2))(1-cosωt)

Dec 1, 2017

vec F = a{\sin\omegat\hati + \cos\omegat\hatj};
Newton's Second Law (constant mass):
vec F = m(dvecv)/(dt);
d\vecv = vecF/mdt;
$\setminus {\int}_{{\vec{v}}_{0}}^{\vec{v} \left(t\right)} \mathrm{dv} e c v = \frac{a}{m} \left\{\left[\setminus {\int}_{0}^{t} \setminus \sin \setminus \omega t \mathrm{dt}\right] \setminus \hat{i} + \left[\setminus \int \setminus \cos \setminus \omega t \mathrm{dt}\right] \setminus \hat{j}\right\}$
$\vec{v} \left(t\right) - {\vec{v}}_{0} = \frac{a}{m} \left\{{\left[- \frac{\setminus \cos \setminus \omega t}{\setminus} \omega\right]}_{0}^{t} \setminus \hat{i} + {\left[\frac{\setminus \sin \setminus \omega t}{\setminus \omega}\right]}_{0}^{t} \setminus \hat{j}\right\}$
It is given that the object is initially at rest: $\setminus \quad {\vec{v}}_{0} = \vec{0}$

$\vec{v} \left(t\right) = \frac{a}{m \setminus \omega} \left\{\left(1 - \setminus \cos \setminus \omega t\right) \setminus \hat{i} + \setminus \sin \setminus \omega t \setminus \hat{j}\right\}$

Work Done: $\setminus \quad W = \setminus {\int}_{{\vec{r}}_{i}}^{{\vec{r}}_{f}} \vec{F} . \mathrm{dv} e c r$
But \quad vecv = (dvecr)/(dt); \qquad dvecr = vecvdt

$W = \setminus {\int}_{0}^{t} \vec{F} . \vec{v} \mathrm{dt}$

$\vec{F} . \vec{v} = a \left\{\setminus \sin \setminus \omega t \setminus \hat{i} + \setminus \cos \setminus \omega t \setminus \hat{j}\right\} . \frac{a}{m \setminus \omega} \left\{\left(1 - \setminus \cos \setminus \omega t\right) \setminus \hat{i} + \setminus \sin \setminus \omega t \setminus \hat{j}\right\}$
$\vec{F} . \vec{v} = {a}^{2} / \left(m \setminus \omega\right) \left\{\setminus \sin \setminus \omega t \left(1 - \setminus \cos \setminus \omega t\right) + \setminus \cos \setminus \omega t \setminus \sin \setminus \omega t\right\}$
$\setminus q \quad \setminus q \quad \setminus \quad = {a}^{2} / \left(m \setminus \omega\right) \setminus \sin \setminus \omega t$

$W = {a}^{2} / \left(m \setminus \omega\right) \setminus {\int}_{0}^{t} \setminus \sin \setminus \omega t \mathrm{dt} = {a}^{2} / \left(m \setminus \omega\right) {\left[- \frac{\setminus \cos \setminus \omega t}{\setminus} \omega\right]}_{0}^{t}$
$W = \left({a}^{2} / \left(m \setminus {\omega}^{2}\right)\right) \left(1 - \setminus \cos \setminus \omega t\right)$